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9. MAGNETISM
THEORY


If v and B are in the plane of paper, then according to

Right-Hand Rule, the direction of F on positively charged
particle will be perpendicular to the plane of paper upwards
as shown in figure (a), and on negatively charged particle will
be perpendicular to the plane of paper downwards, figure (b).
1. MAGNETIC FIELD AND FORCE

In order to define the magnetic field B , we deduce an expression
for the force on a moving charge in a magnetic field.
Consider a positive charge q moving in a uniform magnetic field




B , with a velocity V . Let the angle between V and B be .
(i)

The magnitude of force F experienced by the moving charge
is directly proportional to the magnitude of the charge i.e.
Fq
(ii)

The magnitude of force F is directly proportional to the
component of velocity acting perpendicular to the direction
of magnetic field, i.e.

Definition of B
If v = 1, q = 1 and sin  = 1 or = 90°, the nfrom (1),
F = 1 × 1 × B × 1 = B.
Thus the magnetic field induction at a point in the magnetic
field is equal to the force experienced by a unit charge moving
with a unit velocity perpendicular to the direction of magnetic
field at that point.
F  v sin 

(iii) The magnitude of force F is directly proportional to the
magnitude of the magnetic field applied i.e.,
FB
Combining the above factors, we get
F  qvsin B

Special Cases
Case (i) If  = 0° or 180°, then sin = 0.

or F = kqv B sin 
From (1),
F = qv B (0) = 0.
where k is a constant of proportionality. Its value is found
to be one i.e. k = 1.
It means, a charged particle moving along or opposite to the
direction of magnetic field, does not experience any force.
F = qv B sin 

 
F  q vB
...(1)
Case (ii) If v = 0, then F = qv B sin = 0.
...(2)
It means, if a charged particle is at rest in a magnetic field, it
experiences no force.



The direction of F is the direction of cross-product of

velocity v and magnetic field B , which is perpendicular to

the plane containing v and B . It is directed as given by the
Right-handed-Screw Rule or Right-Hand Rule.
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Case (iii) If = 90°, then sin = 1

F = qv B (1) = qv B (Maximum).

Unit of B . SI unit of B is tesla (T) or weber/(metre)2 i.e. (Wb/m2)
or Ns C–1 m–1
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289
Thus, the magnetic field induction at a point is said to be
one tesla if a charge of one coulomb while moving at right
angle to a magnetic field, with a velocity of 1 ms–1 experiences
a force of 1 newton, at that point.
Dimensions of B 
MLT 2

AT LT 1

speed, velocity, momentum and kinetic energy of charged
particle will change.

 
Case II. When v, E and B are mutually perpendicular to


each other. In this situation if E and B are such that
  
F  Fe  Fm  0 , then acceleration in the particle,

 F
a   0 . It means the particle will pass through the fields
m
without any change in its velocity. Here, Fe = Fm so qE = q v B
or v = E/B.
  MA 1T 2 
2. LORENTZ FORCE
The force experienced by a charged particle moving in space
where both electric and magnetic fields exist is called Lorentz
force.
Force due to electric field. When a charged particle carrying

charge +q is subjected to an electric field of strength E , it
experiences a force given by


...(5)
Fe  qE
This concept has been used in velocity-selector to get a
charged beam having a definite velocity.
3. MOTION OF A CHARGED PARTICLE IN A
UNIFORM MAGNETIC FIELD
Suppose a particle of mass m and charge q, entering a

uniform magnetic field induction B at O, with velocity v ,
making an angle  with the direction of magnetic field acting
in the plane of paper as shown in figure

whose direction is the same as that of E .
Force due to magnetic field. If the charged particle is moving

in a magnetic field B , with a velocity v it experiences a
force given by

 
Fm  q v  B


The direction of this force is in the direction of
perpendicular to the plane contaning v and
directed as given by Right hand screw rule.
 
v  B i.e.

B and is
Due to both the electric and magnetic fields, the total force
experienced by the charged particle will be given by
  

  
 
F  Fe  Fm  qE  q v  B  q E  v  B


  
F  q E  vB


 

Resolving v into two rectangular components, we have :
v cos  (= v1) acts in the direction of the magnetic field and
v sin  (= v2) acts perpendicular to the direction of magnetic
field.

For component velocity v2 , the force acting on the charged
particle due to magnetic field is

 
F  q v2  B
...(6)
This is called Lorentz force.
Special cases

 
Case I. When v, E and B , all the three are collinear.. In
this situation, the charged particle is moving parallel or
antiparallel to the fields, the magnetic force on the charged
particle is zero. The electric force on the charged particle

 qE
a

will produce acceleration
,
m
along the direction of electricl field. As a result of this, there
will be change in the speed of charged particle along the
direction of the field. In this situation there will be no change
in the direction of motion of the charged particle but, the

or

 
F  q v 2  B  qv 2 Bsin 90  q  v sin   B ...(1)

The direction of this force F is perpendicular to the plane


containing B and v2 and is directed as given by Right
hand rule. As this force is to remain always perpendicular to

v2 it does not perform any work and hence cannot change

the magnitude of velocity v2 . It changes only the direction
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MAGNETISM


angle between v1 and B is zero. Thus the charged particle
covers the linear distance in direction of the magnetic field
with a constant speed v cos .
of motion of the particle. Due to it, the charged particle is
made to move on a circular path in the magnetic field, as
shown in figure
Therefore, under the combined effect of the two component
velocities, the charged particle in magnetic field will cover
linear path as well as circular path i.e. the path of the charged
particle will be helical, whose axis is parallel to the direction
of magnetic field, figure
Here, magnetic field is shown perpendicular to the plane of
paper directed inwards and particle is moving in the plane
of paper. When the particle is at points A, C and D the
direction of magnetic force on the particle will be along AO,
CO and DO respectively, i.e., directed towards the centre O
of the circular path.
The force F on the charged particle due to magnetic field


2
provides the required centripetal force = mv 2 / r necessary
for motion along a circular path of radius r.

Bq v 2  mv22 / r or v2  Bq r / m
or
v sin  = B q r/m
...(2)
The angular velocity of rotation of the particle in magnetic
field will be

vsin  Bqr Bq


r
mr
m
The linear distance covered by the charged particle in the
magnetic field in time equal to one revolution of its circular
path (known as pitch of helix) will be
The frequency of rotation of the particle in magnetic field
will be

Bq
v

2 2m
d  v1T  v cos 
...(3)
Important points
The time period of revolution of the particle in the magnetic
field will be
T
1 2m

v
Bq
...(4)
From (3) and (4), we note that v and T do not depend upon
velocity v of the particle. It means, all the charged particles
having the same specific charge (charge/mass) but moving
with different velocities at a point, will complete their circular
paths due to component velocities perpendicular to the
magnetic fields in the same time.
For component velocity v1   vcos   , there will be no force
on the charged particle in the magnetic field, because the
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2m
Bq
1.
2.
3.
If a charged particle having charge q is at rest in a magnetic

field B , it experiences no force; as v = 0 and F = q v B sin  = 0.

If charged particle is moving parallel to the direction of B , it
also does not experience any force because angle  between


v and B is 0° or 180° and sin 0° = sin 180° = 0. Therefore,
the charged particle in this situation will continue moving
along the same path with the same velocity.
If charged particle is moving perpendicular to the direction

of B , it experiences a maximum force which acts


perpendicular to the direction B as well as v . Hence this
force will provide the required centripetal force and the
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291
charged particle will describe a circular path in the magnetic
field of radius r, given by
sufficiently high energy with the help of smaller values of
oscillating electric field by making it to cross the same electric
field time and again with the use of strong magnetic field.
mv 2
 Bqv .
r
4. MOTION IN COMBINED
ELECTRON AND MAGNETIC FIELDS
4.1 Velocity Filter
Velocity filter is an arrangement of cross electric and
magnetic fields in a region which helps us to select from a
beam, charged particles of the given velocity irrespective of
their charge and mass.
A velocity selector consists of two slits S1 and S2 held parallel
to each other, with common axis, some distance apart. In the
region between the slits, uniform electric and magnetic fields
are applied, perpendicular to each other as well as to the
axis of slits, as shown in figure. When a beam of charged
particles of different charges and masses after passing

through slit S1 enters the region of crossed electric field E

and magnetic field B , each particle experiences a force due
to these fields. Those particles which are moving with the
velocity v, irrespective of their mass and charge, the force
on each such particle due to electric field (qE) is equal and
opposite to the force due to magnetic field (q v B), then
Construction. It consists of two D-shaped hollow evacuated
metal chambers D1 and D2 called the dees. These dees are
placed horizontally with their diametric edges parallel and
slightly separated from each other. The dees are connected
to high frequency oscillator which can produce a potential
difference of the order of 104 volts at frequency  107 Hz.
The two dees are enclosed in an evacuated steel box and
are well insulated from it. The box is placed in a strong
magnetic field produced by two pole pieces of strong
electromagnets N, S. The magnetic field is perpendicular to
the plane of the dees. P is a place of ionic source or positively
charged particle figure.
q E = q v B or v = E/B
Working and theory. The positive ion to be accelerated is
produced at P. Suppose, at that instant, D1 is at negative
potential and D2 is at positive potential. Therefore, the ion
will be accelerated towards D1. On reaching inside D1, the
ion will be in a field free space. Hence it moves with a
constant speed in D 1 say v. But due to perpendicular
magnetic field of strength B, the ion will describe a circular
Such particles will go undeviated and filtered out of the
region through the slit S2. Therefore, the particles emerging
from slit S2 will have the same velocity even though their
charge and mass may be different.
The velocity filter is used in mass spectrograph which helps
to find the mass and specific charge (charge/mass) of the
charged particle.
4.2 Cyclotron
A cyclotron is a device developed by Lawrence and
Livingstone by which the positively charged particles like
proton, deutron, alpha particle etc. can be accelerated.
Principle. The working of the cyclotron is based on the fact
that a positively charged particle can be accelerated to a
path of radius r (say) in D1, given by Bqv 
mv 2
where m
r
and q are the mass and charge of the ion.

r
mv
Bq
Time taken by ion to describe a semicircular path is given
by, t 
r m



= a constant.
v Bq B  q / m 
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MAGNETISM
This time is independent of both the speed of the ion and
radius of the circular path. In case the time during which
the positive ion describes a semicircular path is equal to the
time during which half cycle of electric oscillator is completed,
then as the ion arrives in the gap between the two dees, the
polarity of the two dees is reversed i.e. D1 becomes positive
and D2 negative. Then, the positive ion is accelerated
towards D2 and it enters D2 with greater speed which remains
constant in D2. The ion will describe a semicircular path of
greater radius due to perpendicular magnetic field and again
will arrive in a gap between the two dees exactly at the
instant, the polarity of the two dees is reversed. Thus, the
positive ion will go on accelerating every time it comes into
the gap between the dees and will go on describing circular
path of greater and greater radius with greater and greater
speed and finally acquires a sufficiently high energy. The
accelerated ion can be removed out of the dees from window
W, by applying the electric field across the deflecting plates
E and F.
in a conductor is due to motion of electrons, therefore,
electrons are moving from the end Q to P (along X’ axis).

Let, vd drift velocity of electron
– e = charge on each electron.
Then magnetic Lorentz force on an electron is given by

 
f   e vd  B

If n is the number density of free electrons i.e. number of
free electrons per unit volume of the conductor, then total
number of free electrons in the conductor will be given by
Maximum Energy of positive ion
N = n (A) = nA
Let v0, r0 = maximum velocity and maximum radius of the
circular path followed by the positive ion in cyclotron.
Then,

mv02
Bqr0
 Bqv0 or v0 
r0
m
If T is the time period of oscillating electric field then
T = 2t = 2 m/Bq
1
Bq

T 2m
It is also known as magnetic resonance frequency.
The cyclotron angular frequency is given by
c  2v  Bq / m
5. FORCE ON A CURRENT CARRYING CONDUCTOR
PLACED IN A MAGNETIC FIELD
Expression for the force acting on the conductor carrying
current placed in a magnetic field
Consider a straight cylindrical conductor PQ of length ,
area of cross-section A, carrying current I placed in a uniform

magnetic field of induction, B . Let the conductor be placed
along X-axis and magnetic field be acting in XY plane making
an angle  with X-axis. Suppose the current I flows through
the conductor from the end P to Q, figure. Since the current
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

I = n A e vd

Cyclotron Frequency

We know that current through a conductor is related with
drift velocity by the relation
1
1  Bqr0 
B2q 2 r02
Max. K.E.  mv02  m 
 
2
2  m 
2m
The cyclotron frequency is given by v 
Total force on the conductor is equal to the force acting on
all the free electrons inside the conductor while moving in
the magnetic field and is given by


 
 
F  Nf  nA   e vd  B    nAe vd  B ...(7)



2


I  nAevd .

We represent I  as current element vector. It acts in the


direction of flow of current i.e. along OX. Since I  and vd
have opposite directions, hence we can write


...(8)
I    nAevd
From (7) and (8), we have
  
...(9)
F  I  B
 

F  I B
F  IBsin 
...(10)


were  is the smaller angle between I  and B .
Special cases
Case I. If  = 0° or 180°, sin = 0,
From (10), F = IB (0) = 0 (Minimum)
It means a linear conductor carrying a current if placed parallel
to the direction of magnetic field, it experiences no force.
Case II. If  = 90°, sin = q ;
From (10), F = IB × 1 = IB (Maximum)
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293
 

The force on the arm QR is given by F2  I QR  B or

It means a linear conductor carrying current if placed
perpendicular to the direction of magnetic field, it experiences
maximum force. The direction of which can be given by
Right handed screw rule.

F2 = I (QR) B sin  = I b B sin 
The direction of this force is in the plane of the coil directed
downwards.


Since the forces F2 and F4 are equal in magnitude and acting
6. TORQUE ON A CURRENT CARRYING COIL IN
A MAGNETIC FIELD
in opposite directions along the same straight line, they cancel
out each other i.e. their resultant effect on the coil is zero.
Consider a rectangular coil PQRS suspended in a uniform

magnetic field of induction B . Let PQ = RS =  and QR = SP = b.
Let I be the current flowing through the coil in the direction
PQRS and  be the angle which plane of the coil makes with
the direction of magnetic field figure. The forces will be
acting on the four arms of the coil.
Now, the force on the arm PQ is given by
 
 

F1  I PQ  B or F1 = I (PQ) B sin 90° = IB  PQ  B




Direction of this force is perpendicular to the plane of the
coil directed outwards (i.e. perpendicular to the plane of
paper directed towards the reader).
And, force on the arm RS is given by
 
 

F3  I RS  B or F3 = I (PQ) B sin 90° = IB  RS  B




The direction of this force, is perpendicular to the plane of paper
directed away from the reader i.e. into the plane of the coil.
The forces acting on the arms PQ and RS are equal, parallel
and acting in opposite directions having different lines of
action, form a couple, the effect of which is to rotate the coil
in the anticlockwise direction about the dotted line as axis.
The torque on the coil (equal to moment of couple) is given by
 = either force × arm of the couple
The forces F1 and F3 acting on the arms PQ and RS will be as
shown in figure when seen from the top.
Arm of couple = ST = PS cos  = b cos .

  IB  b cos   IBA cos  (  × b = A = area of coil
PQRS)
If the rectangular coil has n turns, then
  nIBA cos 
Note that if the normal drawn on the plane of the coil makes
an angle  with the direction of magnetic field, then +  = 90°
or = 90° – ; And cos = cos (90° – ) = sin 
Then torque becomes,
 
 
  nIBA sin   MBsin   M  B  nIA  B
  

Let F1 , F2 , F3 and F4 be the forces acting on the four current
carrying arms PQ, QR, RS and SP of the coil.
The force on arm SP is given by,
 

F4  I SP  B or F4 = I (SP) B sin (180° – ) = Ib B sin 


 
The direction of this force is in the direction of SP  B i.e.



where, nIA = M = magnitude of the magnetic dipole moment
of the rectangular current loop
 
  
  M  B  nI A  B


This torque tends to rotate the coil about its own axis. Its
value changes with angle between plane of coil and direction
of magnetic field.
in the plane of coil directed upwards.
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MAGNETISM
Special cases 1.
The lower end of the coil is connected to one end of a hair
spring S’ of quartz or phosphor bronze. The other end of this
highly elastic spring S’ is connected to a terminal T2. L is soft
iron core which may be spherical if the coil is circular and
cylindrical, if the coil is rectangular. It is so held within the
coil, that the coil can rotate freely without touching the iron
core and pole pieces. This makes the magnetic field linked
with coil to be radial field i.e. the plane of the coil in all positions
remains parallel to the direction of magnetic field. M is concave
mirror attached to the phosphor bronze strip. This helps us to
note the deflection of the coil using lamp and scale
arrangement. The whole arrangement is enclosed in a nonmetallic case to avoid disturbance due to air etc. The case is
provided with levelling screws at the base.
If the coil is set with its plane parallel to the direction of
magnetic field B, then
  0 and cos   1
 Torque,  = nIBA (1) = nIBA (Maximum)
This is the case with a radial field.
2.
If the coil is set with its plane perpendicular to the direction
of magentic field B, then  = 90° and cos  = 0
 Torque, = nIBA (0) = 0 (Minimum)
7. MOVING COIL GALVANOMETER
Moving coil galvanometer is an instrument used for detection
and measurement of small electric currents.
The spring S’ does three jobs for us : (i) It provides passage
of current for the coil PQRS1 (ii) It keeps the coil in position
and (iii) generates the restoring torque on the twisted coil.
Principle. Its working is based on the fact that when a current
carrying coil is placed in a magnetic field, it experiences a torque.
The torsion head is connected to terminal T 1. The
galvanometer can be connected to the circuit through
terminals T1 and T2.
Construction. It consists of a coil PQRS1 having large
number of turns of insulated copper wire, figure. The coil is
wound over a non-magnetic metallic frame (usually brass)
which may be rectangular or circular in shape. The coil is
suspended from a movable torsion head H by means of
phosphor bronze strip in a uniform magnetic field produced
by two strong cylindrical magnetic pole pieces N and S.
Theory. Suppose the coil PQRS1 is suspended freely in the
magnetic field.
Let,  = length PQ or RS1 of the coil,
b = breadth QR or S1P of the coil,
n = number of turns in the coil.
Area of each turn of the coil, A =  × b.
Let, B = strength of the magnetic field in which coil is
suspended.
I = current passing through the coil in the direction PQRS1
as shown in figure.
Let at any instant,  be the angle which the normal drawn on
the plane of the coil makes with the direction of magnetic field.
As already discussed, the rectangular coil carrying current
when placed in the magnetic field experiences a torque whose
magnitude is given by  = nIBA sin .
If the magnetic field is radial i.e. the plane of the coil is
parallel to the direction of the magnetic field then = 90°
and sin = 1.

= nIBA
Due to this torque, the coil rotates. The phosphor bronze
strip gets twisted. As a result of it, a restoring torque comes
into play in the phosphor bronze strip, which would try to
restore the coil back to its original position.
Let  be the twist produced in the phosphor bronze strip
due to rotation of the coil and k be the restoring torque per
unit twist of the phosphor bronze strip, then total restoring
torque produced = k .
In equilibrium position of the coil, deflecting torque
= restoring torque
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MAGNETISM

nIBA = k
or
I
295
k
 or I  G
nBA
k
 G  a constant for a galvanometer. It is
nBA
known as galvanometer constant.
(b)
The value of B can be increased by using a strong horse
shoe magnet.
(c)
The value of A can not be increased beyond a limit because
in that case the coil will not be in a uniform magnetic field.
Moreover, it will make the galvanometer bulky and
unmanageable.
(d)
The value of k can be decreased. The value of k depends
upon the nature of the material used as suspension strip.
The value of k is very small for quartz or phosphor bronze.
That is why, in sensitive galvanometer, quartz or phosphor
bronze strip is used as a suspension strip.
where
Hence, I  
It means, the deflection produced is proportional to the
current flowing through the galvanometer. Such a
galvanometer has a linear scale.
Current sensitivity of a galvanometer is defined as the
deflection produced in the galvanometer when a unit current
flows through it.
8. AMMETER
An ammeter is a low resistance galvanometer. It is used to
measure the current in a circuit in amperes.
If  is the deflection in the galvanometer when current I is
passed through it, then
A galvanometer can be converted into an ammeter by using
a low resistance wire in parallel with the galvanometer. The
resistance of this wire (called the shunt wire) depends upon
the range of the ammeter and can be calculated as follows :
Current sensitivity,
Is 
 nBA

I
k
k



 I 
nBA 

Let G = resistance of galvanometer, n = number of scale
divisions in the galvanometer,
The unit of current sensitivity is rad. A–1 or div. A–1.
K = figure of merit or current for one scale deflection in the
galvanometer.
Voltage sensitivity of a galvanometer is defined as the
deflection produced in the galvanometer when a unit voltage
is applied across the two terminals of the galvanometer.
Then current which produces full scale deflection in the
galvanometer, Ig = nK.
Let, V = voltage applied across the two terminals of the
galvanometer,
Let I be the maximum current to be measured by galvanometer.
To do so, a shunt of resistance S is connected in parallel
with the galvanometer so that out of the total current I, a
part I g should pass through the galvanometer and the
remaining part (I – Ig) flows through the shunt figure
 = deflection produced in the galvanometer.
Then, voltage sensitivity, VS = /V
If
R = resistance of the galvanometer, I = current through it.
Then V = IR

Voltage sensitivity,
VS 
 nBA IS


IR
kR
R
the unit of VS is rad V–1 or div. V–1.
Conditions for a sensitive galvanometer
A galvanometer is said to be very sensitive if it shows large
deflection even when a small current is passed through it.
From the theory of galvanometer,  
For a given value of I,  will be large if nBA/k is large. It is so
if (a) n is large (b) B is large (c) A is large and (d) k is small.
(a)
VA – VB = IgG = (I – Ig) S
nBA
I
k
The value of n can not be increased beyond a certain limit
because it results in an increase of the resistance of the
galvanometer and also makes the galvanometer bulky. This
tends to decrease the sensitivity. Hence n can not be
increased beyond a limit.
or
 Ig
S
 I  Ig


G


...(20)
Thus S can be calculated.
If this value of shunt resistance S is connected in parallel
with galvanometer, it works as an ammeter for the range 0 to I
ampere. Now the same scale of the galvanometer which was
recording the maximum current Ig before conversion into ammeter
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MAGNETISM
will record the maximum current I, after conversion into ammeter.
It means each division of the scale in ammeter will be showing
higher current than that of galvanometer.
From Ohm’s law, Ig 
or
Initial reading of each division of galvanometer to be used as
ammeter is Ig/n and the reading of the same each division
after conversion into ammeter is I/n.
V
G
Ig
If this value of R is connected in series with galvanometer, it
works as a voltmeter of the range 0 to V volt. Now the same
scale of the galvanometer which was recording the maximum
potential Ig G before conversion will record and potential V
after conversion in two voltmeter. It means each division of
the scale in voltmeter will show higher potential than that of
the galvanometer.
The effective resistance R P of ammeter (i.e. shunted
galvanometer) will be
1
1 1 SG
GS
  
or R P 
RP G S
GS
G S
Effective resistance RS of converted galvanometer into
voltmeter is
As the shunt resistance is low, the combined resistance of
the galvanometer and the shunt is very low and hence
ammeter has a much lower resistance than galvanometer. An
ideal ammeter has zero resistance.
RS = G + R
For voltmeter, a high resistance R is connected in series
with the galvanometer, therefore, the resistance of voltmeter
is very large as compared to that of galvanometer. The
resistance of an ideal voltmeter is infinity.
9. VOLTMETER
A voltmeter is a high resistance galvanometer. It is used to
measure the potential difference between two points of a
circuit in volt.
R
10. BIOT-SAVART’S LAW
According to Biot-Savart’s law, the magnitude of the
magnetic field induction dB (also called magnetic flux
density) at a point P due to current element depends upon
the factors at stated below :
A galvanometer can be converted into a voltmeter by
connecting a high resistance in series with the galvanometer.
The value of the resistance depends upon the range of
voltmeter and can be calculated as follows :
(i) dB  I
(ii) dB  d
n = number of scale divisions in the galvanometer,
(iii) dB  sin 
(iv) dB 
K = figure of merit of galvanometer i.e. current for one scale
deflection of the galvanometer.
Combining these factors, we get
Current which produces full scale deflection in the
galvanometer, Ig = nK.
dB 
Let, G = resistance of galvanometer,

V
V
or G  R 
GR
Ig
Let V be the potential difference to be measured by
galvanometer.
To do so, a resistance R of such a value is connected in
series with the galvanometer so that if a potential difference
V is applied across the terminals A and B, a current Ig flows
through the galvanometer. figure
Now, total resistance of voltmeter = G + R
Mahesh Tutorials Science
or
1
r2
Id sin 
r2
dB  K
Id sin 
r2
where K is a constant of proportionality. Its value depends
on the system of units chosen for the measurement of the
various quantities and also on the medium between point P
MAGNETISM
297
and the current element. When there is free space between
current element and point, then
In SI units, K 
8.
Similarities and Dis-similarities between the Biot-Savart’s law
for the magnetic field and coulomb’s law for electrostatic field
0
and In cgs system K = 1
4
where 0 is absolute magnetic permeability of free space
Similarities
(i)
Both the laws for fields are long range, since in both the
laws, the field at a point varies inversely as the square of the
distance from the source to point of observation.
(ii)
Both the fields obey superposition principle.
and 0  4107 Wb A 1m1  4 107 TA 1m
( 1 T = 1 Wb m–2)

(iii) The magnetic field is linear in the source Id  , just as the
electric field is linear in its source, the electric charge q.

In SI units, dB  0  Id sin  ...(3)
4
r2
In cgs system, dB 
If  = 0° or 180°, then dB = 0 i.e. minimum.
11. MAGNETIC FIELD DUE TO A STRAIGHT
CONDUCTOR CARRYING CURRENT
Id sin 
r2
In vector form, we may write
Consider a straight wire conductor XY lying in the plane of
paper carrying current I in the direction X to Y, figure. Let P
be a point at a perpendicular distance a from the straight
wire conductor. Clearly, PC = a. Let the conductor be made
of small current elements. Consider a small current element


Id  of the straight wire conductor at O. Let r be the
position vector of P w.r.t. current element and  be the angle


between Id  and r. Let CO = .
 
 
 0 I d   r
 0 I d   r
...(4)
dB 
or dB 
4 r 3
4
r3


Direction of dB . From (4), the direction of dB would
obviously be the direction of the cross product vector,
 
d   r . It is represented by the Right handed screw rule or

Right Hand Rule. Here dB is perpendicular to the plane

containing d  and r and is directed inwards. If the point P

is to the left of the current element, dB will be perpendicular

to the plane containing d  and r , directed outwards.


Some important features of Biot Savart’s law
1.
Biot Savart’s law is valid for a symmetrical current distribution.
2.
Biot Savart’s law is applicable only to very small length
conductor carrying current.
3.
This law can not be easily verified experimentally as the
current carrying conductor of very small length can not be
obtained practically.
4.
This law is analogous to Coulomb’s law in electrostatics.


The direction of dB is perpendicular to both Id  and r .
5.
6.
If  = 0° i.e. the point P lies on the axis of the linear conductor
carrying current (or on the wire carrying current) then
dB 

According to Biot-Savart’s law, the magnetic field dB (i.e.
magnetic flux density or magnetic induction) at point P due

to current element Id  is given by
 0 Id sin 0
0
4
r2

 0 Id   r
dB  . 3
4 r
It means there is no magnetic field induction at any point on
the thin linear current carrying conductor.
7.
If  = 90° i.e. the point P lies at a perpendicular position w.r.t.
current element, then
 Id
dB  0 2 , which is maximum.
4 r
or
dB 
 0 Id sin 

4
r2
...(5)
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MAGNETISM
In rt. angled POC, +  = 90° or = 90° – 

 2I
L
0 I
 2I
sin   sin   0 sin   40 a
2
4a
4 a
4a  L2

(iv) When point P lies on the wire conductor, then d  and r for
each element of the straight wire conductor are parallel.

Therefore, d   r  0 . So the magnetic field induction at P = 0.
Then, B 
sin  = sin (90° – ) = cos  ...(6)
Also, cos  
a
a
or r 
...(7)
r
cos 
And, tan  

or   a tan 
a
Direction of magnetic field
Differentiating it, we get
d  a sec  d
2
...(8)
Putting the values in (5) from (6), (7) and (8), we get
dB 


2
0 I a sec  d cos  0 I

cos  d ...(9)
4
4 a
 a2 

2 

 cos  
The magnetic field lines due to straight conductor carrying
current are in the form of concentric circles with the
conductor as centre, lying in a plane perpendicular to the
straight conductor. The direction of magnetic field lines is
anticlockwise, if the current flows from A to B in the straight
conductor figure (a) and is clockwise if the current flows
from B to A in the straight conductor, figure (b). The direction
of magnetic field lines is given by Right Hand Thumb Rule
or Maxwell’s cork screw rule.

The direction of dB , according to right hand thumb rule,
will be perpendicular to the plane of paper and directed
inwards. As all the current elements of the conductor will
also produce magnetic field in the same direction, therefore,
the total magnetic field at point P due to current through the
whole straight conductor XY can be obtained by integrating
Eq. (9) within the limits – 1 and + 2. Thus
B
2

1

 I
dB  0
4 a
2
 I
 cos  d  40 a sin 21

1
0 I
 I
sin 2  sin  1    0  sin 1  sin 2  ...(10)
4 a
4 a
Special cases. (i) When the conductor XY is of infinite length
and the point P lies near the centre of the conductor then
1  2  90
0 I
 2I
sin 90  sin 90  0 ...(11)
4 a
4 a
So,
B
(ii)
When the conductor XY is of infinite length but the point P
lies near the end Y (or X) then 1 = 90° and 2 = 0°.
So,
B
0 I
 I
sin 90  sin 0  0 ...(11 a)
4 a
4 a
Thus we note that the magnetic field due to an infinite long
linear conductor carrying current near its centre is twice
than that near one of its ends.
(iii) If length of conductor is finite, say L and point P lies on
right bisector of conductor, then
1  2   and sin  
Mahesh Tutorials Science
L/2
a 2   L / 2
2

L
4a  L2
2
Right hand thumb rule. According to this rule, if we imagine
the linear wire conductor to be held in the grip of the right
hand so that the thumb points in the direction of current,
then the curvature of the fingers around the conductor
will represent the direction of magnetic field lines, figure
(a) and (b).
MAGNETISM
299
12. MAGNETIC FIELD AT THE CENTRE OF THE
CIRCULAR COIL CARRYING CURRENT

B
0 I
 2I
.2r  0
4 r 2
4 r
If the circular coil consists of n turns, then
Consider a circular coil of radius r with centre O, lying with
its plane in the plane of paper. Let I be the current flowing in
the circular coil in the direction shown, figure (a). Suppose
the circular coil is made of a large number of current elements
each of length d.
B
0 2nI 0 I

 2n ...(13)
4 r
4 r
0 I
× angle subtended by coil at the centre.
4 r

Direction of B
i.e.
B
The direction of magnetic field at the centre of circular current
loop is given by Right hand rule.
Right Hand rule. According to this rule, if we hold the thumb
of right hand mutually perpendicular to the grip of the fingers
such that the curvature of the fingers represent the direction
of current in the wire loop, then the thumb of the right hand
will point in the direction of magnetic field near the centre of
the current loop.
According to Biot-Savart’s law, the magnetic field at the

centre of the circular coil due to the current element Id  is
given by

 0  d   r 
dB 
I

4  r 3 
or
 0 Idr sin   0 Id sin 

4
4 r 2
r3
where r is the position vector of point O from the current

element. Since the angle between d  and r is 90° (i.e.,  = 90°),
therefore,
dB 
 0 Id sin 90
 Id
or dB  0 2
...(12)
2
4
4 r
r

In this case, the direction of dB is perpendicular to the
plane of the current loop and is directed inwards. Since the
current through all the elements of the circular coil will
contribute to the magnetic feild in the same direction,
therefore, the total magnetic field at point O due to current
in the whole circular coil can be obtained by integrating eq.
(12). Thus
dB 

B  dB 
But
 Id  0 I

d
r 2 4 r 2
 40

13. AMPERE’S CIRCUITAL LAW
Consider an open surface with a boundary C, and the current
I is passing through the surface. Let the boundary C be
made of large number of small line elements, each of length

d. The direction of d  of small line element under study is
acting tangentially to its length d. Let Bt be the tangential
component of the magnetic field induction at this element


then Bt and d  are acting in the same direction, angle
between them is zero. We take the product of Bt and d for
that element. Then
 
Bt d  B.d 
 d = total length of the circular coil = circumference of the
current loop = 2r
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MAGNETISM
The relation (19) is independent of the size and shape of the
closed path or loop enclosing the current.
14. MAGNETIC FIELD DUE TO INFINITE LONG
STRAIGHT WIRE CARRYING CURRENT
Consider an infinite long straight wire lying in the plane of
paper. Let I be the current flowing through it from X to Y. A
magnetic field is produced which has the same magnitude
at all points that are at the same distance from the wire, i.e.
the magnetic field has cylindrical symmetry around the wire.
If length d is very small and products for all elements of
closed boundary are added together, then sum tends to be

an integral around the closed path or loop (i.e., ) .
 
Therefore,  of B.d  over all elements on a closed path
 

 B.d  = Line integral of B
around the closed path or
loop whose boundary coincides with the closed path.
According to Ampere’s circuital law,
 
B.d    0 I
...(19)


where I is the total current threading the closed path or loop
0 is the absolute permeability of the space. Thus,
a
n
Let P be a point at a perpendicular distance r from the straight

wire and B be the magnetic field at P. It will be acting
tangentially to the magnetic field line passing through P.
Consider an amperian loop as a circle of radius r, perpendicular
to the plane of paper with centre on wire such that point P
lies on the loop, figure. The magnitude of magnetic field is

same at all points on this loop. The magnetic field B at P
will be tangential to the circumference of the circular loop.
We shall integrate the amperian path anticlockwise. Then


B and d  are acting in the same direction. The line integral

of B around the closed loop is
 
B.d   Bd cos 0  B d  B2r
d
Ampere’s circuital law states that the line integral of magnetic

field induction B around a closed path in vacuum is equal to
0 times the total current I threading the closed path.
The relation (19) involves a sign convention, for the sense
of closed path to be traversed while taking the line integral
of magnetic field (i.e., direction of integration) and current
threading it, which is given by Right Hand Rule. According
to it, if curvature of the fingers is perpendicular to the thumb
of right hand such that the curvature of the fingers represents
the sense, the boundary is traversed in the closed path or
 
loop for B.d  , then the direction of thumb gives the sense


in which the current I is regarded as positive.

As per sign convention, here I is positive,
According to sign convention, for the closed path as shown
in figure, I1 is positive and I2 is negative. Then, according to
Ampere’s circuital law
 
B.d   0  I1  I 2    0 Ie
Using Ampere’s circuital law
 
B.d    0 I or B2r   0 I


where Ie is the total current enclosed by the loop or closed path.

or
B
0 I 0 2I

2r 4 r
...(21)
15. MAGNETIC FIELD DUE TO CURRENT THROUGH
A VERY LONG CIRCULAR CYLINDER
Consider an infinite long cylinder of radius R with axis XY.
Let I be the current passing through the cylinder. A magnetic
field is set up due to current through the cylinder in the form
of circular magnetic lines of force, with their centres lying
Mahesh Tutorials Science
MAGNETISM
301
on the axis of cylinder. These lines of force are perpendicular
to the length of cylinder.
or
B
 0 r Ir
i.e., B  r
2R 2
If we plot a graph between magnetic field induction B and
distance from the axis of cylinder for a current flowing through
a solid cylinder, we get a curve of the type as shown figure
Case I. Point P is lying outside the cylinder. Let r be the
perpendicular distance of point P from the axis of cylinder,

where r > R. Let B be the magnetic field induction at P. It is
acting tangential to the magnetic line of force at P directed


into the paper. Here B and d  are acting in the same direction.
Applying Ampere circuital law we have
 
B.d   0 I or
Bd cos 0   0 I


or
 Bd  0I
or
B
or B2r   0 I
0 I
, i.e., B  1/ r
2r
Here we note that the magnetic field induction is maximum
for a point on the surface of solid cylinder carrying current
and is zero for a point on the axis of cylinder.
16. FORCE BETWEEN TWO PARALLEL CONDUCTORS
CARRYING CURRENT
Consider C 1D 1 and C 2 D 2, two infinite long straight
conductors carrying currents I1 and I2 in the same direction.
They are held parallel to each other at a distance r apart, in
the plane of paper. The magnetic field is produced due to
current through each conductor shown separately in figure.
Since each conductor is in the magnetic field produced by
the other, therefore, each conductor experiences a force.
Case II. Point P is lying inside cylinder. Here r < R. we may
have two possibilities.
(i)
(ii)
r
F1
B

2
F2
B
B2
C1
Applying Ampere’s circuital law, we have
 
B.d    0 r I '
I2
×
×
I1
2
0 r Ir
R2
90°
90°
I
Ir
 r 2  2
2
R
R
2rB  0 r I ' 
D2
B1
If the current is uniformly distributed throughout the crosssection of the conductor, then the current through closed
path L is given by
I' 
or
D1
If the current is only along the surface of cylinder which is
so if the conductor is a cylindrical sheet of metal, then current
through the closed path L is zero. Using Ampere circutal
law, we have B = 0.
C2
Magnetic field induction at a point P on conductor C2D2
due to current I1 passing through C1D1 is given by
B1 
0 2I1
4 r
...(12)
According to right hand rule, the direction of magnetic field

B1 is perpendicular to the plane of paper, directed inwards.
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MAGNETISM
As the current carrying conductor C2D2 lies in the magnetic

field B1 (produced by the current through C1D1), therefore,

PQRS
the unit length of C2D2 will experience a force given by
F2 = B1I2 × 1 = B1I2
Putting the value of B1, we have
 2I I
F2  0 . 1 2
4 r
Thus one ampere is that much current which when flowing
through each of the two parallel uniform long linear
conductors placed in free space at a distance of one metre
from each other will attract or repel each other with a force
of 2 × 10–7 N per metre of their length.




P
Q
R
S
Q
  Q
Here, B.d   Bd cos 0  BL
...(13)
It means the two linear parallel conductors carrying
currents in the same direction attract each other.
  Q  R  S  P 
B.d   B.d   B.d   B.d   B.d 


P
P
R
and

P
  R
 
B.d   Bd cos90  0  B.d 
Q


Q
S
S
 
Also, B.d   0

( outside the solenoid, B = 0)
R

 
B.d   BL  0  0  0  BL
PQRS
17. THE SOLENOID
...(21)
From Ampere’s circuital law
A solenoid consists of an insulating long wire closely wound
in the form of a helix. Its length is very large as compared to
its diameter.
Magnetic field due to a solenoid
Consider a long straight solenoid of circular cross-section.
Each two turns of the solenoid are insulated from each other.
When current is passed through the solenoid, then each
turn of the solenoid can be regarded as a circular loop
carrying current and thus will be producing a magnetic field.
At a point outside the solenoid, the magnetic fields due to
neighbouring loops oppose each other and at a point inside
the solenoid, the magnetic fields are in the same direction.
As a result of it, the effective magnetic field outside the
solenoid becomes weak, whereas the magnetic field in the
interior of solenoid becomes strong and uniform, acting
along the axis of the solenoid.
Let us now apply Ampere’s circuital law.

 
B.d   0 × total current through the rectangle PQRS
PQRS
= 0 × no. of turns in rectangle × current
= 0 n LI
...(22)
From (21) and (22), we have
BL = 0 n LI or B = 0 n I
This relation gives the magnetic field induction at a point
well inside the solenoid. At a point near the end of a solenoid,
the magnetic field induction is found to be 0 n I/2.
18. TOROID
The toroid is a hollow circular ring on which a large number of
insulated turns of a metallic wire are closely wound. In fact, a
toroid is an endless solenoid in the form of a ring, figure.
Let n be the number of turns per unit length of solenoid and
I be the current flowing through the solenoid and the turns
of the solenoid be closely packed.
Consider a rectangular amperian loop PQRS near the middle
of solenoid as shown in figure
S
P
B
R
L
Q
×××××××××××××××××

The line integral of magnetic field induction B over the
closed path PQRS is
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Magnetic field due to current in ideal toroid
Let n be the number of turns per unit length of toroid and I
be the current flowing through it. In case of ideal toroid, the
coil turns are circular and closely wound. A magnetic field
MAGNETISM
303
of constant magnitude is set up inside the turns of toroid in
the form of concentric circular magnetic field lines. The
direction of the magnetic field at a point is given by the
tangent to the magnetic field line at that point. We draw
three circular amperian loops, 1, 2 and 3 of radii r1, r2 and r3 to
be traversed in clockwise direction as shown by dashed
circles in figure, so that the points P, S and Q may lie on
them. The circular area bounded by loops 2 and 3, both cut
the toroid. Each turn of current carrying wire is cut once by
the loop 2 and twice by the loop 3. Let B1 be the magnitude
of magnetic field along loop 1. Line integral of magnetic
field B1 along the loop 1 is

 
B1 .d  
loop 1

19. MAGNETISM & MATTER
19.1 The Bar Magnet
It is the most commonly used form of an artificial magnet.
When we hold a sheet of glass over a short bar magnet and
sprinkle some iron filings on the sheet, the iron filings
rearrange themselves as shown in figure. The pattern
suggests that attraction is maximum at the two ends of the
bar magnet. These ends are called poles of the magnet.
B1d cos 0  B1 2r1 ...(i)
loop 1
Loop 1 encloses no current.
According to Ampere’s circuital law

 
B1 .d   0  current enclosed by loop 1 = 0 × 0 = 0
loop 1
or
B12  r1 = 0 or B1 = 0
Let B3 be the magnitude of magnetic field along the loop 3.
The line integral of magnetic field B3 along the loop 3 is

loop 3
 
B3 .d  

B3d cos 0  B3 2r3
loop 3
From the sectional cut as shown in figure, we note that the
current coming out of the plane of paper is cancelled exactly
by the current going into it. Therefore, the total current
enclosed by loop 3 is zero.
According to Ampere’s circuital law

 
B3 .d   0 × total current through loop 3
loop 3
or
B3 2r3  0  0  0 or B3  0
Let B the magnitude of magnetic field along the loop 2. Line
integral of magnetic field along the loop 2 is

 
B.d   B2r2
1.
The earth behaves as a magnet.
2.
Every magnet attracts small pieces of magnetic substances
like iron, cobalt, nickel and steel towards it.
3.
When a magnet is suspended freely with the help of an
unspun thread, it comes to rest along the north south
direction.
4.
Like poles repel each other and unlike poles attract each
other.
5.
The force of attraction or repulsion F between two magnetic
poles of strengths m1 and m2 separated by a distance r is
directly proportional to the product of pole strengths and
inversely proportional to the square of the distance between
their centres, i.e.,
loop 2
Current enclosed by the loop 2 = number of turns × current
in each turn = 2  r2 n × I
According to Ampere’s circuital law

 
B.d   0  total current
loop 2
or
B2  r2   0  2r2 nI or B   0 nI
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MAGNETISM
m1m 2
mm
or F  K 1 2 2 , where K is magnetic force
2
r
r
constant.
F
In SI units, K 
0
 107 Wb A 1m 1
4
where 0 is absolute magnetic permeability of free space
(air/vacuum).

F
0 m1m 2
4 r 2
...(1)
This is called Coulomb’s law of magnetic force. However, in
cgs system, the value of K = 1.
This corresponds to Coulomb’s law in electrostatics.
SI Unit of magnetic pole strength
Suppose m1 = m2 = m (say),
r = 1 m and F = 10–7 N
From equation (1),
107  107 
m m
or m  1 or m = +1 ampere-metre
12
(Am). Therefore, strength of a magnetic pole is said to be
one ampere-metre, if it repels an equal and similar pole, when
placed in vacuum (or air) at a distance of one metre from it,
with a force of 10–7 N.
6.
2
The magnetic poles always exist in pairs. The poles of a
magnet can never be separated i.e. magnetic monopoles do
not exist.
20. MAGNETIC FIELD LINES
Magnetic field line is an imaginary curve, the tangent to
which at any point gives us the direction of magnetic field

B at that point.
If we imagine a number of small compass needless around a
magnet, each compass needle experiences a torque due to
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the field of the magnet. The torque acting on a compass
needle aligns it in the direction of the magnetic field.
The path along which the compass needles are aligned is
known as magnetic field line.
MAGNETISM
305
Properteis of magnetic field lines
1.
The magnetic field lines of a magnet (or of a solenoid
carrying current) form closed continuous loops.
2.
Outside the body of the magnet, the direction of magnetic
field lines is from north pole to south pole.
3.
At any given point, tangent to the magnetic field line

represents the direction of net magnetic field ( B ) at that
point.
4.
5.
The magnitude of magnetic field at any point is represented
by the number of magnetic field lines passing normally
through unit area around that point. Therefore, crowded
lines represent a strong magnetic field and lines which are
not so crowded represent a weak magnetic field.
No two magnetic field lines can intersect each other.
21. MAGNETIC DIPOLE
A magnetic dipole consists of two unlike poles of equal
strength and separated by a small distance.
We shall show that the SI unit of M is joule/tesla or ampere
metre2.

SI unit of pole strength is Am.
Bar magnet as an equivalent solenoid
We know that a current loop acts as a magnetic dipole.
According to Ampere’s hypothesis, all magnetic phenomena
can be explained in terms of circulating currents.
In figure magnetic field lines for a bar magnet and a current
carrying solenoid resemble very closely. Therefore, a bar
magnet can be thought of as a large number of circulating
currents in analogy with a solenoid. Cutting a bar magnet is
like cutting a solenoid. We get two smaller solenoids with
weaker magnetic properties. The magnetic field lines remain
continuous, emerging from one face of one solenoid and
entering into other face of other solenoid. If we were to
move a small compass needle in the neighbourhood of a bar
magnet and a current carrying solenoid, we would find that
the deflections of the needle are similar in both cases.
To demonstrate the similarity of a current carrying solenoid
to a bar magnet, let us calculate axial field of a finite solenoid
carrying current.
For example, a bar magnet, a compass needle etc. are
magnetic dipoles. We shall show that a current loop behaves
as a magnetic dipole. An atom of a magnetic material behaves
as a dipole due to electrons revolving around the nucleus.
The two poles of a magnetic dipole (or a magnet), called
north pole and south pole are always of equal strength, and
of opposite nature. Further such two magnetic poles exist
always in pairs and cannot be separated from each other.
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magnetic length of the magnet. It is a vector directed from

S-pole of magnet to its N-pole, and is represented by 2  .
In figure, suppose
Magnetic dipole moment is the product of strength of either

pole (m) and the magnetic length ( 2  ) of the magnet.

It is represented by M .
2 = length of solenoid with centre O
Magnetic dipole moment = strength of either pole × magnetic
length
We have to calculate magnetic field at any point P on the
axis of solenoid, where OP = r. Consider a small element of
thickness dx of the solenoid, at a distance x from O.
t
h
e


M  m 2
 
Magnetic dipole moment is a vector quantity directed from
South to North pole of the magnet, as shown in figure
a = radius of solenoid,
n = number of turns per unit length of solenoid,
i = strength of current passed through the solenoid
Number of turns in the element = n dx.
Using equation, magnitude of magnetic field at P due to this
current element is
dB 
0ia 2  n dx 
2  r  x   a 2 


2
3/ 2
...(10)
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MAGNETISM
If P lies at a very large distance from O, i.e., r >> a and r >> x,
then [(r – x)2 + a2]3/2  r3
dB 
0ia 2 ndx
2r 3
U  W   MB  cos 2  cos 1  ...(17)
W
B
0 nia
2r 3

dx 
x  
0 nia
2r 3
2

 x xx 
 
 ni a 2
 2n  2  ia
B 0
2   0
3 
2 r
4
r3
1.
...(19)
When  = 90°
U = – MB cos  = – MB cos 90° = 0
2
...(12)
i.e., when the dipole is perpendicular to magnetic field its potential
energy is zero.
Hence to calculate potential energy of diole at any position
making angle  with B, we use
U = – MB (cos 2 – cos 1) and take 1 = 90° and 2 = .
Therefore,
U = – MB (cos  – cos 90°) = – MB cos 
...(13)
2.
Thus, the axial field of a finite solenoid carrying current is
same as that of a bar magnet. Hence, for all practical purposes,
a finite solenoid carrying current is equivalent to a bar magnet.
Potential energy of a magnetic dipole in a magnetic field
Potential energy of a magnetic dipole in a magnetic field is
the energy possessed by the dipole due to its particular
position in the field.

When a magnetic dipole of moment M is held at an angle 

with the direction of a uniform magnetic field B , the
magnitude of the torque acting on the dipole is
When = 0°
U = – MB cos 0° = – MB
This is the expression for magnetic field on the axial line of
a short bar magnet.
  MBsin 
...(18)
Particular Cases
M = n (2) × i × (a2)
0 2M
4 r 3
1 = 90°, and 2 = , then
 
U   M.B
M = total no. of turns × current × area of cross section
B
n
In vector notation, we may rewrie (18) as
If M is magnetic moment of the solenoid, then

e
W = – MB cos 
As range of variation of x is from x = –  to x = +, therefore
the magnitude of total magnetic field at P due to current
carrying solenoid
2 x  
h
U = W = – MB (cos  – cos 90°)
...(11)
which is minimum. This is the position of stable equilibrium,
i.e., when the magnetic dipole is aligned along the magnetic
field, it is in stable equilibrium having minimum P.E.
3.
When = 180°
U = – MB cos 180° = MB, which is maximum. This is the
position of unstable equilibrium.
...(16)
This torque tends to align the dipole in the direction of the
field. Work has to be done in rotating the dipole against the
action of the torque. This work done is stored in the
magnetic dipole as potential energy of the dipole.
Now, small amount of work done in rotating the dipole
through a small angle d against the restoring torque is
22. MAGNETISM AND GAUSS’S LAW
dW = d= MB sin d
Total work done in rotating the dipole from = 1 to  = 2 is
W
2
 MBsin  d  MB cos 12   MBcos 2  cos 1 

1

Potential energy of the dipole is
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According to Gauss’s law for magnetism, the net magnetic
flux (B) through any closed surface is always zero.
23. EARTH’S MAGNETISM
Magnetic elements of earth at a place are the quantities
which describe completely in magnitude as well as direction,
the magnetic field of earth at that place.
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307
Square (23) and (24), and add
23.1 Magnetic declination
Magnetic declination at a place is the angle between
magnetic meridian and geographic meridian at that place.
H2 + V2 = R2 (cos2  + sin2 ) = R2

R  H2  V2
...(25)
Dividing (24) by (23), we get
R sin  V

or
R cos  H
tan  
V
...(26)
H
The value of horizontal component H = R cos  is different
at different places. At the magnetic poles,  = 90°

H = R cos 90° = zero
At the magnetic equator,  = 0°

H = R cos 0° = R
Horizontal component (H) can be measured using both, a
vibration magnetometer and a deflection magnetometer.
The value of H at a place on the surface of earth is of the
order of 3.2 × 10–5 tesla.
Retain in Memory
1.
The earth’s magnetic poles are not at directly opposite positions
on globe. Current magnetic south is farther from geographic
south than magnetic north is from geographic north.
2.
Infact, the magnetic field of earth varies with position and
also with time. For example, in a span of 240 years from 1580
to 1820 A.D., the magnetic declination at London has been
found to change by 3.5° – suggesting that magnetic poles
of earth change their position with time.
3.
Memory note
Note that the direction of horizontal component H of earth’s
magnetic field is from geographic south to geographic north
above the surface of earth. (if we ignore declination).
24. MAGNETIC PROPERTIES OF MATTER
To describe the magnetic properties of materials, we define
the following few terms, which should be clearly understood
The magnetic declination in India is rather small. At Delhi,
declination is only 0° 41’ East and at Mumbai, the declination
is 0° 58’ West. Thus at both these places, the direction of
geographic north is given quite accurately by the compass
needle (within 1° of the actual direction).
24.1 Magnetic Permeability
It is the ability of a material to permit the passage of magnetic
lines of force through it i.e. the degree or extent to which magnetic
field can penetrate or permeate a material is called relative
magnetic permeability of the material. It is represented by r.
23.2 Magnetic Dip or Magnetic Inclination
Relative magnetic permeability of a mterial is defined as the
ratio of the number of magnetic field lines per unit area (i.e.
flux density B) in that material to the number of magnetic
field lines per unit area that would be present, if the medium
were replaced by vacuum. (i.e. flux density B0).
Magnetic dip or magnetic inclination at a place is defined as
the angle which the direction of total strength of earth’s
magnetic field makes with a horizontal line in magnetic meridian.
23.3 Horizontal Component
It is the component of total intensity of earth’s magnetic
field in the horizontal direction in magnetic meridian. It is
represented by H.
i.e.,
AL = H = R cos 
...(23)
Vertical component along AD is
AM = V = R sin 
...(24)
B
B0
Relative magnetic permeability of a material may also be
defined as the ratio of magnetic permeability of the material
() and magnetic permeability of free space (0)
In figure, AK represents the total intensity of earth’s magnetic
field, BAK = . The resultant intensity R along AK is
resolved into two rectangular components :
Horizontal component along AB is
r 

r 

0
or    r  0
We know that 0 = 4 × 10–7 weber/amp-metre (Wb A–1 m–1)
or henry/metre (Hm–1)
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MAGNETISM
SI units of permeability () are
Hm–1 = Wb A–1 m–1 = (T m2) A–1 m–1 = T m A–1

24.2 Magnetic Intensity ( H )
The degree to which a magnetic field can magnetise a material
is represented in terms of magnetising force or magnetise

intensity ( H ).
But B = H

H  0 H 1   m  or
or
r  1  m

 1  m
0
This is the relation between relative magnetic permeability
and magnetic susceptibility of the material.
25. CLASSIFICATION OF MAGNETIC MATERIALS
24.3 Magnetisation or Intensity of Magnetisation ‘I’
There is a large variety of elements and compounds on earth.
Some new elements, alloys and compounds have been
synthesized in the laboratory. Faraday classified these
substances on the basis of their magnetic properties, into
the following three categories :
It represents the extent to which a specimen is magnetised,
when placed in a magnetising field. Quantitatively,
The magnetisation of a magnetic material is defined as the
magnetic moment per unit volume of the material.
M 
Magnetic moment m

volume
V
There are SI unit of I, which are the same as SI units of H.
Magnetic susceptibility (  m ) of a magnetic material is
defined as the ratio of the intensity of magnetisation (I)
induced in the material to the magnetising force (H) applied
on it. Magnetic susceptibility is represented by  m .
Thus  m 
I
H
Relation between magnetic permeability and magnetic
susceptibility
When a magnetic material is placed in a magnetising field of
magnetising intensity H, the material gets magnetised. The
total magnetic induction B in the material is the sum of the
magnetic induction B0 in vacuum produced by the magnetic
intensity and magnetic induction Bm, due to magnetisation
of the material. Therefore,
B = B0 + Bm
But B0 = 0 H and Bm = m0 I, where I is the intensity of
magnetisation induced in the magnetic material. Therefore,
from above
B  0 H  0 I  0  H  I  ,
i.e.,
B  0  H  I 
Now as  m 
I
 I  m H
H
From above, B  0  H   m H   0 H 1   m 
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(i)
Diamagnetic substances,
(ii)
Paramagnetic substances, and
(iii) Ferromagnetic substances
Their main characteristics are discussed below :
25.1 Diamagnetic Substances
The diamagnetic substances are those in which the
individual atoms/molecules/ions do not possess any net
magnetic moment on their own. When such substances are
placed in an external magnetising field, they get feebly
magnetised in a direction opposite to the magnetising field.
when placed in a non-uniform magnetic field, these
substances have a tendency to move from stronger parts of
the field to the weaker parts.
When a specimen of a diamagnetic material is placed in a
magnetising field, the magnetic field lines prefer not to pass
through the specimen.
Relative magnetic permeability of diamagnetic substances
is always less than unity.
From the relation  r  1   m  , as  r  1,  m is negative.
Hence susceptibility of diamagnetic substances has a small
negative value.
A superconductor repels a magnet and in turn, is repelled
by the magnet.
The phenomenon of perfect diamagnetism in
superconductors is called Meissner effect. Superconducting
magnets have been used for running magnetically leviated
superfast trains.
25.2 Paramagnetic substances
Paramagnetic substacnes are those in which each individual
atom/molecule/ion has a net non zero magnetic moment of
its own. When such substances are placed in an external
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309
inversely proportional to the temperature (T) of the material.
magnetic field, they get feebly magnetised in the direction
of the magnetising field.
When placed in a non-uniform magnetic field, they tend to
move from weaker parts of the field to the stronger parts.
When a specimen of a paramagnetic substance is placed in
a magnetising field, the magnetic field lines prefer to pass
through the specimen rather than through air.
From the SI relation,  r  1   m , as  r  1 , therefore,  m
must be positive. Hence, susceptibility of paramagnetic
substances is positive, though small.
i.e.,
As
B  H , magnetising intensity

I
But
I
 m
H

m 
as the temperature of the substance i.e.  m 
25.3 Ferromagnetic substances
Ferromagnetic substances are those in which each individual
atom/molecule/ion has a non zero magnetic moment, as in a
paramagnetic substance.
When such substances are placed in an external magnetising
field, they get strongly magnetised in the direction of the field.
The ferromagnetic materials show all the properties of
paramagnetic substances, but to a much greater degree. For
example,
(i)
They are strongly magnetised in the direction of external
field in which they are placed.
(ii)
Relative magnetic permeability of ferromagnetic materials is
very large (  103 to 105)
(iii) The susceptibility of ferromagnetic materials is also very
large.
 m  r  1
That is why they can be magnetised easily and strongly.
(iv) With rise in temperature, susceptibility of ferromagnetics
decreases. At a certain temperature, ferromagnetics change
over to paramagnetics. This transition temperature is called
curie temperature. For example, curie temperature of iron is
about 1000 K.
1
T
Combining these factors, we get I 
Susceptibility of paramagnetic substances varies inversely
1
i.e. they
T
lose their magnetic character with rise in temperature.
I  B, and I 
B
T
H
I
1
or

T
H T
1
or
T
m 
C
T
where C is a constant of proportionality and is called Curie
constant.
26. HYSTERISIS CURVE
The hysterisis curve represents the relation between


magnetic induction B (or intensity of magnetization I ) of
a ferromagnetic material with magnetiziing force or magnetic

intensity H . The shape of the hysterisis curve is shown in
figure. It represents the behaviour of the material as it is
taken through a cycle of magnetization.

Suppose the material is unmagnetised initially i.e., B  0

and H  0 . This state is represented by the origin O. Wee
place the material in a solenoid and increase the current

through the solenoid gradually. The magnetising force H

increases. The magnetic induction B in the material
increases and saturates as depicted in the curve oa. This
behaviour represents alignment and merger of the domains

of ferromagnetic material until no further enhancement in B
is possible. Therefore, there is no use of inreasing solenoid
current and hence magnetic intensity beyond this.
25.4 Curie Law in Magnetism
According to Curie law,
Intensity of magnetisation (I) of a magnetic material is (i)
directly proportional to magnetic induction (B), and (ii)
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MAGNETISM
This phenomenon of lagging of I or B behind H when a
specimen of a magnetic material is subjected to a cycle of
magnetisation is called hysteresis.
For example, hysteresis loop for soft iron is narrow and
large, whereas the hysteresis loop for steel is wide and short,
figure
Next, we decrease the solenoid current and hence magnetic

intensity H till it reduces to zero. The curve follows the


path ab showing that when H  0 , B  0 . Thus, some
magnetism is left in the specimen.

The value of magnetic induction B left in the specimen
when the magnetising force is reduced to zero is called
Retentivity or Remanence or Residual magnetism of the
material.
It shows that the domains are not completely randomised
even when the magnetising force is removed. Next, the
current in the solenoid is reversed and increased slowly.
Certain domains are flipped until the net magnetic induction

B inside is reduced to zero. This is represented by the
curve bc. It means to reduce the residual magnetism or
retentivity to zero, we have to apply a magnetising force =
OC in opposite direction. This value of magnetising force is
called coercivity of the material.
As the reverse current in solenoid is increased in magnitude,
we once again obtain saturation in the reverse direction at
d. The variation is represented by the curve cd. Next, the
solenoid current is reduced (curve de), reversed and
increased (curve ea). The cycle repeats itself. From figure,
we find that saturated magnetic induction BS is of the order
of 1.5 T and coercivity is of the order of –90 Am–1.
From the above discussion, it is clear that when a specimen
of a magnetic material is taken through a cycle of
magnetisation, the intensity of magnetisation (I) and
magnetic induction (B) lag behind the magnetising force
(H). Thus, even if the magnetising force H is made zero, the
values of I and B do not reduce to zero i.e., the specimen
tends to retain the magnetic properties.
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The hysterisis loops of soft iron and steel reveal that
(i)
The retentivity of soft iron is greater than the retentivity of
steel,
(ii)
Soft iron is more strongly magnetised than steel,
(iii) Coercivity of soft iron is less than coercivity of steel. It
means soft iron loses its magnetism more rapidly than steel
does.
(iv) As area of I-H loop for soft iron is smaller than the area of
I-H loop for steel, therefore, hysterisis loss in case of soft
iron is smaller than the hysterisis loss in case of steel.
(a)
Permanent Magnets
Permanent magnets are the materials which retain at room
temperature, their ferromagnetic properties for a long time.
The material chosen should have
(i)
high retentivity so that the magnet is strong,
(ii)
high coercivity so that the magnetisation is not erased by
stray magnetic fields, temperature changes or mechanical
damage due to rough handling etc.
(iii) high permeability so that it can be magnetised easily.
Steel is preferred for making permanent magnets.
(b)
Electromagnets
The core of electromagnets are made of ferromagnetic
materials, which have high permeability and low retentivity.
Soft iron is a suitable material for this purpose. When a soft
iron rod is placed in a solenoid and current is passed through
the solenoid, magnetism of the solenoid is increased by a
thousand fold. When the solenoid current is switched off,
the magnetism is removed instantly as retentivity of soft
iron is very low. Electromagnets are used in electric bells,
loudspeakers and telephone diaphragms. Giant
electromagnets are used in cranes to lift machinery etc.
MAGNETISM
311
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MAGNETISM
Specific example
27. HALL EFFECT
In the above circular loop tension in part A and B.
The Phenomenon of producing a transverse emf in a current
carrying conductor on applying a magnetic field perpendicular
to the direction of the current is called Hall effect.
In balanced condition of small part AB of the loop is shown below
Hall effect helps us to know the nature and number of charge
carriers in a conductor.
Consider a conductor having electrons as current carriers.
The electrons move with drift velocity v opposite to the
direction of flow of current
2Tsin
d
d
 dF  Bid  2Tsin
 BiRd
2
2
d d
d

 2T.  BiRd
2
2
2
 
Force acting on electron Fm   e v  B . This force acts
If d is small so, sin
along x-axis and hence electrons will move towards face (2)
and it becomes negatively charged.
T  BiR, if 2R  L so T 


BiL
2
28. STANDARD CASES FOR FORCE ON
CURRENT CARRYING CONDUCTORS
Case 1 : When an arbitrary current carrying loop placed in
a magnetic field (  to the plane of loop), each element of
loop experiences a magnetic force due to which loop
stretches and open into circular loop and tension developed
in it’s each part.
If no magnetic field is present, the loop will still open into
a circle as in it’s adjacent parts current will be in opposite
direction and opposite currents repel each other.
Case 2 : Equilibrium of a current carrying conductor :
When a finite length current carrying wire is kept parallel to
another infinite length current carrying wire, it can suspend
freely in air as shown below
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MAGNETISM
313
Wire is placed along the axis of coil so magnetic field
produced by the coil is parallel to the wire. Hence it will not
experience any force.
Case 4 : Current carrying spring : If current is passed
through a spring, then it will contract because current will
flow through all the turns in the same direction.
In both the situations for equilibrium of XY it’s downward
weight = upward magnetic force i.e. mg 
*
In the first case if wire XY is slightly displaced from its
equilibrium position, it executes SHM and it’s time period
is given by T  2
*
0 2i1i 2
.
.
4 h
h
.
g
If current makes to flow through spring, then spring will
contract and weight lift up.
If direction of current in movable wire is reversed then
it’s instantaneous acceleration produced is 2g.
Case 3 : Current carrying wire and circular loop : If a
current carrying straight wire is placed in the magnetic field
of current carrying circular loop.
If switch is closed then current start flowing, spring will
execute oscillation in vertical plane.
Case 5 : Tension less strings : In the following figure the
value and direction of current through the conductor XY so
that strings becomes tensionless ?
Strings becomes tensionless if weight of conductor XY
Wire is placed in the perpendicular magnetic field due to
coil at it’s centre, so it will experience a maximum force
F  Bi 
balanced by magnetic force (Fm).
0i1
 i 2
2r
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MAGNETISM
In the following situation conducting rod (X, Y) slides at
constant velocity if
Fcos   mg sin   Bi cos   mg sin   B 
mg
tan 
i
TIPS & TRICKS
1.
The device whose working principle based on Halmholtz
coils and in which uniform magnetic field is used called as
“Halmholtz galvanometer”.
2.
The value of magnetic field induction at a point, on the
centre of separation of two linear parallel conductors
carrying equal currents in the same direction is zero.
3.
If a current carrying circular loop (n = 1) is turned into a
coil having n identical turns then magnetic field at the
Hence direction of current is from X  Y and in balanced
condition Fm = mg  Bi = mg  i =
mg
B
Case 6 : A current carrying conductor floating in air such
that it is making an angle  with the direction of magnetic
field, while magnetic field and conductor both lies in a
horizontal plane.
centre of the coil becomes n2 times the previous field i.e.
B
4.
5.
In equilibrium mg = Bi sin  i 
(n turn)
= n2 B(single turn).
When a current carrying coil is suspended freely in earth’s
magnetic field, it’s plane stays in East-West direction.

Magnetic field B produced by a moving charge q is given
 

 0 q  v  r  0 q  v  rˆ 

by B 
; where v = velocity of
4 r 3
4 r 2
mg
B sin 
charge and v < < c (speed of light).
Case 7 : Sliding of conducting rod on inclined rails : When
a conducting rod slides on conducting rails.
6.
If an electron is revolving in a circular path of radius r with
speed v then magnetic field produced at the centre of circular
path B 
7.
0 ev
v
.
.
r
4 r 2
B

The line integral of magnetising field H for any closed
 
path called magnetomotive force (MMF). It’s S.I. unit is amp.
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8.
Ratio of dimension of e.m.f. to MMF is equal to the dimension
of resistance.
9.
The positive ions are produced in the gap between the two
dees by the ionisation of the gas. To produce proton,
hydrogen gas is used; while for producing alpha-particles,
helium gas is used.
MAGNETISM
315
10. Cyclotron frequency is also known as magnetic resonance
frequency.
11.
Cyclotron can not accelerate electrons because they have
very small mass.
12. The energy of a charged particle moving in a uniform magnetic
field does not change because it experiences a force in a
direction, perpendicular to it’s direction of motion. Due to
which the speed of charged particle remains unchanged and
hence it’s K.E. remains same.
13. Magnetic force does no work when the charged particle is
displaced while electric force does work in displacing the
charged particle.
14.
17. If no magnetic field is present, the loop will still open into a
circle as in it’s adjacent parts current will be in opposite
direction and opposite currents repel each other.
Magnetic force is velocity dependent, while electric force
is independent of the state of rest or motion of the charged
particle.
15. If a particle enters a magnetic field normally to the
magnetic field, then it starts moving in a circular orbit.
The point at which it enters the magnetic field lies on the
circumference. (Most of us confuse it with the centre of the
orbit)
16. Deviation of charged particle in magnetic field : If a

charged particle (q, m) enters a uniform magnetic field B
(extends upto a length x) at right angles with speed v as
shown in figure. The speed of the particle in magnetic
field does not change. But it gets deviated in the magnetic
field.
18. In the following case if wire XY is slightly displaced from its
equilibrium position, it executes SHM and it’s time period is
given by T  2
h
.
g
 Bq 
Deviation in terms of time t ;   t  
t
 m
Deviation in terms of length of the magnetic field ;
x
  sin 1   . This relation can be used only when x  r .
r
For x > r, the deviation will be 180° as shown in the following figure
19. In the previous case if direction of currnet in movable wire
is reversed then it’s instantaneous acceleration produced is
2g.
20. Electric force is an absolute concept while magnetic force is
a relative concept for an observer.
21. The nature of force between two parallel charge beams
decided by electric force, as it is dominator. The nature of
force between two parallel current carrying wires decided
by magnetic force.
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22. If a straight current carrying wire is placed along the axis of
a current carrying coil then it will not experience magnetic
force because magnetic field produced by the coil is parallel
to the wire.
23. The force acting on a curved wire joining points a and b as
shown in the figure is the same as that on a straight wire
  
joining these points. It is given by the expression F  iL  B
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MAGNETISM
24. If a current carrying conductor AB is placed transverse to a
long current carrying conductor as shown then force.
Experienced by wire AB
F
0i1i 2
 x
log e 

2
 x 
MAGNETISM
317
SOLVED EXAMPLES
Example - 1
An electron is passing through a field but no force is acting
on it. Under what conditions is it possible, if the motion of
the electron be in the (i) electric field (ii) magnetic field ?
Sol. (i) In electric field, there is always a force on the moving
electron opposite to the direction of field. Thus the force
will be zero only if electric field is zero.
(ii)
In magnetic field, the force acting on a moving electron is
F = qv B sin , it is zero if  = 0º or 180º.
i.e.
the electron is moving parallel to the direction of magnetic
field.
Example - 2
A neutron, a proton an electron and an -particle enter a
region of constant magnetic field with equal velocities.
The magnetic field is along the inward normal to the plane
of paper. The tracks of the particles are shown in figure.
Relate the tracks to the particles.
r
mv
m
or r 
Bq
q
rp

r
mp
m

q   m   2e  1

  
q p  4m   e  2
or
r  2rp i.e. r  rp .
i.e.
track B corresponds to -particle and track A to proton.
Example - 3
Why is ammeter connected in series and voltmeter in
parallel in the circuit ?
Sol. An ammeter is a low resistance galvanometer. It is used to
measure the current in ampere. To measure the current of a
circuit, the ammeter is connected in series to the circuit so
that the current to be measured must pass through it. Since,
the resistance of ammeter is low, so its inclusion in series in
the circuit does not change the resistance and hence the
main current in the circuit.
A voltmeter is a high resistance galvanometer. It is used to
measure potential difference between two points of the
circuit in volt. To measure the potential difference between
the two points of a circuit, the voltmeter is connected in
parallel to the circuit. The voltmeter resistance being high, it
draws minimum current from the main circuit and the potential
difference to be measured is not affected materially.
Sol. We know that force on a charged particle in the magnetic
field is

 
F  q v  B or F  qvBsin , so
(i)
(ii)
For neutral particle i.e. neutron, q = 0, hence F = 0. It means
neutron will go undeflected i.e. track C corresponds to
neutron.
For negatively charged particle i.e. electron, the direction of
force, according to Fleming’s Left hand rule will be towards
right. So track D corresponds to electron.
(iii) For positively charged particle, the direction of force,
according to Fleming’s left hand rule will be towards left. So
both tracks A and B correspond to positively charged
particles (i.e. protons and -particles).
When a moving charged particle is subjected to a
perpendicular magnetic field, it describes a circular path of
radius r given by
Example - 4
A current carrying circular loop is located in a uniform
external magnetic field. If the loop is free to turn, what is its
orientation of stable equilibrium? Show that in this
orientation, the flux of the total field (external field + field
produced by the loop) is maximum.
Sol. The current carrying circular loop behaves as a magnetic

dipole of magnetic moment M acting perpendicular to its
plane. The torque on the current loop of magnetic dipole
moment M in the magnetic field B is
 = MB sin  = IA × B sin ,
( M = AI)


where  is the angle between M and B . The system will be
in stable equilibrium if torque is zero, which is so if  = 0º.



This is possible if B is parallel to A i.e. B is perpendicular
to the plane of the loop. In this orientation, the magnetic
field produced by the loop is in the same direction as that of
external field, both normal to the plane of loop. It is due to
this fact, the magnetic flux due to total field is maximum.
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MAGNETISM
electric field time and again with the use of strong magnetic
field.
Example - 5
Three wires each of length 2.0 m are bent into different
rectangular loops and then suspended in a magnetic field,
figure. If the current in each of them be the same, which
loop shall be acted upon by largest torque ? If any of the
wires be bent into circular loop, then ?
A cyclotron is used (i) to bombard nuclei with high energy
particles and to study the resulting nuclear reaction (ii) to
produce radioactive substances which may be used in
hospitals for diagnosing the diseases in the body.
Example - 9
A charged particle enters into a uniform magnetic field and
experiences upwardforce as indicated in figure. What is
the charge sign on the particle ?
Sol. Torque () on a current loop suspended in a uniform magnetic
field is given by = I AB sin  i.e.   A. Since the area of
loop (c) = 0.5 m × 0.5 m is maximum; hence the largest torque
will be acting on it. When any wire is bent into a circular
loop, the torque will be even more because for a given
perimeter the area of the circle is maximum.
Sol. The particle has a positive charge.
Example - 10
You are given a low resistance R1, a high resistance R2
and a moving coil galvanometer. Suggest how you would
use these to have an instrument that will be able to
measure (i) currents (ii) potential differences.
Example - 6
Sol. (i) To measure currents, the low resistance R1 is connected
in parallel to the moving coil galvanometer.
What is meant by cyclotron frequency ?
Sol. It is the frequency of oscillation of a heavy charged particle
in between two dees of cyclotron, which is equal to the
frequency of high frequency oscillator, creating electric field
between two dees of cyclotron. Cyclotron frequency,
v = Bq/2  m, which is independent of the radius of the
circular path and velocity of the charged particle in the two
dees of cyclotron.
Example - 7


(ii)
To measure potential differences, a high resistance R2 is
connected in series with the moving coil galvanometer.
Example - 11
State properties of the material of the wire used for
suspension of the coil in a moving coil galvanometer.
Sol. The properties of the material of the wire used for suspension
of the coil in a moving coil galvanometer are as follows :

1
A charge 3 coulomb is moving with velocity v  4iˆ  3jˆ ms
1.

2
in a magnetic field B  4iˆ  3jˆ Wbm . Find the force
It should have low torsional constant i.e. restoring torque
per unit twist should be small.
2.
It should have high tensile strength.
acting on the charge.
3.
It should be a non-magnetic substance.
4.
It should have a low temperature coefficient of resistance.
5.
It should be a good conductor of electricity.



 
Sol. F  q v  B  3  4iˆ  3jˆ  4iˆ  3jˆ  = 3 [0] = 0



 

 Cross product of two equal vector is zero.
Example - 8
What is the basic principle of working of cyclotron ? Write
two uses of this machine.
Sol. The working of the cyclotron is based on the fact that a
heavy positively charged particle can be accelerated to a
sufficiently high energy with the help of smaller values of
oscillation electric field, by making it to cross the same
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Example - 12
What is a radial magnetic field ? How has it been achieved
in moving coil galvanometer ?
Sol. Radial magnetic field is that field, in which the plane of the
coil always lies in the direction of the magnetic field. A radial
magnetic field has been achieved by (i) properly cutting the
magnetic pole pieces in the shape of concave faces. (ii)
using a soft iron core within the coil.
MAGNETISM
319
Example - 13
Why is phosphor bronze alloy preferred for the suspension
wire of a moving coil galvanometer ?
Sol. The suspension wire of phosphor bronze alloy is preferred in
moving coil galvanometer because it has several advantages:
(i)
Its restoring torque per unit twist is small. Due to it, the
galvanometer is very sensitive.
(ii)
It has great tensile strength so that even if it is thin, it will not
break under the weight of the coil suspended from its end.
(iii) It is rust resisting. Hence it remains unaffected by the weather
conditions of air in which it is suspended.
Example - 16
An electron and proton enter perpendicularly in a uniform
magnetic field with the same speed. How many times larger
will be the radius of proton’s path than the electron’s ?
Proton is 1840 times heavier than electron.
Sol. The charged particle while moving perpendicular to magnetic
field experiences a force which provides the centripetal force
for its circular motion. The radius r of the circular path traced
by the particle in magnetic field B, is given by Bqv = mv2/r or
r = mv/Bq or r  m if v, B and q are constant.
Since the value of charge on electron and proton is the
same but mass of proton is 1840 times mass of electron,
Example - 14
What is the main function of a soft iron core used in a
moving coil galvanometer ?
hence
rp
re

mp
me

1840m e
 1840 or r = 1840 r .
p
e
me
Example - 17
Sol. (i) This makes the magnetic field radial. In such a magnetic
field the plane of the coil is always parallel to the direction
of magnetic field. Due to which the galvanometer scale
becomes linear.
(ii)
This increases the strength of magnetic field due to the
crowding of the magnetic lines of force through the soft
iron core, which in turn increases the sensitiveness of the
galvanometer.
Example - 15
Define current sensitivity and voltage sensitivity of a
galvanometer. Increase in the current sensitivity may not
necessarily increase the voltage sensitivity of a
galvanometer. Justify.
Sol. For definition of current sensitivity and voltage sensitivity
refer to Art. 3(b).11.
Let  be the deflection produced in the galvanometer on
applying voltage V, then
current sensitivity 
voltage sensitivity 
 nBA

I
k
 nBA

V kR
Thus, the current sensitivity can be increased by increasing,
n, B, A and by decreasing k. If n is increased, it will increase
the resistance of conductor.
The voltage sensitivity can be increased by increasing n, B,
A and by decreasing k and R.
Therefore, the increase in current sensitivity of galvanometer
may not necessarily increase the voltage sensitivity of the
galvanometer.
Two parallel wires carrying current in the same direction
attract each other while two beams of electrons travelling
in the same direction repel each other. Why ?
Sol. Two parallel wires carrying currents in the same direction
attract each other due to magnetic interaction between two
wires carrying currents because the current in a wire
produces a magnetic field and the magnetic interaction is of
attractive nature when current is the two parallel wires is in
the same direction.
The two beams of electrons travelling in the same direction
will be a source of both an electric and magnetic fields. Due
to magnetic interaction, there will be force of attraction
between the two moving electrons but due to electrostatic
interaction, there will be a force of repulsion between them.
If the beams of electrons are moving slowly, the electrostatic
force of repulsion between the electrons dominates the
magnetic attraction between them.
Example - 18
An electron beam moving with uniform velocity is
gradually diverging. When it is accelerated to a very high
velocity, it again starts converging. Why ?
Sol. Moving electrons, apart from electrical repulsion experience
magnetic attraction also. If the electron beam is moving
under normal conditions, the electrical repulsive force is
much stronger than the magnetic attraction and hence the
beam diverges. When the electron beam is moving at very
high velocity, the magnetic force of attraction becomes more
effective than electrical repulsion and the beam starts
converging.
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MAGNETISM
Example - 19
Distinguish between Biot Savart’s law and Ampere’s
circuital law.
Sol.
1.
2.
3.
Biot-Savart’s Law
Ampere’s Circuital Law
This law is based on the
principle of magnetism.
This law is based on the
principle of electromagnetism.
This law is valid for
asymmetrical current
distribution.
This law is valid for
symmetrical current
distributions.
This law is the differential This law is the integral form


form of magnetic field
of B or H .

induction B or

magnetising force H .
Example - 20
Two small circular loops, marked (1) and (2), carrying equal
currents are placed with the geometrical axes perpendicular
to each other as shown in figure. Find the magnitude and
direction of the net magnetic field produced at the point O.
Example - 21
Two parallel coaxial circular coils of equal radius R and
equal number of turns N carry equal currents I in the same
direction and are separated by a distance 2 R. Find the
magnitude and direction of the net magnetic field
produced at the mid-point of the line joining their centres.
Sol. Magnetic field induction at the mid-point due to current
loop 1 is
B1 
0
0 I R 2
2I R 2
, acting towards right.

4 R 2  R 2 3/ 2 2 2R 2 3/ 2




Magnetic field induction at the mid point due to current
loop 2 is
B2 

0 I R 2
2 R2  R2

3/ 2
0 I R 2


2 2R 3

3/ 2
, acting towards right.
Total magnetic field induction is
B  B1  B2 
0 I R 2

2 2R

2 3/ 2

0 I R 2

2 2R

2 3/ 2

0 I R 2
2 2R
3

0 I
2 2R
Example - 22
Magnetic field lines can be entirely confined within the
core of a toroid, but not within a straight solenoid. Why ?
Sol. It is so because the magnetic field idnuction outside the
toroid is zero.
Example - 23
Name the physical quantity whose unit is tesla. Hence
define a tesla.
Sol. Magnetic field induction at O due to current loop 1 is
B1 

0 I R 2
2 x2  R 2

3/ 2
, acting towards left.
Magnetic field induction at O due to current loop 2 is
B2 

0 I R 2
2 x2  R2
Example - 24

3/ 2
acting vertically upwards.
Resultant magnetic field induction at O will be
B  B12  B22  2 B1
 2

Sol. Tesla is the SI unit of magnetic field induction or magnetic
flux density at a point in the magnetic field. The magnetic
field induction at a point in a magnetic field is said to be 1
tesla if one coulomb charge while moving with a velocity of
1 m/s, perpendicular to the magnetic field experiences a force
of 1 N at that point.
0 I R 2
2 x2  R

2 3/ 2
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
 B1  B2 

0 I R 2
2 x2  R2

3/ 2
What is meant by a magnetic field ? How is it produced ?
Sol. A magnetic field is the space around a magnet or the space
around a wire carrying current, in which its magnetic effect
can be felt.
A magnetic field may be produced in many ways. For
example, (i) by a magnet (ii) by a current carrying conductor
(iii) by a moving charge (iv) by a varying electric field.
(displacement current)
MAGNETISM
Example - 25
321
Example - 30
What is the potential energy of a dipole when it is
perpendicular to a magnetic field ?
State two methods to destroy the magnetism of a magnet.
Sol. (i) By heating the magnet.
Sol. P.E. = –MB cos  = –MB cos 90º = zero.
Example - 26
(ii) By applying magnetic field in the reverse direction.
Example - 31
An electron of energy 2000 eV describes a circular path in
magnetic field of flux density 0.2 T. What is the radius of
the path ? Take e = 1.6 × 10–19 C, m = 9 × 10–31 kg.
What is the basic difference between magnetic and electric
lines of force ?
Sol. Magnetic lines of force are closed, continuous curves, but
electric lines of force are discontinuous.
Sol. Here, energy of electron, E’ = 2000 eV
= 2000 × 1.6 × 10–19 J = 3.2 × 10–16 J.
Example - 27
A magnetic needle free to rotate in a vertical plane, orients
itself with its axis vertical at a certain place on the earth.
What are the values of
B = 0.2 T ; r = ?
As,
E' 
1
mv 2
2
 v
2E '
m
(a) Horizontal component of earth’s field ?
(b) angle of dip at this place.
Also, Bev 
Sol. H = 0 and  = 90º. The place will be magnetic pole of earth.
Example - 28
Why do magnetic lines of force prefer to pass through iron
than air ?
Sol. This is because permeability of soft iron is much greater
than that of air.
Example - 29
Define the term : magnetic dipole moment of a current loop.
Write the expression for the magnetic moment when an
electron revolves at a speed v around an orbit of radius r in
hydrogen atom.
or
mv2
r
r
mv m 2E '
2E 'm


Be Be m
Be

2  3.2  1016  9  1031
 7.5  104 m
0.2  1.6  1019
Example - 32
A long straight wire AB carries a current of 4 A. A proton P
travels at 4 × 106 ms–1 parallel to the wire, 0.2 m from it and
in a direction opposite to the current as shown in figure.
Calculate the force which the magnetic field of current
exerts on the proton. Also specify the direction of the force.
Sol. A current carrying loop behaves as a system of two equal
and opposite magnetic poles separated by a distance. Hence
it behaves as a magnetic dipole. Magnetic dipole moment of
current loop is the product of current I and area A enclosed
by the loop of current, i.e. M = IA.
In a hydrogen atom, when an electron revolves at a speed v
around an orbit of radius r, the magnetic moment is given by
 eh 
M  n

 4m 
where e is charge on electron, m is mass of electron ;
n denotes the number of orbit and h is Plack’s constant.
Sol. Here, I = 4A ; v = 4 × 106 ms–1 ; a = 0.2 m.
Magnetic field induction at P is
0 2I 107  2  4

 4  106 T
4 r
0.2

The direction of B , according to Right Hand Thumb rule is
perpendicular to the plane of paper directed inwards.
B
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MAGNETISM
Since proton is moving in opposite direction to the current
carrying straight wire, hence the proton is moving
perpendicular to the direction of magnetic field due to current
through straight wire. The force on moving proton of charge
q due to magnetic field is
F = qvB sin 90º = (1.6 × 10–19) × (4 × 106) × (4 × 10–6)
= 2.56 × 10–18 N
The direction of force on proton, according to Fleming’s
Left Hand Rule acts in the plane of paper towards right.
eE  evB or v 
E 2  104

 5  106 m / s
B 4  103
When electron moves perpendicular to magnetic field, the
radius r of circular path traced by electron is


 


9.1 1031  5  106
mv
r

 7.11 103 m  7.11 mm
eB
1.6  1019  4  103
Example - 35
Example - 33
A cyclotron oscillator frequency is 10 M Hz. What should be
the operating magnetic field for accelerating -particle ? If
the radius of the dees is 50 cm, what is the kinetic energy in
MeV of the -particle beam produced by the accelerator?
(e = 1.6 × 10–19 C ; m = 4.0028 a.m.u. ; 1 a.m.u. = 1.66 × 10–27 kg)
Figure shows a rectangular current-carrying loop placed
2 cm away from a long, straight, current carrying conductor.
What is the direction and magnitude of the net force acting
on the loop ?
Sol. Here, v = 10 MHz = 107 Hz ; r0 = 50 cm = 0.50 m ; B = ?
m = 4.0028 × 1.66 × 10–27 kg = 6.645 × 10–27 kg,
q = 2 e = 2 × 1.6 × 10–19 = 3.2 × 10–19 C.
As,
v
Bq
2m
or
B
2m v
q
 2
22 6.645  1027  107

 1.305 T
7
3.2  1019
Sol. Here, I1 = 15 A ; I2 = 25 A ;
Maximum kinetic energy is




2
19
  0.50 
B2 q 2 r 2 1.305  3.2  10


J
2m
2  6.645  1027
2
E max
r1 = 2 × 10–2 m ; r2 = (2 + 10) × 10–2 m
1.3052   3.2 2 1038  0.25
2  6.645  1027  1.6  1013
2

 107 
MeV  20.5 MeV
Example - 34
An electron beam passes through a magnetic field of 4 × 10–3
weber/m2 and an electric field of 2 × 104 Vm–1, both acting
simultaneously. The path of electron remaining undeviated,
calculate the speed of the electrons. If the electric field is
removed, what will be the radius of the electron path ?
Sol. Here, B = 4 × 10–3 weber/m2 ; E = 2 × 104 V/m.
As the path of moving electron is undeviated, so force on
moving electron due to electric field is equal and opposite
to the force on moving electron due to magnetic field i.e.
Mahesh Tutorials Science
Force on BC, F1 
0 2I1I 2
 length BC
4 r1
2  15  25

2  102


 25  102

= 9.375 × 10–4 N (repulsive, away from XY)
Force on DA, F2 
 107 
0 2I1I 2
 length DA
4 r2
2  15  25
 25  102
2
 2  10  10
= 1.5625 × 10–4 N (attractive towards XY)
Net force on the loop F = F1 – F2 = (0.375 – 1.5625) × 10–4
= 7.8175 × 10–4 N
(respulsive, away from XY)
MAGNETISM
323
Example - 36
A long straight conductor PQ, carrying a current of 60 A, is
fixed horizontally. Another long conductor XY is kept
parallel to PQ at a distance of 4 mm, in air. Conductor XY is
free to move and carries a current I. Calculate the magnitude
and direction of current I for which the magnetic repulsion
just balances the weight of conductor XY. (Mass per unit
lengths for conductor XY is 10–2 kg/m).
(iii)
 

  M  B  MBsin  where  is the angle between M



and B or between A and B .
Initially,  = 0º,  = MB sin 0º = 0.
Finally,  = 90º,  = MB sin 90º = MB = 10 × 2 = 20 Nm.
d
d d
d
 I   I    MBsin 
dt
d dt
d
(iv)
  I  I

Id  MBsin d .
Integrating it within the given conditions,

/2
0
0
I   d 
Sol. Here, I1 = 60 A ; I2 = I A, r = 4 mm = 4 × 10–3 m ;
I
Mass per unit length of conductor XY, m = 10–2 kg/m.
As magnetic repulsion is balancing the weight of conductor
XY
so,
or
0 2I1I 2
 mg
4 r
I
or
 2ni
22 100  3.2
B 0
 107  2  
 2  103 T
4 r
7
0.10
(ii)
2
 22 
M = niA = nir2 = 100   3.2       0.10   10 Am2
 7 
1/ 2
 2  20 


 0.1 
 20 rad / s.
Change in KE of rotation = work done in rotation
1 2
I  MB  cos 1  cos 2  where
2
The current in XY must flow opposite to that in PQ, because
only then the force will be repulsive.
Sol. (i) Here, n = 100, r = 0.10 m, i = 3.2 A, B = 2 T, I = 0.1 kg m2
 2MB 


 I 
Second Method for (iv)
4  105  9.8
 32.67 A
2  107  60
A 100 turn closely wound circular coil of radius 10 cm
carries a current of 3.2 A. (i) What is the field at the centre of
the coil ? (ii) What is the magnetic moment of this
arrangement ? The coil is placed in a vertical plane and is
free to rotate about a horizontal axis which coincides with
its diameter. A uniform magnetic field of 2 T in the horizontal
direction exists such that initially the axis of the coil is in
the direction of the field. The coil rotates through an angle
of 90º under the influence of the magnetic field. (iii) What
are the magnitudes of the torques on the coil in the initial
and final position ? (iv) What is the angular speed acquired
by the coil when it has rotated by 90º ? The moment of
inertia of the coil is 0.1 kg m2.
2

/2


 MB   cos  0   MB cos  cos 0º   MB
2
2


1/ 2
or
107  2  60  I
 102  9.8
4 103
Example - 37
 MBsin  d
1  0º ; 2  90º , I  0.1 kg m2 ; MB  20 Nm
 2MB  cos 1  cos 2  
 

I


1/ 2

 2  20   cos 0º  cos90º  


0.1


1/ 2
 20 rad / s
Example - 38
A circular coil of 100 turns, radius 10 cm carries a current of
5 A. It is suspended vertically in a uniform horizontal
magnetic field of 0.5 T, the field lines making an angle of
60º with the plane of coil. Calculate the magnitude of the
torque that must be applied on it to prevent it from turning.
Sol. Here, n = 100 ; I = 5 A ; B = 0.5 T ;  = 90º – 60º = 30º ; r = 10
cm = 0.10 m ;
A  r 2 
22
2
  0.10  m 2
7
Torque,  = nIBA sin  = 100 × 5 × 0.5 ×
22
× (0.10)2 × sin 30º
7
= 3.927 N-m
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MAGNETISM
Example - 39
 % decrease in voltage sensitivity 
Compare the current sensitivity and voltage sensitivity of
the following moving coil galvanometers :
3
Vs  Vs
5  100  40%

Vs
Meter A : n = 30, A = 1.5 × 10–3 m2, B = 0.25 T, R = 20 
Meter B : n = 35, A = 2.0 × 10–3 m2, B = 0.25 T, R = 30 
You are given that the springs in the two metres have the
same torsional constants.
Example - 41
A galvanometer having 30 divisions has a current
sensitivity of 20 A/division. It has a resistance of 25 .
How will you convert it into an ammeter upto 1 ampere ?
How will you convert this ammeter into a voltmeter up to 1
volt ?
Sol. For metre A, n1 = 30, A1 = 1.5 × 10–3 m2, B1 = 0.25 T, R1 = 20 .
For metre B, n2 = 35, A2 = 2.0 × 10–3 m2, B2 = 0.25 T, R2 = 30 .
Current sensitivity of a meter 

nBA
k
Sol. Current sensitivity = 20 A/div. = 20 × 10–6 A/div.
Current sensitivity of meter A
Current sensitivity of meter B
Current for full scale deflection, ig = 20 × 10–6 × 30
= 6 × 10–4 A
nBA
k2
nBA
 1 1 1
 1 1 1
k1
n 2 B2 A 2 n 2 B2 A 2  k1  k 2 
For converting galvanometer into ammeter the shunt required
30  0.25  1.5  103 45 9



35  0.25  2.0  103 70 14
Now, voltage sensitivity of a meter 

S
I  Ig
.G 
6  104  25
 0.1050 
1  6  104
G' 
GS
0.015  25

 0.015 
G  S 25  0.015
Conversion of ammeter into voltmeter
Here, Ig = 1 ampere, V = 1 volt, G’ = 0.015 

n1B1A1
k R
nBAR
 2 2  1 1 1 2
k1R1 n 2 B2 A 2 n 2 B2 A 2 R1

30  0.25  1.5  103  30 9 30 27
 

35  0.25  2.0  103  20 14 20 28
Resistance to be used in series,
R
V
1
 G '   0.015  0.985 
Ig
1
Example - 42
Example - 40
A resistance of 1980  is connected in series with a
voltmeter, after which the scale division becomes 100 times
larger. Find the resistance of voltmeter.
The current sensitivity of a moving coil galvanometer
increases by 20% when its resistance is increased by a factor
2. Calculate by what factor the voltage sensitivity changes.
Sol. Given,
Ig
Resistance of ammeter formed,
nBA
kR
Voltage sensitivity of meter A
Voltage sensitivity of meter B
Is'
Vs  Vs'
 100
Vs
20
120
 Is 
Is 
Is ; R '  2R
100
100
Then, initial voltage sensitivity, Vs 
Sol. Let R be the resistance of voltmeter. Let n be the number of
divisions in the voltmeter. The voltage recorded by each
division of voltmeter when current ig flows through it is
ig R/n = V
Is
R
when resistance is connected in series of voltmeter then
ig (R + 1980)/n = 100 V
New voltage sensitivity,
Dividing (ii) by (i), we get
I'  120  1
3
Vs'  s  
Is  
 Vs
R '  100  2R 5
Mahesh Tutorials Science
...(i)
R + 1980 = 100 R
or
R = 1980/99 = 20 
...(ii)
MAGNETISM
Example - 43
325
Example - 46
In the magnetic meridian of a certain place, the horizontal
component of the earth’s magnetic field is 0.26 G and dip
angle is 60º. What is the magnetic field of earth at this
location ?
43. A magnetised steel wire 31.4 cm long has a pole strength of
0.2 Am. It is then bent in the form of a semicircle. Calculate
magnetic moment of the needle.
Sol. Here, L = 31.4 cm. m = 0.2 Am, M = ?
Sol. Here, H = 0.26 G,  = 60º, R = ?
When the wire is bent in the form of a semicircle of radius r,
then L = r = 3.14 r
As
H = R cos 
L
31.4
r

 10 cm
3.14 3.14

R
Distance between the two ends of wire,
Example - 47
A magnetic needle has magnetic moment of 6.7 × 10–2 Am2
and moment of inertia of 7.5 × 10–6 kg m2. It performs 10
complete oscillations in 6.70 s. What is the magnitude of
the magnetic field ?
2 = 2r = 20 cm = 0.2 m
M = m × 2 = 0.2 × 0.2 = 0.04 Am2
Example - 44
A magnetised needle of magnetic moment 4.8 × 10–2 J T–1 is
placed at 30º with the direction of uniform magnetic field of
magnitude 3 × 10–2 T. What is the torque acting on the
needle ?
Sol. Here, M = 4.8 × 10–2 J T–1 ;  = 30º ; B = 3 × 10–2 T
torque,  = ?
As
 = MB sin 

 = 4.8 × 10–2 × 3 × 10–2 sin 30º
H
0.26
0.26


 0.52 G
cos  cos 60º 1/ 2 
Sol. Here, M = 6.7 × 10–2 Am2, I = 7.5 × 10–6 kg m2
Time for one oscillation, T 
From T*  2
4   22 / 7   7.5  10 6
6.7  102  0.67 
2
 0.01 T
Example - 48
A ship is to reach a place 10º south of west. In what
direction should it be steered if declination at the place is
17º west ?
Sol. As the ship is to reach a place 10º south of west i.e. along
OA, figure, therefore, it should be steered west of (magnetic)
north at an angle of (90 – 17 + 10) = 83º.
I
4 2 I
;B
MB
MT 2
2

= 7.2 × 10–4 N-m
Example - 45
6.70
 0.67 s ; B  ?
10
The core of a toroid having 3000 turns has inner and outer
radii 11 cm and 12 cm respectively. Calculate relative
permeability of its core, given that a current of 0.7 amp.
produces a magnetic field of intensity 2.5 T in the core.
Sol. Here, total number of turns = 3000
Average radius, r 
11  12
 11.5 cm
2
= 11.5 × 10–2 m ; i = 0.7 amp. and B = 2.5 T
No. of turns/length,
n
As
3000
3000
3  105


2r 2 11.5  102
23
B=ni
B = 0 r n i

r 
B
2.5  23

 684.5
0 ni 4 107  3  105  0.7
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326
MAGNETISM
Example - 49
(a) 2iBL directed along the negative Z-axis
–5
The susceptibility of magnesium at 300 K is 1.2 × 10 .
At what temperature will the susceptibility be equal to
1.44 × 10–5 ?
m T '
C
Sol. As  m  T  '  T
m
or
T' 
(b) 5iBL directed along the positive Z-axis
(c) iBL direction along the positive Z-axis
(d) 2iBL directed along the positive Z-axis
Sol : (c)
As PQ and UT are parallel to Q, therefore FPQ = FUT = 0
m
1.2  105
T 
 300  250 K
'
m
1.44  105
The current in TS and RQ are in mutually opposite direction.
Hence, FTS – FRQ = 0
Therefore the force will act only on the segment SR whose
value is Bil and it’s direction is +z.
Example - 50
A solenoid has a core of a material with relative permeability
400. The windings of the solenoid are insulated from the
core and carry a current of 2 A. If the number of turns is
1000 per metre, calculate (i) H (ii) B (iii) Intensity of
magnetisation I, and the magnetising current.
Example - 52
An electron moves straight inside a charged parallel plate
. The space between
the plates is filled with constant magnetic field of induction

B. Time of straight line motion of the electron in the
c
Sol. Here, r = 400, I’ = 2A, n = 1000 per metre
(i)
H = nI’ = 1000 × 2 = 2 × 103 Am–1
(ii)
B = H = 0 r H = 4× 10–7 × 400 (2 × 103) = 1.0 T
a
p
a
c
i
t
o
r
t
u
n
i
f
o
r
m
c
h
a
r
g
e
d
e
n
s
i
t
y
capacitor is
×
× –
e
×
×
(iii) From B = 0 (H + I), where I is intensity of magnetisation,
I
a
B
1.0
H 
 2  103
0
4 107
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
= 7.95 × 105 – 0.02 × 105 = 7.93 × 105 Am–1.
(iv) The magnetising current Im is the additional current that
needs to be passed through the windings of the solenoid in
the absence of the core, which would produce a B value as
in the presence of the core. Thus,
B = 0 n (I’ + Im)
1.0 = 4 × 10–7 × 1000 (2 + Im)
Im 
1.0
 2  796  2  794 A
4 104
 ve 
B
P
Y
T
U
X
Mahesh Tutorials Science
0B
(d) e 
 eE = evB
R
Q
e
(c)  B
0
 0 B

The net force acting on the electron is zero because it moves
with constant velocity, due to it’s motion on straight line.

 


 Fnet  Fe  Fm  0  | Fe |  | Fm |
A conductor PQRSTU, each side of length L, bent as shown
in the figure, carries a current i and is placed in a uniform
magnetic induction B directed parallel to the positive Y-axis.
The force experience by the wire and its direction are
S
(b)
Sol. (b)
Example - 51
Z
i
e
(a)  B
0

E


B 0B


E  

0

The time of motion inside the capacitor . t 
  0 B

.
v

MAGNETISM
327
Sol. (d)
Example - 53
A proton of mass m and charge +e is moving in a circular
orbit of a magnetic field with energy 1MeV. What should
be the energy of -particle (mass = 4 m and charge = +2e),
so that it can revolve in the path of same radius
(a) 1 MeV
(b) 4 MeV
(c) 2 MeV
(d) 0.5 MeV
Sol. (a)
By using r 
2 mK
qB
; r  same, B  same
B
0ni; where n 
=
 B  4  10  7 
N
2R
500
 0.5  5  10  4 T.
2   0 .1
Example - 56
Figure shows a square loop ABCD with edge length a.
The resistance of the wire ABC is r and that of ADC is 2r.
The value of magnetic field at the centre of the loop
assuming uniform wire is
q2
m
B

K

 q  m p  2q p
K

Hence      
K p  q p  m   q p
K = Kp = 1meV.
2
2

m
  p 1= 1
 4m p

C
A
O
i
Example - 54
For the solenoid shown in figure. The magnetic field at
point P is
D
n turn
30°
(a)
2  0i
3 a
(b)
2  0i

3 a
(c)
2  0i
a
(d)
2  0i

a
60°
P
 0 ni
4
 3  1
 0 ni
(c)
2
 3  1
(a)
(b)
3  0 ni
4
 0 ni
(d)
4
 3  1
Sol. (b)
According to question resistance of wire ADC is twice that
of wire ABC. Hence current flows through ADC is half
Sol. (a)
B
that of ABC i.e.
0
. 2 ni sin   sin  .
4
From figure  = (90o – 30o) = 60o and  = (90o – 60o) = 30o

B
 0 ni
sin 60  sin 30   0 ni
2
4
 3  1.
 i1 
i
2i
and i 2 
3
3
Magnetic field at centre O due to wire AB and BC
(part 1 & 2) B1 
Example - 55
The average radius of a toroid made on a ring of nonmagnetic material is 0.1 m and it has 500 turns. If it carries
0.5 ampere current, then the magnetic field produced along
its circular axis inside the toroid will be
(a) 25 × 10–2 Tesla
(b) 5 × 10–2 Tesla
(c) 25 × 10–4 Tesla
(d) 5 × 10–4 Tesla
i2 1
 . Also i1 + i2 = 1
i1 2
 0 2i1 sin 45
.

4
/2
 0 2 2 i1 and magnetic field at centre O due to wires
.

4

AD and DC

(i.e. part 3 and 4) B3  B 4 
0 2 2 i2
4

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MAGNETISM
Square coil
Also i1 = 2i2. So (B1 = B2) > (B3 = B4)
i
Hence net magnetic field at centre O
i
Bnet = (B1 + B2) – (B3 + B4)
45°
45°
O
i
B
a/2
(1)
(2)
Length L = 4a
i1
A
C
B
O
i
i2
(3)
(4)
0 2 2 i
.
4 a
B net  4B 
0 8 2 i
.
4 a
B circular
2

Hence B
8 2
square
D
Example - 58
What is the net force on the coil
2 
i
2 2  i
2 2  2

 3   0 .
3
 2 0 .
4
a
4
a

1A
15 cm
2 cm
0 4 2 i
2  1  2  0i 
.
4 3 a
3a
(a) 25 × 10–7 N moving towards wire
Example - 57
(b) 25 × 10–7 N moving away from wire
The ratio of the magnetic field at the centre of a current
carrying circular wire and the magnetic field at the centre
of a square coil made from the same length of wire will be
(a)
10 cm
2A
2
(b)
4 2
2
8 2
(c) 35 × 10–7 N moving towards wire
(d) 35 × 10–7 N moving away from wire
Sol. (a)
Force on sides BC and CD cancel each other.
Force on side AB

(c)

(d)
2 2
4 2
Sol. (b)
FAB  10  7 
2  2 1
2  10  2
 15  10  2  3  10  6 N
Force on side CD
Circular coil
i
FAB  10  7 
2  2 1
12  10  2
 15  10  2  0.5  10  6 N
r
2A
1A
FAB
i
Length L = 2 r
 2i  0 42i
 .
Magnetic field B  0 .
4 r
4 r
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2 cm
B 10 cm C
15 cm
FCD
A
D
Hence net force on loop = FAB – FCD = 25  10–7 N (towards
the wire).
MAGNETISM
329
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
6.
Magnetic Field due to point charge
1.
Equal current I flows in two segments of a circular loop in
the direction shown in figure
A moving charge will produce
(a) no field
(b) an electric field
(c) a magnetic field
(d) both ‘b’ and ‘c’
Magnetic Field due to Current

2.
An element d   dxiˆ (where dx = 1 cm) is placed at the
origin and carries a large current I = 10A. What is the
magnetic field on the y-axis at a distance of 0.5 m ?
ˆ
(a) 2  108 kT
ˆ
(b) 4 108 kT
ˆ
(c) 2  108 kT
ˆ
(d) 4  108 kT
Radius of the loop is r. The magnitude of magnetic field
induction at the centre of the loop is
(a) zero
Right hand rule
3.
A current carrying power line carries current from west to
east. The direction of magnetic field 1m above the power
line will be
(a) east to west
(b) west to east
(c) south to north
(d) north to south
(c)
7.
Current loop
4.
5.
A circular coil A of radius r carries current I. Another circular
coil B of radius 2r carries current of I. The magnetic fields at
the centres of the circular coils are in the ratio of
(a) 3 : 1
(b) 4 : 1
(c) 1 : 1
(d) 2 : 1
A circular conducting ring of radius R is connected to two
exterior straight wires ending at two ends of a diameter.
The current I split into unequal portions while passing
through the ring as shown. What is magnetic field induction
at the centre of the ring?
8.
0 i
   
2 r
(b)
 0 i
4r
(d)
0 i
 2   
2 r
Ratio of magnetic field at the centre of a current carrying
coil of radius R and a distance 3R on its axis is
(a) 10 10
(b) 20 10
(c) 30 10
(d) 5 10
Three rings each having equal radius R are placed mutually
perpendicular to each other and each having centre at the
origin of coordinate axes system .If current I is flowing
through each ring then the magnitude of the magnetic field
at the common centre is
z axis
x axis
I/4
O
R
I
I
y axis
3I/4
I
(a) 0
4R
(c)
0 I
3R
I
(b) 0
8R
(d) zero
3 0 I
2R
(a)
(c)


3  2 0 I
2R
(b)

(d)


3  1 0 I
2R

2  1 0 I
2R
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9.
MAGNETISM
A coil of 50 turns and 10 cm diameter has resistance of 10
ohm. What must be potential difference across the coil
so as to nullify the earth’s magnetic field B = 0.314 G at
the centre of the coil.
(a) 0.5 volt
(b) 1.0 volt
(c) 1.5 volt
(d) 2.5 volt
14. The magnetic flux density B at a distance r from a long
straight rod carrying a steady current varies with r as shown
in figure
B
B
Straight Current Wire
10. Two very long straight parallel wires carry currents I and 2I
in opposite directions. The distance between the wires is r.
At a certain instant of time a point charge q is at a point
equidistant from the two wires in the plane of the wires. Its
instantaneous velocity v is perpendicular to this plane.
The magnitude of the force due to the magnetic field acting
on the charge at this instant is
(a) zero
(c)
11.
 0 Iqv
 r
(b)
30 Iqv
2 r
(d)
 0 Iqv
2 r
The magnetic field at the point of intersection of the
diagonals of a square loop of side length L carrying
current I is
(a)
2 20 I
L
(b)
2 0 I
L
(c)
2 0 I
L
(d)
4 20 I
L
12. A current I flowing through the sides of an equilateral
triangle of side a. The magnitude of the magnetic field at
the centroid of the triangle is
2 0 I
a
(a)
(c)
9 0 I
2a
(b)
3 3 0 I
2a
(d)
2 20 I
2a
(a)
(b)
O
O
r
B
B
(c)
(d)
O
O
r
r
15. A current of i ampere flows along an infinitely long straight
thin walled tube, then the magnetic induction at any point
inside the tube is
(a) infinite
(c)
0 2i
 tesla 
4r
(b) zero
(d)
0i0
tesla
2r
Solenoid
16. A solenoid of 1.5 metre length and 4.0 cm diameter possesses
10 turn per cm. A current of 5 ampere is flowing through it.
The magnetic induction at axis inside the solenoid is
(a) 2 × 10–3 tesla
(b) 2 × 10–5 tesla
(c) 2 × 10–2 gauss
(d) 2× 10–5 gauss
17. At the mid point along the length of a long solenoid, the
magnetic field is equal to B. If the length of solenoid is
doubled and the current is reduced to half, the magnetic
field at the new mid point will nearest to
Inside and outside wire
(a) 2B
(b) B
13. A long, straight, solid metal wire of radius 2 mm carries a
current uniformly distributed over its circular cross-section.
The magnetic field induction at a distance 2 mm from its axis
is B. Then the magnetic field induction at distance 1 mm
from axis will be
(c) B/4
(d) B/2
18. A long solenoid is formed by winding 20 turns/cm. The
current necessary to produce a magnetic field of 20 milli
tesla inside the solenoid will be aproximately
(a) B
(b) B/2
(a) 1.0 A
(b) 2.0 A
(c) 2B
(d) 4B
(c) 4.0 A
(d) 8.0 A
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MAGNETISM
331
19. A long solenoid has 800 turns per metre length of solenoid.
A current of 1.6 A flows through it. The magnetic induction
at the end of the solenoid on its axis is
(a) 16 × 10–4 tesla
(b) 8 × 10–4 tesla
(c) 32 × 10–4 tesla
(d) 4 × 10–4 tesla
20. A toroidal solenoid has 3000 turns and a mean radius of
10 cm. It has soft iron core of relative permeability 2000.
What is the magnitude of magnetic field in the core when
a current of 1 A is passed through the solenoid.
(a) 1.2 T
(b) 12 T
(c) 5.6 T
(d) 4.5 T
Magnetic field
21. A magnetic field
(a) always exerts a force on a charged particle
(b) never exerts a force on a charged particle
(c) exerts a force, if the charged particle is moving across
the magnetic field lines
(d) exerts a force, if the charged particle is moving along the
magnetic field lines
Motion Circular
22. Imagine that you are seated in a room and there is a uniform
magnetic field pointing vertically downwards. At the center
of the room, an electron is projected horizontally with a
certain speed. Discuss the speed and the path of the electron
in this field.
(a) electron moves in anticlockwise path
(b) electron moves in clockwise path
(c) electron moves left wards
(d) electron moves right wards
23. A charged particle moving in a uniform magnetic field
penetrates a layer of lead and thereby loses one half of its
kinetic energy. How does the radius of curvature of its path
change ?
(a) The radius increases to r 2

25. A uniform magnetic field B  B0ˆj exists in space. A particle
of mass m and charge q is projected towards x-axis with
speed v from a point (a, 0, 0). The maximum value of v for
which the particle does not hit the y-z plane is
(a)
Bqa
m
(b)
Bqa
2m
(c)
Bq
am
(d)
Bq
2am
26. A charge +q is moving upwards vertically. It enters a
magnetic field directed to the north. The force on the charged
will be towards
(a) north
(b) south
(c) west
(d) east
27. An electron has a circular path of radius 0.01 m in a
perpendicular magnetic induction 10–3 T. The speed of the
electron is nearly
(a) 1.76 × 104 m/s
(b) 1.76 × 106 m/s
(c) 3.52 × 106 m/s
(d) 7.04 × 106 m/s
28. A charged particle enters a uniform magnetic field with
velocity vector at an angle of 45º with the magnetic field. The
pitch of the helical path is p. The radius of the helix will be
(a)
p

(b)
p
2
p
(c)
2p
(d)
2 
29. A deutron of kinetic energy 50 keV is describing a circular
orbit of radius 0.5 metre in a plane perpendicular to magnetic

field B . The kinetic energy of the proton that describes a
circular orbit of radius 0.5 metre in the same plane with the

same B is
(a) 200 keV
(b) 100 keV
(c) 50 keV
(d) 25 keV
(b) The radius reduces to r / 2
Lorentz force
(c) The radius remains the same
30. An electron and a proton travel with equal speed in the
same direction at 90º to a uniform magnetic field as this is
switched on. They experience forces which are initially
(d) The radius becomes r/2
24. If a charged particle is describing a circle of radius r in a
magnetic field with a time period T, then
(a) T 2  r 3
(b) T 2  r
(c) T  r 2
(d) T  r 0
(a) identical
(b) equal but in opposite direction
(c) in the same direction but differing by a factor of about 1840
(d) in opposite direction and differing by a factor of about 1840
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MAGNETISM
31. The mass of a proton is 1840 times that of electron. If an
electron and a proton are injected in a uniform electric field
at right angle to the direction of the field, with the same
kinetic energy, then
(a) the proton trajectory will be less curved than that of
electron
(b) both the trajectories will be straight
(c) both the trajectories will be equally curved
(d) the electron trajectory will be less curved than that of
proton
32. An electron is moving along positive x axis. A uniform
electric field exists towards negative y axis. What should
be the directions of the magnetic field of suitable
magnitude so that net force on the electron is zero?
(a) positive y axis
(b) positive z axis
(c) negative z axis
(d) negative y axis.
Force on straight current wire
36. The current in wire is directed towards east and the wire is
placed in magnetic field directed towards north. The force
on the wire is
(a) vertically upwards
(b) vertically downwards
(c) due south
(d) due east
37. A current of 3 A is flowing in a linear conductor having a
length of 40 cm. The conductor is placed in a magnetic field
of strength 500 gauss and makes an angle of 30° with the
direction of the field. It experiences a force of magnitude
(a) 3 × 10–4 N
(b) 3 × 10–2 N
(c) 3 × 102 N
(d) 3 × 104 N
38. A charged particle is whirled in a horizontal circle on a
frictionless table by attaching it to a string fixed at one
point. If the magnetic field is switched on in the vertical
direction the tension in the string
Parallel Fields
(a) will increase
33. A uniform electric field and a uniform magnetic field are
pointed in the same direction. If an electron is projected in
the same direction, the electron
(b) will decrease
(a) velocity will increase in magnitude
(b) velocity will decrease in magnitude
(c) will turn to its left
(d) will turn to its right
Under uniform magnetic field
34. A metal wire of mass m slides without friction on two rails
placed at a distance  apart. The track lies in a uniform vertical
magnetic field B. A constant current I flows along the rails
across the wire and back down the other rail. The acceleration
of the wire is
(a)
BmI

(b) mBI
(c)
BI
m
(d)
(d) will remain unchanged
39. A current of 10 ampere is flowing in a wire of length 1.5
metre. A force of 15 newtons acts on it when it is placed in a
uniform magnetic field of 2 tesla. The angle between the
magnetic field and the direction of the current is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
40. A current I1 carrying wire AB is placed near an another long
wire CD carrying current I2. Figure. If free to move, wire AB
will have
mI
B
35. A straight horizontal wire of mass 10 mg and length 1 m
carries a current of 2 ampere .What minimum magnetic field
B should be applied in the region so that the magnetic
force on the wire may balance its weight.
-4
(c) may increase or decrease
-4
(a) rotational motion only
(b) translational motion only
(a) 2.45 × 10 T
(b) 4.9 × 10 T
(c) rotational as well as translational motion
(c) 4.9 × 10-5 T
(d) 9.8 × 10-4 T
(d) neither rotational nor translational motion
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MAGNETISM
333
Torque on Current loop
41. Four wire loops each of length 2.0 metres are bent into four
loops P, Q, R and S and then suspended in a uniform
magnetic field. Same current is passed in each loop. Which
statement is correct ?
46. A current of 2 ampere is passed in a coil of radius 0.5 m and
number of turns 20. The magnetic moment of the coil is
(a) 0.314 Am2
(b) 3.14 A–m2
(c) 314 A–m2
(d) 31.4 A–m2
47. The area of cross-section of three magnets of same length
are A, 2A and 6A respectively. The ratio of their magnetic
moments will be
(a) Couple on loop P will be the highest
(b) Couple on loop Q will be the highest
(c) Couple on loop R will be the highest
(a) 6 : 2 : 1
(b) 1 : 2 : 6
(c) 1 : 4 : 36
(d) 36 : 4 : 1
48. Magnetic field at the centre of the circular loop of area A
is B. Then the magnetic moment of the loop will be
BA A
2
(b) 0
2BA A

(d) none of these
(a) 0
BA A

(d) Couple on loop S will be the highest
42. A circular loop of area 1 cm2 carrying a current of 10 ampere
is placed in a magnetic field of 0.1 T perpendicular to plane
of the loop.The torque on the loop due to magnetic field is
(a) 10-4 N.m
(b) 10-2 N.m
(c) 10 N.m
(d) zero
43. A wire of length  in formed into a circular loop of one turn
only and is suspended in a magnetic field B. When a current
I is passed through the loop, the torque experienced by it is
2
(a) (1/4)BI
2
(a) 0.01 T
(b) 0.2 T
(c) 0.5 T
(d) 0.9T
Current Sensitivity, Voltage
(d) (1/4)BI 
50. The sensitivity of a galvanometer does not depend upon
44. A conducting ring of mass 2 kg and radius of 0.5 m is placed
on a smooth horizontal plane .The ring carries a current of
4 A. A horizontal magnetic field B=10 T is switched on at
t=0 as shown in diagram. What is initial angular acceleration
of the ring ?
I
(a) a very strong magnetic field in the permanent magnet
(b) the current it measures
(c) a very thin, weak suspension
(d) a large number of turns in the coil
Between Parallel Currents
R
B
(a) 40π rad/s
(c) 10 π rad/s
49. A magnetic needle has magnetic moment of 6.7×10-2 A.m2
and moment of inertia 7.5 × 10-6 kgm2.It performs 10 complete oscillations in 6.7 seconds .What is the magnitude of
the magnetic field.
(b) (1/4) IB
2
(c) (1/4)B I
(c) 
0
(b) 20π rad/s
(d) zero
Magnetic moment
45. A wire of length L metre carrying a current I ampere is bent
in the form of a circle. Its magnitude of magnetic moment
will be
51. The forces existing between two parallel current carrying
conductors is F. If the current in each conductor is doubled,
then the value of force will be
(a) 2F
(b) 4F
(c) 5F
(d) F/2
52. Two parallel wires carry currents of 20 A and 40 A in opposite
directions. Another wire carrying current of 20 A and antiparallel to 20A is placed midway between the two wires
.The magnetic force on this wire will be
(a) towards 20 A
(b) towards 40 A
(a) IL/4
(b) I2L2/4
(c) perpendicular to plane of wires
(c) I2L/8
(d) IL2/4
(d) zero
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MAGNETISM
53. Through two parallel wires A and B, 10A and 2A of currents
are passed respectively in opposite directions. If the wire A
is infinitely long and the length of the wire B is 2m, then
force on the conductor B, which is situated at 10 cm distance
from A, will be
(a) 8 × 10–7 N
–7
(c) 4 × 10 N
(b) 8 × 10–5 N
–5
(d) 4 × 10 N
54. If a current is passed in a spring, it
(a) gets compressed
(b) gets expanded
Magnetic Moment
59. A steel wire of length  has a magnetic moment M. It is bent
into L shape from the middle. The new magnetic moment is
(a) M
(b) M / 2
(c) M/2
(d) 2M
Magnetic Field
60. A bar magnet of length 3 cm has a point A and B along axis
at a distance of 24 cm and 48 cm on the opposite ends. Ratio
of magnetic fields at these points will be
(c) oscillates
(d) remains unchanged
55. Choose the correct statement. There will be no force
experienced if
(a) 8
(b) 3
(a) Two parallel wires carry current in same direction
(c) 4
(d) 1/ 2 2
(b) A positive charge is projected along the axis of the
solenoid
(c) A positive charge is projected between the pole pieces
of a bar magnet
(d) Two protons move parallel to each other with same
speed
Conversion
56. The deflection in a galvanometer falls from 50 division to
20 when a 12 ohm shunt is applied. The galvanometer
resistance is
(a) 18 ohms
(b) 36 ohms
(c) 24 ohms
(d) 30 ohms
57. A galvanometer of resistance 100  gives a full scale
deflection for a current of 10–5 A. To convert it into a ammeter
capable of measuring upto 1 A, we should connect a
resistance of
(a) 1 in parallel
61. A short bar magnet of length 4 cm has a magnetic moment
1
of 4JT .What is the magnitude of the magnetic field at a
distance 2 m from the centre of the magnet on its equatorial
line.
(a) 2 105 T
(b) 5 × 10–8 T
(c) 1.2  107 T
(d) 3.4  105 T
62. What is the magnetic field due to a dipole of magnetic
moment 1.2Am2 at a point 1 m away from it .The observation
point is in a direction making an angle of 60o with the
dipole axis.
(a) 1.6 × 10–7 T
(b) 1.2 × 10–6 T
(c) 1.2 × 10–4 T
(d) 1.73 × 10–5 T
63. Two identical dipoles each of magnetic moment 1 A m2 are
placed at a separation of 2 m with their axes perpendicular
to each other. What is the magnetic field at a point midway
between the dipoles ?
(b) 10–3  in parallel
(c) 105  in series
(d) 100  in series
58. We have a galvanometer of resistance 25 . It is shunted by
a 2.5  wire. The part of total current I0 that flows through
the galvanometer is given as
(a) (I/I0) = (1/11)
(b) (I/I0) = (1/10)
(c) (I/I0) = (1/9)
(d) (I/I0) = (2/11)
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M2
2m
M1
(a) 1.2×10-4 T
(b) 3.1×10-3T
(c) 7.6×10-5T
(d)
5×10-7 T
MAGNETISM
335
64. Of the following figure, the lines of magnetic induction due
to a magnet SN, are given by
69. The angles of dip at the poles and the equator respectively are
(a) 30°, 60°
(b) 90°, 0°
(c) 30°, 90°
(d) 0°, 0°
70. At a certain place, the horizontal component of the earth’s
magnetic field is B0 and the angle of dip is 45°. The total
intensity of the field at that place will be
(1)
(2)
(a) B0
(b)
(c) 2B0
(d) B02
2 B0
71. At a certain place on earth, a magnetic needle is placed
along the magnetic meridian at an angle of 60° to the
horizontal. If the horizontal component of the magnetic
field at the place is found to be 2 105 T . What is the
magnitude of total earth’s field at that place.
(3)
(4)
(a) 2 104 T
(b) 4 105 T
(c) 105 T
(d)
3  10 5 T
72. Agonic line is that curve at which
(a) 1
(b) 2
(a) total intensity of earth ‘s magnetic field is same
(c) 3
(d) 4
(b) the angle of dip is same
65. A thin rectangular bar magnet suspended freely has period
of oscillation of 4 seconds. What will be period of oscillation
if the magnet is broken into two halves; each having length
half of original; and one piece is made to oscillate in the
same field.
(c) angle of declination is same
(d) magnetic declination is zero
73. The magnetic lines of force due to horizontal component of
earth’s magnetic field will be
(a) 2 s
(b) 3 s
(a) elliptical
(c) 1 s
(d) 4 s
(b) circular
Earth Magnetism
(c) horizontal and parallel
66. The total intensity of the Earth’s magnetic field at equator is
5 units. What is its value at the poles ?
(d) curved
(a) 5
(b) 4
(c) 3
(d) 2
67. At a certain place, horizontal component of Earth’s field is
3 times the vertical component. The angle of dip at this
place is
(a) 0
(b) /3
(c) /6
(d) none of the above
68. In a magnetic meridian of a certain place, horizontal component of earth’s field is 0.25G and the angle of dip is 60o.What
is the magnetic field of the earth at this location.
74. The magnetic induction along the axis of an air cored
solenoid is 0.03 T. On placing an iron core inside the
solenoid the magnetic induction becomes 1.5T .The relative
permeability of iron core will be
(a) 12
(b) 40
(c) 50
(d) 300
Magnetizing Field Intensity
75. An iron rod of length 20 cm and diameter 1 cm is placed
inside a solenoid on which the number of turns is 600. The
relative permeability of the rod is 1000. If a current of 0.5 A is
placed in the solenoid, then the magnetisation of the rod
will be
(a) 0.5G
(b) 0.25 G
(a) 2.997 × 102 A/m
(b) 2.997 × 103 A/m
(c) 0. 25 3 G
(d) none of these
(c) 2.997 × 104 A/m
(d) 2.997 × 105 A/m
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336
76.
77.
MAGNETISM
The mass of iron rod is 80 gm and its magnetic moment is
10A.m2.If the density of iron is 8 gm/cc, then the value of
intensity of magnetization will be
(a) 106A/m
(b) 3000 A/m
(c) 105 A/m
(d) 1A/m
A solenoid has core of a material with relative permeability 400.The winding of the solenoid are insulated from
the core and carry a current of 2 ampere. If the number of
turns is 1000 per meter, what is magnetic flux density inside the core?
(a) 0.4T
(c) 0.7 T
79.
84.
(b) 0.5 T
(d) 1.0T
Susceptibility
78.
83.
85.
The magnetic susceptibility of a material of a rod is 499.
Permeability of vacuum is 4 × 10 –7 H/m. Absolute
permeability of the material of the rod in henry/meter is
(a) × 10–4
(b) 4 × 10–4
(c) 3 × 10–4
(d) 2 × 10–4
86.
Magnetic susceptibility is negative for
(a) Paramagnetic material only
87.
(b) Diamagnetic material only
(c) Ferromagnetic material only
(d) Paramagnetic and Ferromagnetic materials
Magnetic permeability
80.
81.
A magnetising field of 2 × 103 amp/m produces a magnetic
flux density of 8 tesla in an iron rod. The relative
permeability of the rod will be
(a) 102
(b) 100
(c) 103
(d) 104
The main difference between electric lines of force and
magnetic lines of force is
(a) 1000 A/m
(b) 1400 A/m
(c) 2000 A/m
(d) 2400A/m
In the above problem, magnetizing field in the presence
of core will be
(a) 1000 A/m
(b) 2000 A/m
(c) 2400 A/m
(d) 3200 A/m
The intensity of magnetization in the presence of core
will be
(a) 1000 A/m
(b) 2.3 × 104 A/m
(c) 7.94 × 105 A/m
(d) 4.3 × 10-5 A/m
The magnetization in the absence of the core will be
(a) 2400 A/m
(b) 2.3 × 104 A/m
(c) 7.94 × 105 A/m
(d) zero
The relative permeability of the material will be
(a) 397.7
(b) 448.5
(c) 533
(d) 657
The coercivity of a bar magnet is 4000A/m .In order to
demagnetize it is placed inside a solenoid of length 12 cm
and having 60 turns. What current should be passed
through the solenoid?
(a) 2A
(b) 4A
(c) 8A
(d) 16A
Ferromagnetic
88.
A uniform magnetic field parallel to the plane of paper,
existed in space initially directed from left to right. When a
bar of soft iron is placed in the field parallel to it, the lines
of force passing through it will be represented by figure
(a) Electric lines of force are closed curves whereas
magnetic lines are open curve
(b) Electric lines of force are open curve and magnetic
lines are closed curve
(A)
(B)
(c) Magnetic field lines cut each other whereas
electric lines don’t
(d) Electric lines of force cut each other whereas magnetic
lines of force don’t cut
82.
There are 1000 turns /m in a Rowland’s ring and a current
of 2A is flowing in the windings .The value of magnetic
induction produced is found to be 1.0T.When no core is
present then magnetizing field produced in the ring will
be
Mahesh Tutorials Science
(C)
(D)
(a) A
(b) B
(c) C
(d) D
MAGNETISM
89.
337
A sensitive magnetic instrument can be shielded very
effectively from outside magnetic field by placing it
inside a box of
(a) Teak wood
(b) plastic material
(c) A metal of low magnetic permeability
90.
91.
Curies Law
94. Curie’s law states that
(a) magnetic susceptibility is inversely proportional to the
absolute temperature
(b) magnetic susceptibility is inversely proportional the
square root of the absolute temperature
(d) A metal of high magnetic permeability
(c) magnetic susceptibility is directly proportional to the
absolute temperature
When a Ferromagnetic substance is heated to a
temperature above its Curie temperature it
(d) magnetic susceptibility does not depend on temperature
Hysteris Curve
(a) behaves like Diamagnetic material
95. The hysterisis curve is studied generally for
(b) behaves like Paramagnetic material
(a) ferromagnetic materials
(c) is permanently demagnetized
(b) paramagnetic materials
(d) remains Ferromagnetic
(c) diamagnetic materials
A Ferromagnetic material is placed in an external magnetic
field. The magnetic domains
(d) all of these
96. The B–H curve (i) and (ii) shown in figure associated with
(a) increase in size
(b) decrease in size
(c) may increase or decrease in size
(d) have no relation with field
Diamagnetic
92.
93.
For a diamagnetic material
(a)  r  1,  m  1
(b)  r  1,  m  1
(a) (i) diamagnetic and (ii) paramagnetic substance
(c)  r  1,  m  0
(d)  r  1,  m  0
(b) (i) paramagnetic and (ii) ferromagnetic substance
(c) (i) Soft iron and (ii) Steel respectively
Water is
(a) diamagnetic
(b) paramagnetic
(c) ferromagnetic
(d) none of these
(d) (i) steel and (ii) Soft iron respectively
Permanent Magnets
97. The most suitable metal for permanent magnet is
(a) copper
(b) aluminium
(c) steel
(d) iron
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MAGNETISM
EXERCISE - 2 : PREVIOUS YEAR COMPETITION QUESTIONS
PREVIOUSYEARSAFMC QUESTIONS
1.
7.
A current carrying wire in the neighbourhood produces
(AFMC 1999)
(a) electric and magnetic fields
8.
(b) magnetic field only
(c) no field
–6
(b) 2 × 10 N
–3
(d) 2 × 10 N
(c) 2 × 10 N
3.
9.
–4
–2
A long hollow copper pipe carries a current, then magnetic
field produced is
(AFMC 1999)
(a) both inside and outside the pipe
10.
(b) neither inside nor outside the pipe
(c) outside the pipe only
(d) inside the pipe only
4.
2
A coil having 100 turns and area of 0.001 m is free to
rotate about an axis, the coil is placed perpendicular to a
2
magnetic field of 1.0 Wb/m . If the coil is rotated rapidly
through an angle of 180°, how much charge will flow
through the coil ? The resistance of the coil is 10.
11.
(AFMC 2002)
5.
6.
(b) Temperature
(c) Flux density
(d) Magnetic field intensity
A wire carrying current I and other carrying 2 I in the same
direction produce a magnetic field B at the mid-point. What
will be the field when 2 I wire is switched off ?
(b) 2B
(c) B
Two parallel wires in free space are 10 cm apart and each
carries a current of 10A in the same direction. The force
exerted by one wire on other per metre of length of the
wire is
(AFMC 1999)
(a) 2 × 10 N
(a) 0.02 C
(b) 0.04 C
(c) 0.08 C
(d) 0.01 C
2
Wb/m is equal to
12.
(AFMC 2001)
(a) dyne
(b) tesla
(c) watt
(d) henry
R
B
(b) repel each other
(d) any of the above
Mahesh Tutorials Science
(AFMC 2005)
(d) 4B
When a charged particle moving with velocity v is

subjected to a magnetic feild of induction B , the force on
it is non-zero. This implies that
(AFMC 2006)


(a) angle between v and B is necessarily 90°


(b) angle between v and B can have any value other
than 90°


(c) angle between v and B can have any value other
than zero and 180°


(d) angle between v and B is either zero or 180°
The path of an electron in a uniform magnetic field may be
(a) circular but not helical
(AFMC 2007)
(b) helical but not circular
(c) neither helical nor circular
(d) either helical or circular
A straight wire of mass 200 g and length 1.5 m carries a
current of 2 A. It is suspended in mid-air by a uniform
horizontal magnetic field B. The magnitude of B (in tesla)
–2
is (assume g = 9.8 ms )
(AFMC 2008)
(a) 2
(b) 1.5
(c) 0.55
(d) 0.65
A wire PQR is bent as shown in fig.and is placed in a region
of uniform magnetic field B. The length of PQ = QR = l. A
current I ampere flows through the wire as shown. The
magnitude of the force on PQ and QR will be
(AFMC 2008)
Two wires carry current in different directions. They will
(AFMC 2003)
(a) attract each other
(c) create gravitational field
(AFMC 2003)
(a) Time
(a) B/2
(d) electric field
2.
Which one is a vector quantity ?
P
I
I
Q
(a) BIl, 0
(b) 2BIl, 0
(c) 0, BIl
(d) 0, 0
MAGNETISM
13.
339
When a positively charged particle enters a uniform
magnetic field with uniform velocity, its trajectory can be
(AFMC 2008)
(1) a straight line
20.
(2) a circle
(3) a helix
(a) (1) only
(b) (1) or (2)
(c) (1) or (3)
(d) any one of (1), (2) and (3)
21.
PREVIOUS YEARS CBSE–PMT QUESTIONS
14.
A current carrying coil is subjected to a uniform magnetic
field. The coil will orient so that its plane becomes
(CBSE–PMT 1988)
(a) inclined at 45° to the magnetic field
22.
(b) inclined at any arbitrary angle to the magnetic field
(c) parallel to the magnetic field
(d) perpendicular to magnetic field
15.
16.
17.
Tesla is the unit of
(CBSE–PMT 1988)
(a) magnetic flux
(b) magnetic field
(c) magnetic induction
(d) magnetic moment
(b) 0.5 cm
(c) 4.0 cm
(d) 1.0 cm
The magnetic field at a distance ‘r’ from a long wire
carrying current ‘i’ is 0.4 Tesla. The magnetic field at a
distance ‘2r’ is
(CBSE–PMT 1992)
(a) 0.2 Tesla
(b) 0.8 Tesla
(c) 0.1 Tesla
(d) 1.6 Tesla
A straight wire of length 0.5 metre and carrying a current
of 1.2 ampere is placed in uniform magnetic field of
induction 2 Tesla. The magnetic field is perpendicular to
the length of the wire. The force on the wire is
(CBSE–PMT 1992)
(a) 2.4 N
(b) 1.2 N
23.
(d) 2.0 N
To convert a galvanometer into an ammeter, one needs to
connect a
(CBSE–PMT 1992)
Energy in a current carrying coil is stored in the form of
(CBSE–PMT 1989)
(a) low resistance in parallel
(a) electric field
(b) magnetic field
(c) low resistance in series
(c) dielectric strength
(d) heat
(b) high resistance in parallel
(d) high resistance in series
The total charge induced in a conducting loop when it is
moved in magnetic field depends on (CBSE–PMT 1990)
24.
A coil carrying electric current is placed in uniform magnetic
field
(CBSE–PMT 1993)
(a) torque is formed
(b) e.m.f. is induced
(b) initial magnetic flux only
(c) both (a) and (b) are correct
(c) the total change in magnetic flux
(d) none of these
(d) final magnetic flux only
25.
The magnetic induction at a point P which is at the distance
–3
of 4 cm from a long current carrying wire is 10 T. The
field of induction at a distance 12 cm from the current will
be
(CBSE–PMT 1990)
–4
(a) 3.33 × 10 T
–3
(c) 3 × 10 T
19.
(a) 2.0 cm
(c) 3.0 N
(a) the rate of change of magnetic flux
18.
A uniform magnetic field acts right angles to the direction
of motion of electrons. As a result, the electron moves in a
circular path of radius 2 cm. If the speed of electrons is
doubled, then the radius of the circular path will be
(CBSE–PMT 1991)
(CBSE–PMT 1993)
(a) remain unaffected
(b) start moving in a circular path Y-Z plane
–4
(b) 1.11 × 10 T
(c) retard along X-axis
–3
(d) 9 × 10 T
A deuteron of kinetic energy 50 keV is describing a circular
orbit of radius 0.5 metre in a plane perpendicular to magnetic
field B. The kinetic energy of the proton that describes a
circular orbit of radius 0.5 metre in the same plane with the
same B is
(CBSE–PMT 1991)
(a) 25 keV
(b) 50 keV
(c) 200 keV
(d) 100 keV
A charge moving with velocity v in X-direction is
subjected to a field of magnetic induction in negative
X-direction. As a result, the charge will
(d) moving along a helical path around X-axis
26.
A electron enters a region where magnetic (B) and electric
(E) fields are mutually perpendicular, then
(CBSE–PMT 1994)
(a) it will always move in the direction of B
(b) it will always move in the direction of E
(c) it always possesses circular motion
(d) it can go undeflected also
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340
27.
MAGNETISM
A straight wire of diameter 0.5 mm carrying a current of 1A
is replaced by another wire of 1mm diameter carrying same
current. The strength of magnetic field far away is
(CBSE–PMT 1995)
(c) both inside and outside the pipe
(d) no where
34.
(a) twice the earlier value
A straight wire of diameter 0.5 mm carrying a current of 1A
is replaced by another wire of diameter 1 mm carrying the
same current. The strength of magnetic field far away is
(b) same as the earlier value
28.
(a) twice the earlier value
(d) one-quarter of the earlier value
(b) one-half of the earlier value
At what distance from a long straight wire carrying a
–6
current of 12A will the magnetic field be equal to 3 × 10
2
Wb/m ?
(CBSE–PMT 1995)
(c) one quarter of the earlier value
–1
(a) 8 × 10 m
–2
(c) 18 × 10 m
29.
30.
(CBSE–PMT 1999)
(c) one-half of the earlier value
–2
(b) 12 × 10 m
(d) same as earlier value
35.
–2
(d) 24 × 10 m
The magnetic field (dB) due to a small element (dl) at a

distance ( r ) and element carrying current i is
(CBSE–PMT 1996)


0 2  dl  r 
  dl  r 
dB

i
(a) dB  0 i 
(b)



4  r 2 
4  r 


0  dl  r 

 dl  r 
dB

i
(c) dB  0 i 2 
(d)



4  r 3 
4  r 
A 10eV electron is circulating in a plane at right angles
–4
2
to a uniform field at magnetic induction 10 Wb/m
(= 1.0 gauss). The orbital radius of the electron is
(CBSE–PMT 1996)
(a) 12 cm
(b) 16 cm
Magnetic field due to 0.1A current flowing through a
circular coil of radius 0.1m and 1000 turns at the centre of
the coil is
(CBSE–PMT 1999)
–4
(a) 0.2 T
(b) 2 × 10 T
–8
(c) 4.9 × 10 T
36.
–4
(d) 9.8 × 10 T
3
An electron moves with a velocity 1 × 10 m/s in a magnetic
field of induction 0.3 T at an angle 30°. If e/m of electron is
11
1.76 × 10 C/kg, the radius of the path is nearly
(CBSE–PMT 2000)
–8
(b) 2 × 10 m
–6
–10
(a) 10 m
37.
–8
(c) 10 m
(d) 10 m
Current is flowing in a coil of area A and number of turns
N, then magnetic moment of the coil M is equal to
(CBSE–PMT 2001)
(a) NiA
(b) Ni/A
2
8 2
(c) 2 cm

31.
(d) 18 cm
38.
Two parallel wires in free space are 10 cm apart, and each
carries a current of 10A, in the same direction. The force,
one wire exerts on the other, per metre of length, is
(CBSE–PMT 1997)
–7
(b) 2 × 10 N, attractive
–4
(d) 2 × 10 N, attractive
(a) 2 × 10 N, repulsive
(c) 2 × 10 N, repulsive
32.
–7
For protecting a sensitive equipment from the external
magnetic field, it should be
(CBSE–PMT 1998)
39.
(b) placed inside an iron can
(c) wrapped with insulation around it when passing current
through it
(d) surrounded with fine copper sheet
If a long hollow copper pipe carries a current, then
magnetic field is produced
(CBSE–PMT 1999)
(a) inside the pipe only
Mahesh Tutorials Science
(a)
qB
m
(b)
qB
2m
(c)
qBE
2m
(d)
qB
2E
–4
(a) placed inside an aluminium can
33.
(c) Ni / A
(d) N Ai
A charged particle of charge q and mass m enters

perpendicularly in a magnetic field B . Kinetic energy of
the particle is E; then frequency of rotation is
(CBSE–PMT 2001)
(b) outside the pipe only
40.
The magnetic field of a given length of wire carrying a
current for a single turn circular coil at centre is B, then its
value for two turns for the same wire when same current
passing through it is
(CBSE–PMT 2002)
(a) B/4
(b) B/2
(c) 2B
(d) 4B

A charge q moves in a region where electric field E and

magnetic field B both exist, then the force on it is
(CBSE–PMT 2002)
MAGNETISM
41.
42.
341
 
(a) q v  B

 
(b) q E  q v  B

 
(c) q B  q B  v

 
(d) q B  q E  v


The magnetic flux through a circuit of resistance R changes
by an amount  in a time t. Then the total quantity of
electric charge Q that passes any point in the circuit during
the time t is represented by
(CBSE–PMT 2004)
(a) Q 
1 
.
R t
(b) Q 
(c) Q 

t
(d) Q  R .
46.

R
2   
 
3  Bi 
(b)
2   
 
3  Bi 
(d)
1 
3 Bi
1/ 2
1/ 2
  

(c) 2 
 3 Bi 
(c) opposite to oy
44.
(d) along oy


If the angle between the vectors A and B is , the value
  
of the product B  A .A is equal to


(c)
(d)
v/B
When a charged particle moving with velocity v is

subjected to a magnetic field of induction B , the force on
it is non-zero. This implies that
(CBSE–PMT 2006)


(a) angle between v and B is necessarily 90°
zero and 180°


(d) angle between v and B is either zero or 180°
47.
Under the influence of a uniform magnetic field a charged
particle is moving in a circle of radius R with constant
speed v. The time period of the motion
(CBSE–PMT 2007)
(a) depends on v and not on R
(b) depends on both R and v
(c) is independent of both R and v
(d) depends on R and not on v
49.
A charged particle (charge q) is moving in a circle of radius
R with uniform speed v. The associated magnetic moment
 is given by
(CBSE–PMT 2007)
(a)
qvR
2
(b) qvR
(c)
qvR 2
2
(d) qvR
2
A closed loop PQRS carrying a current is placed in a uniform
magnetic field. If the magnetic forces on segments PS, SR
and RQ are F1, F2 and F3 respectively and are in the plane
of the paper and along the directions shown, the force on
the segment QP is
(CBSE–PMT 2008)
Q
(CBSE–PMT 2005)
(a) BA cos 
(b) BA sin 
(c) BA sin  cos 
(d) zero
2
2
45.
P
2
An electron moves in a circular orbit with a uniform speed
v. It produces a magnetic field B at the centre of the circle.
The radius of the circle is proportional to
B/ v


(c) angle between v and B can have any value other than
48.
(b) along ox
(b) v/B
90°
A very long straight wire carries a current I. At the instant
when a charge +Q at point P has velocity v , as shown,
the force on the charge is
(CBSE–PMT 2005)
(a) opposite to ox
(a) B/v


(b) angle between v and B can have any value other than

t
A coil in the shape of an equilateral triangle of side l is
suspended between the pole pieces of a permanent magnet

such that B is in plane of the coil. If due to a current i in
the triangle a torque  acts on it, the side l of the triangle is
(CBSE–PMT 2005)
(a)
43.
(CBSE–PMT 2005)
F3
F1
S
R
F2
Lakshya Educare
342
MAGNETISM
(a) F3 – F1 – F2
(c)
50.
51.
 F3  F1 
2
(b)
F
2
2
 F3  F1 
2
(b) putting in series a resistance of 15 
 F22
(c) putting in series a resistance of 240 
(d) putting in parallel a resistance of 15 
(d) F3 – F1 + F2
A particle mass m, charge Q and kinetic energy T enters a

transverse uniform magnetic field of induction B . After 3
s the kinetic energy of the particle will be
(CBSE–PMT 2008)
(a) 3T
(b) 2T
(c) T
(d) 4T
56.
(a) repelled by both the pole
(b) repelled by the north pole and attracted by the south
pole
(c) attracted by the north pole and repelled by the south
pole
A circular disc of radius 0.2 m is placed in a uniform
1  Wb 

 in such a way that
  m2 

its axis makes an angle of 60° with B . The magnetic flux
linked with the disc is
(CBSE–PMT 2008)
magnetic field of induction
52.
53.
(a) 0.02 Wb
(b) 0.06 Wb
(c) 0.08 Wb
(d) 0.01 Wb
(b) 5550 
(c) 6050 
(d) 4450 
(d) attracted by both the poles.
57.
4
free to rotate in a horizontal plane. A horizontal magnetic
–4
taking the magnet slowly from a direction parallel to the
field to a direction 60° from the field is :
58.
(a) 0.6 J
(b) 12 J
(c) 6 J
(d) 2 J
Charge q is uniformly spread on a thin ring of radius R.
Hz. The magnitude of magnetic induction at the center of
the ring is :
Under the influence of a uniform magnetic field, a charged
particle moves with constant speed v in a circle of radius
R. The time period of rotation of the particle
0 q
2fR
(b)
0 qf
2R
(c)
0 qf
2R
(d)
0 q
2fR
(b) depends on R and not on v
(c) is independent of both v and R
59.
(d) depends on both v and R
The magnetic force acting on a charged particle of charge
– 2C in a magnetic field of 2 T acting in y direction, when

6
–1
the particle velocity is 2iˆ  3jˆ × 10 ms is
55.
(a) 8 N in –z direction
(CBSE 2009)
(b) 4 N in z direction
(c) 8 N in y direction
(d) 8 N in z direction
A galvanometer having a coil resistance of 60  shown
full scale deflection when a current of 1.0 A passes through
it.It can be converted into an ammeter to read currents
upto 5.0 A by
(CBSE 2009)
(a) putting in parallel a resistance of 240 
Mahesh Tutorials Science
60.
(CBSE 2011)
(a)
(CBSE 2009)

(CBSE 2009)
The ring rotates about its axis with a uniform frequency f
(a) depends on v and not on R
54.
–1
A bar magnet having a magnetic moment of 2 × 10 JT is
field B = 6 × 10 T exists in the space. The work done in
A galvanometer of resistance 50  is connected to a battery
of 3 V alongwith a resistance of 2950 in series. A full scale
deflection of 30 divisions is obtained in the galvanometer.
In order to reduce this deflection to 20 divisions, the
resistance in series should be
(CBSE 2008)
(a) 5050 
If a diamagnetic substance is brought near the north or
the south pole of a bar magnet, it is
(CBSE 2009)
Two similar coils of radius R are lying concentrically with
their planes at right angles to each other. The currents
flowing in them are I and 2I, respectively. The resultant
magnetic field induction at the centre will be(CBSE 2012)
(a)
0 I
R
(b)
(c)
3 0 I
2R
(d)
5 0 I
2R
0 I
2R
A compass needle which is allowed to move in a horizontal
plane is taken to a geomagnetic pole. It : (CBSE 2012)
(a) will stay in east-west direction only
MAGNETISM
343
(b) will become rigid showing no movement
65.
(c) will stay in any position
A bar magnet of length ‘’ and magnetic dipole moment
‘M’ is bent in the form of an arc as shown in figure. The
new magnetic dipole moment will be
(CBSE 2013)
61.
An alternating electric field, of frequency v, is applied
across the dees (radius = R) of a cyclotron that is being
used to accelerate protons (mass = m). The operating
magnetic field (B) used in the cyclotron and the kinetic
energy (K) of the proton beam, produced by it, are given
by
(CBSE 2012)
60°
r
(d) will stay in north-south direction only
(a) B 
mv
and K  m 2 vR 2
e
(a)
M
2
(b) M
(b) B 
mv
and K  2m2 v 2 R 2
e
(c)
3
M

(d)
2
M

PREVIOUSYEARSAIIMS QUESTIONS
2 mv
and K  m 2 vR 2
(c) B 
e
(d) B 
62.
63.
66.
2 mv
and K  2m  2 v 2 R 2
e
A proton carrying 1 MeV kinetic energy is moving in a
circular path of radius R in uniform magnetic field. What
should be the energy of an –particle to describe a circle
of same radius in the same field ?
(CBSE 2012)
(a) 4 MeV
(b) 2 MeV
(c) 1 MeV
(d) 0.5 MeV
67.
3 J of work to turn it through 60°. The torque
needed to maintain the needle in this position will be
64.
3J
A current loop in a magnetic field :
(d) 50 cm
Which one of the following statement is not correct about
the magnetic field ?
(AIIMS 2000)
(d) Magnetic lines of force do not cut each other
68.
(b) 2 3 J
(d)
(c) 25 cm
(c) The magnetic lines form a closed loop
(CBSE 2012)
(c) 3 J
(b) 75 cm
(b) Tangents to the magnetic lines give the direction of
the magnetic field
reqiures
3
J
2
(a) 100 cm
(a) Inside the magnet the lines go from north pole to south
pole of the magnet
A magnetic needle suspended parallel to a magnetic field
(a)
–14
An electron moving with kinetic energy 6.6 × 10 J enters
–3
a magnetic field 4 × 10 T at right angle to it. The radius of
its circular path will be nearest to
(AIIMS 1997)
69.
(CBSE 2013)
(a) Can be in equilibrium in two orientations, one stable
while the other is unstable.
What should be amount of current through the ring of
radius of 5 cm so that field at the centre is equal to the
–5
2
earth’s magnetic field 7 × 10 Wb/m is ? (AIIMS 2000)
(a) 0.28 A
(b) 5.57 A
(c) 2.8 A
(d) none of these
Which one of the following are used to express intensity
of magnetic field in vacuum ?
(AIIMS 2000)
(a) Oersted
(b) Tesla
(c) Gauss
(d) None of these
(b) Experiences a torque whether the field is uniform or
non uniform in all orientations
(c) Can be in equilibrium in one orientation
(d) Can be in equilibrium in two orientations, both the
equilibrium states are unstable.
Lakshya Educare
344
70.
MAGNETISM
the separation H is
An electron is travelling along the x-direction. It encounters
a magnetic field in the y-direction. Its subsequent motion
will be
(AIIMS 2003)
(AIIMS 2006)
R
Wire
Ic
(a) straight line along the x-direction
(b) a circle in the xz-plane
H
(c) a circle in the yz-plane
(d) a circle in the xy-plane
71.
Straight
A rectangular loop carrying a current i1, is situated near a
long straight wire carrying a steady current i2. The wire is
parallel to one of the sides of the loop and is in the plane
of the loop as shown in the figure. Then, the current loop
will
(AIIMS 2003)
i1
75.
i2
Ie
Ie R
(a) I 
c
Ic R
(b) I 
e
Ic
(c) I R
e
Ie 
(d) I R
c
The figure shows three situations when an electron with

velocity v travels through a uniform magnetic field B . In
each case, what is the direction of magnetic force on the
electron ?
(AIIMS 2007)
(a) move away from the wire
(b) move towards the wire
(c) remain stationary
(d) rotate about an axis parallel to the wire
72.
A circular coil of radius R carries an electric current. The
magnetic field due to the coil at a point on the axis of the
coil located at a distance r from the centre of the coil, such
that r > > R, varies as
(AIIMS 2004)
(a) 1/r
(b) 1/r
2
(c) 1/r
73.
(a) +ve z-axis, –ve x-axis, +ve y-axis
(b) –ve z-axis, –ve x-axis and zero
(c) +ve z-axis, +ve y-axis and zero
3/2
3
(d) 1/r
(d) –ve z-axis, +ve x-axis and zero
76.
The magnetic field due to a straight conductor of uniform
cross-section of radius a and carrying a steady current is
represented by
(AIIMS 2004)
77.
(a)
(b)
A long straight wire of radius a carries a steady current I.
The current is uniformly distributed across its crosssection. The ratio of the magnetic field at a/2 and 2a is
(a) 1/4
(b) 4
(c) 1
(d) 1/2
(AIIMS 2008)
Statement–1 : The magnetic field produced by a current
carrying solenoid is independent of its length and crosssectional area.
Statement–2 : The magnetic field inside the solenoid is
uniform.
(AIIMS 2008)
(c)
74.
(d)
Circular loop of a wire and a long straight wire carry
currents I c and I e, respectively as shown in figure.
Assuming that these are placed in the same plane. The
magnetic field will be zero at the centre of the loop when
Mahesh Tutorials Science
(a) If both Statement–1 and Statement–2 are true and the
Statement–2 is the correct explanation of the Statement–1.
(b) If both Statement–1 and Statement–2 are true but the
Statement–2 is not the correct explanation of the
Statement–1.
(c) If Statement–1 is true but Statement–2 is false.
(d) If both Statement–1 and Statement–2 are false.
MAGNETISM
78.
345
If M be the mass of the charged particle, which enters with
velocity v normal to the magnetic field B, it will revolve
with angular speed given by ?
(AIIMS 1996)
B
(a) q M
0
(c)
q0 M
B
(b)
q0 B
M
(d) q0BM
Lakshya Educare
346
MAGNETISM
ANSWER KEY
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
1. (d)
11. (a)
21. (c)
31. (c)
41. (d)
51. (b)
61. (b)
71. (b)
81. (b)
91. (c)
2. (b)
12. (c)
22. (b)
32. (c)
42. (d)
52. (b)
62. (a)
72. (d)
82. (c)
92. (c)
3. (d)
13. (b)
23. (b)
33. (b)
43. (b)
53. (b)
63. (d)
73. (c)
83. (b)
93. (a)
4. (d)
14. (d)
24. (d)
34. (c)
44. (a)
54. (a)
64. (a)
74. (c)
84. (c)
94. (a)
5. (b)
15. (b)
25. (a)
35. (c)
45. (d)
55. (b)
65. (a)
75. (d)
85. (d)
95. (a)
6. (c)
16. (a)
26. (c)
36. (a)
46. (d)
56. (a)
66. (a)
76. (a)
86. (a)
96. (c)
7. (a)
17. (d)
27. (b)
37. (b)
47. (b)
57. (b)
67. (c)
77. (d)
87. (c)
97. (c)
8. (a)
18. (d)
28. (b)
38. (c)
48. (c)
58. (a)
68. (a)
78. (d)
88. (b)
9. (a)
19. (b)
29. (b)
39. (a)
49. (a)
59. (b)
69. (b)
79. (b)
89. (d)
10. (a)
20. (b)
30. (b)
40. (c)
50. (b)
60. (a)
70. (b)
80. (d)
90. (b)
8. (c)
18. (a)
28. (a)
38. (b)
48. (a)
58. (c)
68. (b)
78. (b)
9. (c)
19. (d)
29. (d)
39. (d)
49. (b)
59. (b)
69. (a)
10. (d)
20. (c)
30. (c)
40. (b)
50. (c)
60. (b)
70. (b)
EXERCISE - 2 : PREVIOUS YEAR COMPETITION QUESTIONS
1. (b)
11. (d)
21. (a)
31. (d)
41. (b)
51. (a)
61. (d)
71. (b)
2. (b)
12. (c)
22. (b)
32. (b)
42. (c)
52. (d)
62. (c)
72. (d)
3. (c)
13. (d)
23. (a)
33. (b)
43. (d)
53. (c)
63. (c)
73. (a)
4. (a)
14. (c)
24. (a)
34. (d)
44. (d)
54. (a)
64. (a)
74. (a)
5. (b)
15. (b)
25. (a)
35. (c)
45. (c)
55. (d)
65. (c)
75. (b)
6. (b)
16. (b)
26. (d)
36. (a)
46. (c)
56. (a)
66. (d)
76. (c)
Dream on !!

Mahesh Tutorials Science
7. (d)
17. (c)
27. (b)
37. (a)
47. (c)
57. (c)
67. (a)
77. (b)