Download workshop on ME - Kendriya Vidyalaya SAC, Vastrapur, Ahmedabad

KENDRIYA VIDYALAYA SANGATHAN, AHMEDABAD REGION
II WORKSHOP OF PHYSICS ON 23rd JULY-2016
QUESTION BANK ON MOVING CHARGE AND MAGNETISM (MAGNETIC EFFECT OF CURRENT)
1 MARK QUESTIONS
1
Where is the magnetic field at a current element (i) minimum and (ii) maximum?
ANS i) Minimum
- along axis
ii) Maximum - in a plane passing throw’ element and perpendicular to axis
2
Consider the circuit shown, where APB and AQB are semicircles. What will be the magnetic field at
the centre C of the circular loop?
P
A
I

C
B
I
Q
Zero
What will be the path of a charged particle moving in a uniform magnetic field at any arbitrary angle?
Helical
In a certain arrangement, a proton does not get deflected while passing through a magnetic field
region. State the condition under which it is possible.
ANS v is parallel or antiparallel to B
5
A galvanometer gives full scale deflection with the current Ig. Can it be converted into an ammeter of
range I < Ig ?
ANS NO, Hint : Use S = Ig G / (I – Ig)
2 MARKS QUESTIONS
1
Define the term magnetic dipole moment of a current loop. Write the expression for the
ANS
3
ANS
4
magnetic moment when an electron revolves at a speed ‘v’, around an orbit of radius ‘ r’ in
hydrogen atom
ANS Magnetic moment of a current loop:
M = NIA
i.e., magnetic moment of a current loop is the product of number of turns, current flowing in
the loop and area of loop. Its direction is perpendicular to the plane of the loop.
Magnetic moment of Revolving Electron, M = evr/2
2
A wire of length L is bent round in the form of a coil having N turns of same radius. If a
steady current I flows through it in a clockwise direction, find the magnitude and direction of
the magnetic field produced at its centre
ANS
L
L  N  2 r 
r
2 N
B
0 N I

0  N 2 I
2r
L
Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the current
sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. Justify.
ANS Current sensitivity :It is defined as the deflection of coil per unit current flowing in it.
Current Sensitivity, S=NAB/C
Voltage sensitivity: It is defined on the deflection of coil per unit potential difference across
its ends.
Voltage Sensitivity, SV=NAB/GC
where is resistance of galvanometer.
3
4
Justification: When number of turns N is doubled, then the current sensitivity (µN) is
doubled; but at the same time, the resistance of galvanometer coil (G) will also be doubled, so
voltage sensitivity Swill remain unchanged; hence increasing current sensitivity does not
necessarily increase the voltage sensitivity.
An electron of kinetic energy 25 keV moves perpendicular to the direction of a uniform
magnetic field of 0.2 milli-Tesla. Calculate the time period of rotation of the electron in the
magnetic field.
ANS T= 2 π m / Bq ,
= 2x3.14 x 9.1 x10- 31 / 0.2 x 10-3 x 1.6 x 10-19
= 178.6 x 10-9 = 1.78 x 10-7 sec
5
Derive the relation between µ0 ,ε0 and c,symbols has their usual meanings.
ANS
= 9x 109 Nm2/C2 and
=10-7Tm/A
µ0 ε0=
/
=10-7/9x 109 =1/(3x108)2 =1/c2
c = 1/
3 MARKS QUESTIONS
1
Derive a mathematical expression for the force acting on a current carrying straight conductor
kept in a magnetic field. Under what conditions is this force (i) zero and (ii) maximum?
ANS consider a conductor of length l, area of cross-section A, carrying current I lies perpendicular

to a magnetic field B .
Each electron moving with drift velocity vdexperiences a magnetic Lorentz force

 
f   e (vd  B)
If n is the number of free electrons per unit volume, then total number of electrons in the
conductor is
N = n x volume = nAl
Total force on the conductor is


 
 
F  Nf  nAl[e(vd B)]  enA[l vd  B]


If I l represents a current element vector in the direction of current, then vectors l

and vd will have opposite directions and we can take


l vd =  v d l
 


F  enAvd (l  B)
 

F  I (l  B)
[ enAvd  I ]
The magnitude of the force is
F = I l B sin
(i) If =0 or 180 , then F = I l B(0) = 0
Thus a current carrying conductor placed parallel to the direction of the magnetic field
does not experience any force.
(ii) If =90 , then F = I l B sin90o or Fmax= I l B
Thus a current carrying conductor placed perpendicular to the direction of a magnetic
field experiences a maximum force.
2
How will you convert a galvanometer into an ammeter of range 0 - I amperes? What is the
effective resistance of an ammeter?
ANS An ammeter is connected in series to a circuit. So it must have very small resistance so that it
does not affect the current. Therefore to convert a galvanometer into an ammeter, a low
resistance, called shunt, is connected in parallel with the galvanometer coil.
Let RG be the resistance of galvanometer and Ig be the current with which
galvanometer gives full scale deflection. A shunt of resistance S is connected in
parallel with it. To measure maximum current I, the maximum current through
galvanometer must be Ig and hence rest current I - Ig should pass through the shunt.
As galvanometer and shunt are in parallel, the potential difference across them is
equal. So,
I g G  I  I g S
or
S
I gG
I  I 
g
The effective resistance of the ammeter becomes
GS
RA 
GS
3
How can a galvanometer be converted into a voltmeter to read a maximum potential
difference V? Discuss with related mathematical expression.
ANS A voltmeter is connected in parallel with a circuit element. So it must draw a very small
current, otherwise the voltage to be measured would decrease. To insure it, a large resistance
is connected in series with the galvanometer.
Let RG be the resistance of galvanometer and Ig be the current with which galvanometer gives
full scale deflection. To measure a maximum potential difference V, a high resistance R is
connected in series with it.
Total resistance of the device = R + RG
Therefore by Ohm’s law
V  I g ( R  RG )
V
V
or R 
 RG
Ig
Ig
4
Derive a mathematical expression for the force per unit length acting on each of the two
straight parallel metallic conductors carrying current in the same direction and kept near each
other. Hence define an ampere. Why do such current carrying conductors attract each other?
ANS Consider two long parallel wires AB and CD carrying currents I1 and
I2 in the same direction. Let r be the separation between them.
The magnetic field produced by current I1 at any point on wire CD is
R  RG 
B1 
 0 I1
2 r
This field acts perpendicular to the wire CD and points into the plane
of paper. It exerts a force on current carrying wire CD. The force
acting on length l of the wire CD will be
 I
 II
F2  I 2 IB1 sin 90  I 2 l. 0 1  0 1 2 l
2 r
2 r
Force per unit length,
 II
F
f  2  0 1 2
l
2 r
According to Fleming’s left hand rule, this force acts at right angles to CD, towards AB in the
plane of the paper. Similarly, an equal and opposite force is exerted on the wire AB by the
field of wire CD. Thus when the currents in the two wires are in the same direction, the forces
between them are attractive.
Definition of ampere:
If I1=I2 =1A and r =1m, then

f  0  2  10 7 Nm1
2
One ampere is that value of steady current, which on flowing in each of the two parallel
infinitely long conductors of negligible cross-section placed in vacuum at a distance of 1 m
from each other, produces between them a force of 2 X 10-7 newton per metre of their length.
5
Using Biot-Savart law, deduce an expression for the magnetic field on the axis of a circular
current loop. Hence obtain the expression for the magnetic field at the centre of the loop.
ANS Consider a circular loop of wire of radius a and carrying current I, as shown in figure.

Let the plane of the loop be perpendicular to the plane of paper. We wish to find field B at an
axial point P at a distance r from the centre C.

Consider a current element dl at the top of the loop. It has an outward coming current.


If s be the position vector of point P relative to the element dl , then from Biot-Savart law,
the field at point P due to the current element is
 I dl sin 
dB  0
4
r2
 
Since dl  s , i.e., = 90 , therefore
 Idl
dB  0 2
4 s


The field dB lies in the plane of paper and is perpendicular to s , as shown by PQ.Let  be the
angle between OP and CP. Then dB can be resolved into two rectangular components.
1. dBsin  along the axis,
2. dB cos  perpendicular to the axis.
For any two diametrically opposite elements of the loop, the components
perpendicular to the axis of the loop will be equal and opposite and will cancel out.
Their axial components will be in the same direction, i.e., along CP and get added up.
Therefore, total magnetic field at the point P in the direction CP is
=  dB sin 
 Idl
a
and dB  0 2
s
4 s
 0 Idl a
B= 
.
.
4 s 2 s
Since  0 and I are constant, and s and aare same for all points on the circular loop, we
have
=
 0 Ia
 0 Ia
0 I a 2
dI

.
2

a

[  dI  Circumference  2 a]
4 s 3 
4 s 3
2 s3
sin  
As,
 B
s  (r 2  a 2 )
1
2
0 I a 2
2 (r 2  a 2 )
3
2
If the coil consists on N turns, then B 
0 N I a 2
.
3
2 (r 2  a 2 ) 2
Magnetic field at the centre: For the field at the centre of the loop, r = 0. Therefore
 NI
B 0
2a
6
Derive an expression for the torque on a rectangular coil of area A, carrying a current I and
placed in a magnetic field B. The angle between the direction of B and vector perpendicular to
the plane of the coil is

ANS Consider a rectangular coil PQRS suspended in a uniform magnetic field B , with its axis
perpendicular to the field.
Let I be the current flowing
through the coil PQRS, a and b be the sides of the coil PQRS,

A = ab = area of the coil and is the angle between the direction of B and normal to
the plane of the coil.
According to Fleming’s left hand rule, the magnetic forces on sides PS and QR are equal,
opposite and collinear (along the axis of the loop),so their resultant is zero.
The side PQ experiences a normal inward force equal to IbB while the side RS experiences an
equal normal outward force. These two forces form a couple which exerts a torque given by
 = Force x perpendicular distance
= IbB x a sin = IBA sin
If the rectangular loop has N turns, the torque increases N times i.e.,
 = NIBA sin
ButNIA = m, the magnetic moment of the loop, so
 = mB sin

In vector notation, the torque  is given by



 =mxB

The direction of the torque t is such that it rotates the loop clockwise about the axis of
suspension.
5 MARKS QUESTIONS
1
With the help of a labeled diagram, explain the principle, construction, theory and working oa
cyclotron.
ANS It is a device used to accelerate charged particles like protons, deuterons, - particles, etc., to
very high energies.
Principle: A charged particle can be accelerated to very high energies by making it pass
though a moderate electric field a number of times. This can be done with the help of a
perpendicular magnetic field which throws the charged particle into a circular motion, the
frequency of which does not depend on the speed of the particle and the radius of the circular
orbit.
Construction: As shown in figure, a cyclotron consists of two small, hollow, metallic halfcylinders D1 and D2, called dees. An alternating voltage is applied across the gap between the
two dees. The dees are placed between the poles of a strong electromagnet. A source of
charged particles is placed near the centre of the dees. These ions move on a circular path in
the dees, D1 and D2, on account of the uniform perpendicular magnetic field B. The whole
arrangement is evacuated to minimize collisions between the ions and the air molecules.
2
Theory: As a particle of charge q and mass m follows a circular path under the effect of
perpendicular magnetic field B, so
Magnetic force on charge = Centripetal force
mv2
mv
Or
qv Bsin 90 
or
r
r
qB
Period of revolution of the charged particle is given by
2 r 2 mv 2 m
T=

.

v
v qB
qB
1
qB
Hence frequency of revolution of the particle will be f c  
T 2 m
Clearly, this frequency is independent of both the velocity of the particle and the radius of the
orbit and is called cyclotron frequency of magnetic resonance frequency.
Working: Suppose a positive ion enters the gap between the two dees and finds dee D1 to be
negative. It gets accelerated toward dee D1. As it enters the dee D1, it does not experience any
electric field due to shielding effect of the metallic dee. The perpendicular magnetic field
throws it into a circular path. At the instant the ion comes out of dee D1, it finds dee D2
negative. It now gets accelerated towards deeD2 . It moves faster through D2 describing a
larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the
same as the frequency of revolution of the ion, then every time the ion reaches the gap
between the two dees, the electric field is reversed and ion receives a push and finally it
acquires very high energy.
With the help of a neat and labeled diagram, explain the underlying principle, construction
and working of a moving coil galvanometer. What is the function of (i) uniform radial field
(ii) soft iron core in such a device?
A galvanometer is a device to detect current in a circuit, the magnitude of which depends on
the strength of current.
Construction: A pivoted-type galvanometer consists of a rectangular coil of fine insulated
copper wire wound on a light aluminium frame. The motion of the coil is controlled by a pair
of hair springs of phosphor-bronze. The springs provide the restoring torque. A light
aluminium pointer attached to the coil measures its deflection on a suitable scale.
The coil is placed symmetrically between the concave poles of a permanent horseshoe magnet. There is a cylindrical soft iron core which not only makes the field radial
but also increases the strength of the magnetic field.
Theory and working: As the field is radial, the plane of the coil always remains parallel to

the field B . When a current flows through the coil, a torque acts on it. It is
 = Force x perpendicular distance = NIbB x a sin 90 = NIB(ab) = NIBA

Here =90 , because the normal to the plane of coil remains perpendicular to the field B in
all positions.
The torque  deflects the coil through an angle . A restoring torque is set up in the coil due
to the elasticity of the springs such that
 restoring  k
Where k is the torsion constant of the springs i.e., torque required to produce unit angular
twist.
In equilibrium position,
Restoring torque = Deflecting torque
k = NIBA
NBAI

k
Thus the deflection produced in the galvanometer coil is proportional to the current flowing
through it.
Functions:
(i) A uniform magnetic field provides a linear current scale.
(ii) A soft iron core makes the field radial. It also increases the strength of the magnetic
field and hence increases the sensitivity of the galvanometer.
HOT QUESTIONS
1
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. What is the magnitude of the field?
ANS For mid-air suspension, weight of the wire must be balanced by the magnetic force. Hence
IlB sin 900  mg
mg 0.2  9.8
B

 0.65 T
Il
1.5  2
2
A uniform conducting wire of length 12 a and resistance R is wound up as a current carrying
coil in the shape of (i) an equilateral triangle of side a, (ii) a square of side a and (iii)
a regular hexagon of side a. The coil is connected to a source of potential difference
V0. Find magnetic moments of the coils in each case.
ANS As source voltage and resistance are same, current through the coils is same. It is
V
I 0
R
(i) There will be 4 turns in the triangle.
3 2
Area of equilateral triangle, A1 
a
4
3 2
M1  4  I 
a  3a 2 I
Thus
4
(ii) There will be 3 turns in the square.
Area of square, A2  a 2
Thus
M 2  3  I  a 2  3a 2 I
(iii)There will be 2 turns in the regular hexagon.
3 3 2
a
Area of hexagon, A3 
2
3 3 2
M3  2 I 
a  3 3a 2 I
Thus
2
3
A circular coil of 16 turns and radius 0.1 m carrying a current of 0.75 A rests with its plane
normal to an external field of magnitude 5x10-2T.The coil is free to turn about an axis
in its plane perpendicular to the field direction. When the coil is turned slightly and
released it oscillates about its stable equilibrium with a frequency of 2/s .What is the
moment of inertia of the coil about its axis of rotation?
ANS M = NIA; ν =
I = MB/4π2ν2 ; I = 1.2 x 10-4 kg m2
4
A charge particle ‘q’ moving in a straight line is accelerated by a potential difference V. It
enters a uniform magnetic field B perpendicular to its path. Deduce an expression for the
radius of the circular path of the particle in terms of V.
ANS The kinetic energy gained by the particle
1
m v 2  qV
2

v
2 qV
m
Thus the radius of the circular path is
mv
m 2 qV
1 2 mV
r


qB qB
m
B
q