Download MAGNETIC EFFECT OF CURRENT

MAGNETIC EFFECT OF CURRENT
1. Concept of Magnetic Field
If we place or suspend a small magnetic needle near a bar-magnet, the needle
rests in a definite direction if we place this needle at some other point; it rests in
some other direction. This shows that the magnetic needle near the bar-magnet
experiences a torque which turns the needle to a define direction. The region
near a magnet, where a magnetic needle experiences a torque and rests in a
definite direction, is called ‘magnetic field.’ The line drawn from the south to the
north pole of a magnetic needle freely-suspended at a point in the magnetic field
is the direction of the field at that point.
Earth is also a magnet and has a magnetic field. This is why a freely-suspended
magnetic needle always rests in the north-south direction. The north pole of the
needle points towards north and the South Pole towards south. This shows that
the earth’s magnetic field acts from south to north.
2. Oersted’s Experiment
Oersted, in 1820, found experimentally that a magnetic field is established
around a current-carrying conductor just as it occurs around a magnet. His
experiment. Conducting wire AB is connected to the poles of a battery through a
key. The wire is kept above a magnetic needle parallel to it (in north-south
direction). So long there is no current in the wire, the magnetic needle remains
parallel to the wire. As soon as the key is pressed, a current flows through the
wire and the needle is deflected if the current is reversed, the needle is deflected
in the opposite direction the deflection of the needle indicates that with the
passage of current in the wire, a magnetic field is established around it. On
increasing the current in the wire, or on bringing the needle closer to the wire,
the deflection of the needle increases. This experiment shows that the magnetic
field is produced due to the electric current. Since electric current is ‘charge in
motion’, it is concluded that moving charges produced magnetic field in the
surrounding space.
(A charge, whether stationary or in motion, produces an electric field around it. If
it is in motion, then, in addition to the electric field, it also produces a magnetic
field.)
3. Force on a Charge moving in a Magnetic Field: Definition of B
A charge moving in a magnetic field experiences a deflecting force. In fact, if a
charge moving through a point experience a deflecting force, then a magnetic
field is said to exist at that point. This field is denoted by a vector quantity B,
called the ‘magnetic field’ or ‘magnetic induction’. B is defined in terms of the
force experienced by the charge moving in the field.
If a particle carrying a positive charge q and moving with velocity v through a
point p in a magnetic field experiences a deflecting for F then the magnetic field
B at point P is defined by the equation F=qv x B ….(i)
The force F on the charged particle is perpendicular to the plane formed by v and
B. Its sense (the way it points along the direction line) is determined by righthand screw rule of vector product. When the moving charge is positive, the
direction of F is the direction of advance of a right-hand screw whose axis is
perpendicular to the plane formed by V and B and which is rotated from V to B.
The direction of F will be opposite to the direction of advance of the screw if the
charge is negative but moving in the same direction.
The direction of the force on the charged particle moving in magnetic field can
also be determined by Fleming’s left-hand rule, taking ‘direction of motion of
positive charge’ instead of ‘direction of current’. (This rule is given ahead in this
chapter.)
The magnitude of the force on the charge particle is F=qvB sin,
Where
……(ii)
is the angle between v and B the force is maximum when v and B are at
right angles,
= 900 (Fig4) It is then given by Fmax=quB …….(iii)
If v is parallel to B ( =0 or 1800), then F=0. This means that if the charged particle
is moving parallel to the magnetic field, it does not experience any force. In fact,
this defines the direction of magnetic field B.
If a charged particle moving through a point P in magnetic field B does not
experience a deflecting force, then the direction of motion of the particle is the
direction of B at Point P.
Again, in eq. (ii) if u=0, then F=0. This means that if the charged particle is
stationary in the magnetic field, then it does not experience any force. (It is
important to note that a ‘stationery’ charged particle in an electric field does
experience a force.)
The Force experienced by a moving charged particle in a magnetic field can be
verified by an electron-tube. At one end of this tube is a filament, heated by
electric current, and at the other end is a fluorescent screen. Hot filament emits
electrons (negative charge) which fall on the screen in the form of a thin beam
and so a bright spot is observed on the screen. When this tube is placed in a
magnetic field, the spot is displaced to one side. This shows that a force is acting
on the electrons due to which the spot has been displaced from its initial
position. If we place the electron tube in different directions in the magnetic
field, we see that the displacement of the spot changes with direction and
becomes zero in one particular direction. In this direction, no force acts on the
electrons. This is the direction of the magnetic field.
Lorentz Force: if a particle carrying a charge q is moving in space where both an
electric field E and a magnetic field B are Present, then the force on the particle
will be given by:
F=q(E+v B)
This is called ‘Lorentz force’.
Unit of Magnetic Field B: The magnitude of the magnetic field is given by
B=
If F = 1 newton, q=1 coulomb, u=1 meter/sec and
B=
=900, then
=1
Because coulomb-sec-1 – ampere.
Thus, the SI unit of magnetic field B is newton/(ampere-meter), also written as
N/(A-m). it is also known as ‘tesla’ (T)
1T = 1 N A-1 M-1
Thus, if a charge of 1 coulomb moving with a velocity of 1 meter/sec
perpendicular to a uniform magnetic field experiences a force of 1 newton,
then the magnitude of the field is 1 tesla.
Another SI unit of magnetic field is ‘weber/meter2’
Thus
1 T = 1 N A -1M-1 =1 Wbm-2
In the CGS system, the magnetic field is expressed in ‘gauss’
1 tesla = 104 gauss.
The earth’s magnetic field is about 0.5 gauss. The electromagnets used in the
laboratories produce magnetic fields of the order of 1 tesla. Fields of the order of
200-400 tesla can be produced by special devices for a very short spell of time
( 10-6 sec). It is believed that much stronger magnetic field are found in some
stars which are called neutron-stars.
Dimensions of B : The unit of B is
=
=kg-sec-2 –amp-1
Dimension of B are [MT-2A-1]
6.1 MAGNETIC FIELD OF MOVING CHARGE
We know that a point charge q, at rest in the observer’s inertial frame, produces
an electric field along the radius vector and it given by
E=
1
4 %!
0
q
r
r3
If the charge is moving relative to the observer’s inertial frame, it produces a
magnetic field in addition to electric field. The magnitude of which is proportional
to the speed of the charge relative to the observer provided (v<c). The magnetic
field vector B at the point P, a distant r from the charge q moving a velocity v is
found to be
n0 q
B = 4r 3 (v # r )
r
The direction of B is thus perpendicular to the plane of v and r. It is in the
direction of advance of a right handed screw rotated from v to r. its magnitude is
given by
n 0 qvsin i
B= ( r )
4
r2
……………………(2)
The following points should be remembered regarding with the magnetic field.
1.
The magnetic field B is zero at all points on a line on which charge moves.
That is when
= 0 or
2.
It is maximum in the plane perpendicular to the v and through the charge,
as sin
3.
= 1800, B=0.
= 1, at all points in this plane:
B remain unaltered in magnitude at all points on the circumference of
circle passing through P and lying in a plane perpendicular to v with its centre on
the velocity direction.
4.
The Direction of B is given by either
(a) Maxwell’s right hand screw rule: if the direction of current through the
conductor or the direction of velocity of positive charge is represented by
the linear motion of the screw motion, then the direction of the magnetic
field can be represented by the direction of rotation of the screw.
(b) Right hand claps rule: If a current carrying conductor is clapsed in the right
hand so that the thumb indicates the current direction, then the direction
of the magnetic field is represented by the finger tips round the wire.
Electromagnetic field
So far, we have considered electric and magnetic fields separately, without
establishing any clear relation between them. This could be done only because
the two field were static. In other cases, however, it is impossible. It will be
shown that electric and magnetic fields must always be considered together as a
single total electromagnetic field. In other words, it turns out that electric and
magnetic fields are in certain sense the components of a single physical object
which we call the electromagnetic field. The division of the electromagnetic field
into electric and magnetic fields is to relative nature since it depends to a very
large extent to the reference system in which the phenomenon are considered.
Force between moving charges
The force acting on a charge q2, moving with velocity v2 in magnetic field
produced by charge q1 moving with a velocity v1 is
F21 = q (v 2 # B1)
6
@
n 0 q1
= q2 (v 2 #
3 (v # r )
4r r
6
@
n 0 q1 q2
=
2 # (v1 # r )
v
3
4r r
2
The magnitude of force which they exert on each other
For
n 0 q1 q2 v1 v2
4r
r2
v1 = v2 = v
n 0 q1 q2 2
Fm = r
v
4 r2
Fm =
……………………………(i)
In addition to the magnetic force, there is an electric force between them, whose
magnitude is given by
1 q1 q2
Fe = r e
4 0 r2
………………………..(ii)
This Force is of repulsive nature. On dividing equation (i) by (ii), we have
Fm = 2 n e
v 0 0
Fe
As
c=
1
n 0 e0
Therefore,
Fm = v2 ……………(3)
Fe c2
Since v < c, and so Fm < Fe. As Fm < Fe, so the net force between the charges is of
repulsive nature.
6.2 THE BIOT SAVARTS LAW
Biot-Savarts law is fundamental law of the magnetic. It gives the magnetic field by
a small current element and is based on the experimental facts. Let us consider a
small element dl of a wire of cross-sectional area A carrying current i. If n is the
number of charge carries per unit volume, each of charge q=e and moving with a
drift velocity vd, then
i = neAdv
The quantity of charge flowing in time dt,
dq = idt=neAvd(dt)
Magnetic Effect of Current
Since vd(dt)=dl,
therefore,
dq = neAdl.
The magnetic field due to the current element dl at any point P at a distance r
dB
n 0 dq(vd # r )
4r
r3
n 0 (neAdl) (vd # r )
=
4r
r3
n0
(dl # r )
=
(neAdl)
4r
r3
=
Or
n 0 idl # r
dB = r
4
r3
n 0 idl Sini
dB = r
4
r2
……………………(1)
This is called Bio-Sarvarts Law
The field due to the whole current carrying conductor is given by
B=
#dB
Or
n0
B = 4r
# i (dl
# r)
r
3
=
n0 il # r
4r
r3
……………..(2)
Magnetic field due to a straight current carrying conductor
Consider a straight wire carrying current i. Its ends subtend angles
1 and
2
at
point P at which field is to be determined.
Take a small element dl of the wire at a distance l from O. The magnetic field due
to element is
dB =
n 0 (idl sin(90o + i )
4r
x2
…………………..(i)
The direction of field at P is perpendicular to the plane of the diagram and going
into it. The direction of the field is the same for all elements of the wire and
hence net field due to the wire is obtained by integrating equation (i).
From the diagram,
= tan
Or l = r tan
And dl = r sec2 d
Putting in equation (i),
dB =
n 0 i (r sec2 i di )
cos i
4r
x2
x = sec i
or x = r sec i
r
n0 i
cos i
dB = r
4 r
i1
Total field
n0
B = r i #cosi di
4 r - i2
Or
B = n 0 i (sin i 1 + sin i 2)
4r r
We can draw magnetic field lines in the same way that of electric field lines. A
tangent to a magnetic field line gives the direction of the magnetic field existing
at that point. For a long straight wire, the field lines are circles with their centre’s
on the wire.
Special cases:
(a)
Field due to a long straight wire
r
and i
2
n0
` B= r i
2 r
i 1=
2=
r
2
(b)
Field along the end of the wire
i 1 = 0 and i 2 = r
2
n0 i
` B= r
4 r
Force Between two parallel current carrying wires
Consider two long straight wires kept parallel to each other at a distance r and
carrying currents i1 and i2 respectively in the same direction.
Magnetic force on the small element dl of the wire 2 due to the magnetic field of
l is:
dF = i2(dlxBl)
The magnitude of the force
dF = i2(dl)B1 sin 900
Where
n 0 i1
B1 = r
2 r
n 0 i1 i 2 (dl)
dF = r
2
r
dF = n 0 i1 i 2
2r r
dl
Thus force per unit length of wire 2 due to 1 is n 0 i1 i 2 . The same amount of force
2r r
1 exerts on 2.
The direction of force on wires are towards each other is attraction (By FLHR).
If direction of currents in wires are opposite, the force between them will be
repulsive.
NOTE: There is no electrical interaction between the wires, because they have
no net charges.
Field due to a circular current carrying coil
Consider a circular loop of radius a carrying current i. we have to find the
magnetic field at a point P on the axis of the loop at distance x from the centre of
the loop. Consider a current element idl of the wire. The magnetic field at P due
to this element is
n 0 dl # rc
dB = r i
4
r2
As dl is perpendicular to the plane of the paper, and so dl x r must lie in the plane
of paper, as shown in figure. The magnitude of the field
n0
dB = r i dl2
4
r
Another element on the diametrically opposite point produces the same amount
of field on the point P. The resultant field due to these elements = 2dB sin .
The effective field of one element at P,
=dB sin
Field due to complete loop
B=
#dBsin a
n0
=
4r
=
Or
2r a
# idl
0
r2
sin a
n 0 i # 2r a a
#
4r
r
r2
n 0 2i r a2
B= r
4
r3
a2 + x2
n 0 ia2 ………………………………….………(1)
B=
2(a2 + x2) 3/ 2
r=
As i (r a2) = iA = M, where A is the area of the loop
= magnetic moment of the loop,
B = n 0 2 2M 2 3/ 2
4r (a + x )
………………………. (2)
Field at a point far away from the Centre
X > > a,
B = n 0 2M3 …………………………………………………….(3)
4r x
From equation (1), we can get field at the centre of loop, x = 0
B=
n 0i
2a
If there are N turns in the loop
B=
n 0 Ni
…………………………………………………….(4)
2a
NOTE:
1.
Equation (3) and (4) are derived for circular loop, but they can be used for
other shape of the loop also which has symmetry about the axis of the loop.
2.
In equation (4), N may be a whole number or fraction or decimal number.
3.
For the following:
N=( i )
2r
n 0( i )i
2r
B=
2a
4.
The magnetic field at a point not on the axis of the loop is mathematically
difficult to calculate. The field lines for the circular loop are not circles, but they
are closed curves that link the conductor.
Magnetic field due to a circular loop
Field between two similar coaxial circular loops
Let us consider two loops, each having N turns and carrying current i are placed
at a distance 2d apart
Assuming the current is following in the same direction in each coil, the magnetic
field at a short distance x from midway point O
n 0 Nia2
1
1
B=
2 (a2 + x1 2) 3/ 2 (a2 + x22) 3/ 2
n 0 Nia2
1
1
=
B
2 (a2 + (d + x2)) 3/ 2 (a2 + (d - x2)) 3/ 2
The field will be uniform between the loops, if
= 0 i.e.,
n 0 Nia2
(d + x)
1
+3
(- 3) 2
2 5/ 2 = 0
2 6
+ x) 2) 5/ 2 6 (a2 + (d - x)@
(a + (d@
)
2 5/ 2
2 5/ 2
Or Now, (d + x) a + (d - x)
= (d - x) a + (d + x)
............... (i)
6
@
6
@
5/
2
5/
2
= a2 + d2 + x2 + 2xd
a + (d + x) 2
2
2
Since x is small, so neglecting x2, we have
2
6
@
= a2 + d2 + 2xd 5/ 2
5/ 2
;
E
= (a2 + d2) 5/ 2 1 + 22xd 2
(a + d )
;
E
= (a2 + d2) 5/ 2 1 + 25xd 2
(a + d )
= 22xd 2 << 1
a + d
Similarly,
6
@
;
5/ 2
= (a2 + d2) 5/ 2 1a + (d - x) 2
5xd E
a + d2
2
Substituting these values in equation (i), we get
(d+x)[1-5xd ] = (d-x)[1+5xd ]
a2+d2
or
a2+d2
5d2 = 1, ` = a
d
or a = 2d
2
(a + d2
2
2
n
And B = 2 > 0 Nia 8
2
2
H
1
`a j2 B3/ 2
a +
2
n
B= 8 0 Ni …………………………………………….(2)
5 5a
Variation of magnetic field between the loops.
EXAMPLE : A wire carrying current I has the configuration shown in fig.6.16 two
semi-infinite straight sections, both tangent to the same circle, are connected
by a circular arc, or central angle , along the circumference of the circle, with
all sections lying in the same plane. What must
be in order for B to be zero at
the centre of the circle?
SOLUTION: The field due to the straight parts of the wire
;
E
B1 = 2 n i
0
4r R
Out of the plane
And field due to curved part of the wire
n0
B2 =
ci m
i
2r
, into the plane of the wire.
2R
The resultant field at the centre to be zero
B1=B2
Or
i
6 n 0 @ n 0 2r i
=
= 2 i = 2rad
2
4r
2R
rad Ans
SOLENIOD
A solenoid is a wire would in a closely spaced over a hollow cylindrical
non-conducting core. The wire coated with an insulating material so that
the adjacent turns physically touch each other, but they are electrically
insulated.
dN = ndx
If n is the number of turns per unit length, each carrying a current i,
uniformly wound round a cylinder of radius a, then the number of turns
in length dx is ndx. Thus the magnetic field at the axial point P due to the
element.
dB =
0 ndx i

2 a2  x2

3 /2
The direction of magnetic field is along the axis of the solenoid and in the
sense of advance of a right handed screw. From geometry, we have
x = a cot 180    acot 
and dx = a cosec2  d and hence
dB =
0nisin d
2
Total field
 ni
B= 0
2
=
2
 sin d
1
0ni

 cos  2
1
2
0ni
 cos 1  cos 2  ….(1)
2
or B =
Special cases:
Case-1:
Solenoid is of infinite length and the point chosen is at the middle
1  0, 2  
Ө B = 0ni
Case-2:
Solenoid is of infinite length and the point is at the end of the solenoid
1 

, 
2 2
B 
0ni
2
NOTE:
1.
If the length of the solenoid is large compared with its radius, the
internal field near its center is very nearly uniform and parallel to axis,
and the external field near the center is very small. In practice we take
the magnetic field inside very tightly wound long solenoid is uniform
everywhere and zero outside it.
2.
If some material medium is present inside solenoid, the magnetic field
inside it is B = r 0ni; where r is the relative permeability of the
medium.
TOROID OR ANCHOR RING
It is a solenoid of small radius bent round to form a toroid. In an ideal
toroid, the field is confined entirely within the core and is uniform. The
value of magnetic field at any point on the mean circumferential line is
given by
B = 0ni
If N is the total turns in the toroid, then
n=
N
,
2R
 N 
i
2R 
= 0 

 N 
 B  0 
i
 2R 
Or B =
 0Ni
2R
Where R is the mean radius of the ring.
(b)
Field due to a current loop (or its segment) at its center:
In case of a circular loop:
(i)
Each element is at the same distance from the center, i.e., r = R =
constant.
(ii)
 
The angle between element dL and r is always   as in a circle
 2
tangent and normal are always perpendicular to each other.
(iii)
Contribution of each element of B is in the same direction, i.e., out of
the page if the current is anticlockwise and into the page if clockwise.
So Biot-Savart law
B
0
4

IdL  r
r3
for this situation yields:
B=
0
4

IdL
R2
R d,
But as dL =
so


B= 0
4

i.e., B =
0 I
4 R
IR d
R2
0
….(1)
This is the required result and from this it is clear that:
(1)
If the loop is semicircular i.e., ф = π,
B=
0 I
4 R
…(2)
and will be out of the page for anticlockwise current while into the page
for clockwise current respectively.
(2)
If the loop is a full circle with N turns, i.e.,   2N :
B=
0 2 NI
4 R
…(3)
Field for this situation is depicted respectively.
(d)
Field at an axial point of a Solenoid:
If many turns of an insulated wire are wound around a cylinder the
resulting coil is called a solenoid [Fig. 13.13 (A)]. The field at a point on
the axis of a solenoid can be obtained by superposition of fields due to a
large number of identical coils all having their centre on the axis of the
solenoid.
This if the desired result and from this it is clear that:
(1)
If the solenoid is of infinite length and the point is well inside the
solenoid,
 
 
 2
So, B 
0
2nI 1  1 ,
4
i.e., 3  0nI
(2)
…(2)
If the solenoid is of infinite length and the point is near one end,
0
 
and    
 2
(3)
So, B 
0
2nI 1  0 ,
4
i.e., B 
1
  nI
2 0
…(3)
If the solenoid is of finite length and the point in on the perpendicular
bisector of its axis,

So, B 
0
2nI sin 
4
with sin  
(4)
L
L2  4R2
…(4)
If the solenoid is of finite length and the point is on its but near one end
0
So, B 
0
2nI sin 
4
with sin α =
(5)
L
2
L R
…(5)
2
The field and its variation with distance along the axis of a solenoid
NOTE:
If the solenoid had been of infinite length, there will be no ends and
hence everywhere inside the solenoid the field will be uniform, equal to
  nI which outside it will be zero. This actually happens in a 'toroid'
0
which is a solenoid bent in a circular shape so that its two ends joint each
other.
Question. A vertical straight conductor carries a current vertically upwards.
Points P and Q lie respectively to the east and west of the conductor at
same distance from it. Is the field at P lesser or greater than at Q?
Answer: As shown in Fig. 13.15 (A), for point P (situated east of the wire), the
field of the current-carrying wire will be directed north while for point Q
(situated west of the wire), it will be directed south and both will have


same magnitude BW  
0I 
.
2d
However, as the horizontal component of earth's magnetic field BH is
along the direction from south to north so if the effect's magnetic field is
considered,
BP  BW  BH
while BQ  BW  BH
And so, in this situation, field at P(which is east of the vertical wire) will
be greater than at Q.
NOTE:
In this situation there will be only one neutral point towards the west of
the wire and if Q is the neutral point.
0 2I
 BH
4 d
i.e., d 
0 2I
4 BH
Question: Equal currents are flowing through two infinitely along parallel wires.
Will there be a magnetic field at a point exactly halfway between the
wires when the currents in them are (a) in the same direction, (b) in
opposite directions?
Answers: If the current in both the wires is in the same direction, the field at P
due to A will be into the page while due to B out of the page and as the
point P is equidistant from the wires and the wires carry equal currents,
BP  BA  BB
0 2I
 2I
  0
4 d
4 d
0
However, if the wires carry equal currents in opposite directions, the
field at P due to both the wires will be in the same direction; so
  2I 
BP  2  0
0
 4 d 
and will be into the page for the situation. If the directions of I in wires A
and B are interchanged BP will still have the same magnitude but
directed out of the page]. The lines of force for the given situations.
Question. A steady current is set up in a network composed of wires of equal
resistances as shown in Fig. 13.18 (A) and (B). What is the magnetic
field at the centre P in each case:
Answer: By symmetry of the problem for each current-carrying wire there is
another wire such that it cancels the field of the first; e.g., the field at P
due to wire AB is cancelled by the field of wire DC in Fig. 13.18 (A) and
field of AB is cancelled by the field of wire GF in Fig. 13.18 (B). So the
resultant field at P due to the given network is zero in both the cases.
Question. Three sections of a circuit, each section consisting of a wire curved
along a circular arc (all with same radius) and two long straight wires
that are tangential to the arc. In (A) the straight wires pass each other
without touching. The sections carry equal currents. In which case is
the magnitude of the magnetic field produced at the centre of
curvature (a) greatest (b) least?
Answer: Assuming that each section is composed of three elements, viz., a, b and
c with anticlockwise current in the arc, the field due to each element at P
will be out of the page. Now as elements a and c together in each case


produce the same field at P  
0I 
, so the field at P will be greatest
2R 
in the case in which the field due to arc b is greatest and as for a circular
 0I 
, i.e., B  , and here
4R 
arc, for point at its centre, B = 

 3 
 
A  
 B      C    , so the field will be greatest in

 2
 2
case A and least in case C.
Problem. Two infinitely long, thin, insulated straight wires lie in the x-y plane
along the x and y-axes respectively. Each wire carries a current I,
respectively in the positive x-direction and the positive y-direction.
Find the locus of a point in this plane where the magnetic field is zero.
Solution: As field due to a long straight current-carrying wire at a distance d from
it is given by
B=
0 2I
4 d
so the field at due to currents in wires along x and y axes respectively
will be:
BA 
0 2I
k
4 y
and BB 
 
0 2I
k
4 x
According to given problem,
BA  BB  0,
i.e.,
0
1 1 
2I     0
4
y x
This is possible only if
1 1
  0,
y x
i.e., y = x
So the locus is a straight line passing through the origin and subtending
an angle of 45° with the positive x-axis, i.e., line AB.
Problem. A pair of stationary and infinitely long bent wires are placed in the x-y
plane as shown in Fig. 13.21. The wires carry currents of 10 ampere
each as shown. The segments L and M are along the x-axis. The
segments P and Q are parallel to the y-axis such that OS = OR = 0.02 m.
Find the magnitude and direction of the magnetic induction at the
origin O.
Solution: As point O is along the length of segments L and M so the field at O due
to these segments will be zero. Further, as the point O is near one end of
a long wire,
BR  BP  BQ
=
0 I
4 RO

0 I
4 SO
so BR 
0  2I 
4  d 
[as RO = SO = d]
Substituting the given data,
BR  107 
= 10
4
2  10
0.02
Wb
m2
i.e., B is 10–4 T and out of the page. Ans.
Problem. A current of 1 ampere is flowing in the sides of an equilateral triangle
of side 4.5 × 10–2 m. Find the magnetic field at the centroid of the
triangle.
Solution: As shown in Fig. 13.22, the field at the centroid
BO  BAB  BBC  BCA
And as in accordance with RHSR at O, all the fields are out of the page
and point O is symmetrically situated with respect to wires:
BAB  BBC  BCA
and hence,
B0  3BAB  3BAB
Now as in case of current-carrying wire of finite length for a point at a
distance d from the wire,
B=
0 I
sin   sin 
4 d
But as here
 
 
 3
and d =
1
a
a sin 60 
,
3
2 3
BO  3 

0
I
3
2
4  a 
2


2 3
0 18 I
4 a
Substituting the given data,
BO  107 
5
= 4  10
18  1
4.5  102
Wb
m2
i.e., resultant field is 4 × 10–5 T and out of the page if current in the loop
is anticlockwise (and will be into the page if clockwise). Ans.
Problem. Two straight infinitely long and thin parallel wires and spaced 0.1 m
apart and carry a current of 10 ampere each. Find the magnetic field at
a point distance 0.1 m from both wires in the two cases when the
currents are in the (a) same and (b) opposite directions.
Solution: As the point P is equidistant from both the wires A and B and the
distance between the wires is equal to the distance of point P from a
wire, so that
BA = BB = B
=
0 2I
2  10

 107 
4 d
0.1
= 2 × 10–5 T
Now as lines of force in case of a current-carrying wire are circles
encircling the wire so BA will perpendicular to AP while BB to BP. So
(a) If the wire carry current in the same direction (say out of the page), B A and
 
BB will have directions with angle   between them. So BR = 2B cos 30 =
 3
2 3  105 T along negative x-axis. Ans.
(b)
If the wires carry current in opposite direction (say A out of the page
while B into the page), BA and BB will have directions as shown in Fig.
 2 
between them. So
3 
13.23 (B) with angle 

BR = 2B cos 60
= 2 × 10–5 along positive y-axis. Ans.
Problem. Two long straight parallel wires are 2 m apart, perpendicular to the
plane of the paper. The wire A carries a current of 9.6 ampere, directed
into the plane of the paper. The wire B carries a current such that the
 10 
m from
11 
magnetic field of induction at the point P at a distance 

the wire B is zero. Find (a) the magnitude and direction of the current
in B, (b) the magnitude of the magnetic field of induction at the point S,
and (c) the force per unit length on the wire B.
Solution: (a) As the field due to a long current-carrying wire at a point distant d
from it and not near its ends is given by
B=
0 2I
4 d
so the field at P will be
BR  BA  BB
=
0
4
 2IA 2IB 



d
dB 
 A
According to given problem
BR = 0
So,
IB
I
 A
dB
dA
i.e., IB =  IA
dB
dA
So, IB = 9.6 
10
1

11 
 10  
2   11 


= – 3A
i.e., current in the wire B is 3 ampere and opposite to that in A, i.e., it will
be out of the page.
(b)
As lines of force for a current-carrying wire are circles encircling the wire,
so BA will be perpendicular to AS while BB to BS and as in this problem
AS2 + BS2 = AB2, i.e., AS is perpendicular to BS, BA and BB for given
directions of current will be along SB and SA respectively.
So, BA =
0 2IA
along SB
4 dA
and BB 
0 2IB
along SA
4 dB
and hence,
B=
B2A  BB2
[as SA is perpendicular to SB]
1/2
i.e., B = 10
7
2
  9.6  2
 3.0  
 4 
  4 
 

1.6
1.2


= 1.3 × 10–6 T Ans.
(c)
As force per unit length between two parallel current-carrying wires
separated by a distance d is given by
 2I1I2
dF
 0
dL 4 d
and is attractive if current in the wires is in same direction otherwise
repulsive, so here
dF
2  9.6  3
 10
dL
2
6
= 2.88  10
N
and repulsive Ans.
m
Problem. Two long parallel wires carrying currents 2.5 ampere and I ampere in
the same direction (directed into the plane of the paper) are held at P
and Q respectively such that they are perpendicular to the plane of the
paper. The points P and Q are located at distance of 5 m and 2 m
respectively from a collinear point R.
(a)
An electron moving with velocity of 4 × 105 m/s along the positive xdirection experiences a force of magnitude 3.2 × 10–20 N at the point R.
Find the value of I.
(b)
Find all the positions at which a third long parallel wire carrying a
current of magnitude 2.5 ampere may be placed so that the magnetic
induction at R is zero.
Solution:
(a)
As force on a charged particle in a magnetic field is given by

F  q v B

The field due to both current-carrying wires will be along negative y-axis.
Now as the particle is moving along positive x-axis, the angle between
 
the direction of motion of charged particle and field,     rad.
 2
So, F = qvB sin 90
i.e., B =
F
qv
Substituting the given data,
3.2  1020
B=
1.6  1019  4  105
= 5 × 10–7 T
Now, as this field is produced by wires P and Q and point R is on the
same side of both,
0
4
 2IP 2IQ 



dQ 
 dP
= 2 × 10–7 T
i.e.,
2  2.5 2I

5
5
2
i.e., I = 4A Ans.
(b)
If the distance of point R from third current-carrying wire is x, then BR = 0
implies
0 2  2.5
 5  107  0
4
x
Now there are two possibilities:
(i)
If the current in third wire is into the page, I = +2.5 A, so x = –1, i.e., the
point R must be to the left of the wire, i.e., the wire must be to the right
of R, i.e., at A with RA = 1 m.
(ii)
If the current in third wire is out of the page, I = –2.5 A, so x = +1, i.e., the
point r must be to the right of the wire, i.e., the wire must be to the left
of R, i.e., at B with RB = 1 m.
Problem: A very long thin strip of metal of width 'b' carries a current I along its
length as shown in Fig. 13.26. Find the magnitude of magnetic field in
the plane of the strip at a distance 'a' from the edge nearest to the
point.
Solution: Assuming the strip to be made up of a large number of elements
parallel to its length, consider an element of length dx at a distance x
from the point P. Treating the element as a long current-carrying wire,
dB =
0 2I'
4 x
Now as the current in the strip of width b is I, so the current in the
element of width dx will be
I' =
I
dx
b
and hence,
dB =
0 2I dx
4 b x
So, B =
 a  b

a
0 2I
4 b
dx 0 2I
b


loge 1   Ans.
x
4 b
a

NOTE:
If the point is distant from the strip, i.e., a > b.
b  b b2

loge 1     2  ....
a  a 2a

So, B =
=
b
a
0 2I b

4 b a
0 2I
4 a
i.e., for a distant point the strip behaves as a current carrying wire which
justifies out result.
Problem. In the Bohr model of hydrogen atom, the electron circulates around
the nucleus in the path of radius 5.1 × 10–11 m at a frequency of 6.8 ×
1015 rev/s. Calculate the magnetic induction B at the centre of the
orbit. What is the equivalent dipole moment?
Solution: The revolving electron in equivalent to a circular current
I = ef = 1.6 × 10–19 × 6.8 × 1015
= 1.088 mA
So the field at the centre of the path
B=
0 2I
4 R
= 10
7
2    1.088  103

5.1  1011
= 13.4 Wb/m2 Ans.
Further, as a current-carrying loop is equivalent to a magnetic dipole of
moment
M = NIS = R2I as N  1

So, M =   5.1  1011

2
 1.088  103
i.e., M = 8.88 × 10–24 A-m2 Ans.
Problem. A charge of 1 coulomb is placed at one end of a non-conducting rod of
length 0.6 m. The rod is rotated in a vertical plane about a horizontal
axis passing through the other end of the rod with angular frequency
104 π rad/s. Find the magnetic field at a point on the axis of rotation at
a distance of 0.8 m from the centre of the path.
Now half of the charge is removed from one end and placed on the
other end. The rod is rotated in a vertical plane about a horizontal axis
passing through the mid point of the rod with the same angular
frequency. Calculate the magnetic field at a point on the axis at
distance of 0.4 m from the centre of the rod.
Solution: As revolving charge q is equivalent to a current
I = qf
= q

2
=1
104 
2
= 5 × 103 A
and field at a distance x from the centre on the axis of a current-carrying
coil is given by
0 2 NIR 2
B=
4 R 2  x2 3 / 2

So, B = 10

7

2  1  5  103  0.6
0.62  0.82 


2
3 /2
= 1.13 × 10–3 T Ans.
Now, if half of the charge is placed at the other end and the rod is
rotated at the same frequency,
 q
 q
I' =   f    f
 2
 2
= qf = I
= 5 × 103 A
However, according to the given problem in the situation R' = 0.3 m and
x' = 0.4 m, so,
B' = 10
7
2  1  5  103  0.3
0.32  0.42 


2
3 /2
= 2.26 × 10–3 T Ans.
Problem. Two circular coils each of 100 turns are held such that one lies in the
vertical plane and the other in the horizontal plane with their centres
coinciding. The radii of the vertical and horizontal coils are respectively
20 cm and 30 cm. If the directions of the currents in them are such that
the earth's magnetic field at the centre of the coils is exactly
neutralized, calculate the currents in each coil [H = 27.8 A/m; angle of
dip = 30°]
Solution: As the field due to a current-carrying coil is along its axis, the vertical
coil will produce horizontal field and horizontal coil vertical, i.e.,
0 2NVIV
 BH
4
RV
and
0 2NHIH
 BV
4 RH
But as tan  
BV
,
BH
BV  BH tan  
BH
3

  
as


  

6 

and, 1
A
Wb
 4  107 2
m
m
So, 10
7
i.e., IV =
2  100  IV
 4  107  27.8
0.2
0.2  2  27.8
 0.1112 A Ans.
100
and, 107
i.e., IH =
2  100  IH
27.8
 4  107
0.3
3
2  3  27.8
1000
= 0.0963 A Ans.
Problem. A circular coil of radius 0.157 m has 50 turns. It is placed such that its
axis is in the magnetic meridian. A dip needle is supported at the centre
of the coil with its axis of rotation horizontal and in the plane of the
coil. The angle of dip is 30° when a current flows through the coil. The
angle of dip becomes 60° on reversing the current. Find the current in
the coil assuming that the magnetic field due to the coil is smaller than
the horizontal component of earth's magnetic field [BH = 3 × 10–5 T]
B 
Solution: An angle of dip, tan    V  and in the setting specified in the
 BH 
problem,
BH  BH  BC
with BH > BC
So that,
tan 30 =
BV
BH  BC 
and tan 60 =
i.e.,
=
BH  BC
tan60

BH  BC
tan30
3
3
 1 


3
So, BC =
=
BV
BH  BC 
1
B
2 H
1
 3  105
2
= 1.5 × 10–5 T
Now as field at the centre of a current-carrying coil is given by
B=
0 2NI
4 R
so 107
i.e., I =
2NI
 1.5  105
R
0.157  150
2  3.14  50
= 0.075 A Ans.
Problem. Calculate the field at the centre of a semicircular wire of radius R in
situations (A), (B) and (C) if the straight wire is of infinite length.
Solution: Keeping in mind that the field due to a straight wire of infinite length
for a point at distance d from one of its ends is zero if the point is along
 0I 
if the point is on a line perpendicular to its length
4d
its length and 

 0I 
and here
4R 
while at the centre of a semicircular coil is 

BR  Ba  Bb  Bc
(A)
BR  0 
=
0 I
 0
4 R
0 I
   into the page
4 R
(B)
BR 
=
(C)

0  I
4 R

0 I
4 R
0 I
   2 out of the page
4 R
BR 
=
0 I
4 R
0 I
4 R

0 I
 I
 0
4 R
4 R
0 I
   2 into the page
4 R
Problem. The wire loop PQRSP formed by joining two semi-circular wires of
radii R1 and R2 carries a current I as shown in Fig. 13.31. What is the
magnetic induction at the centre O and magnetic moment of the loop
in cases (A) and (B)?
Solution: As the point O is along the length of the straight wires, so the field at O
due to them will be zero and hence
(A)
B
0  I
I



4  R 2
R1
i.e., B 



1
0
1
I 

 out of the page
4  R1 R 2 
and, M  NIS
1
1
R 22   R12
2
2
= 1 I



i.e., M 
(B)
1
I R 22  R12  into the page
2
Following as in case (A), in this situation,
1
1


 into the page
 R1 R 2 
B
0
I
4
and,
1
 I R 22  R12  into the page
2
Problem. A battery is connected between two points A and B on the
circumference of a uniform conducting ring of radius r and resistance R.
One of the arc AB of the ring subtends an angle Ө at the centre. What is
the value of the magnetic field at the centre due to the current in the
ring?
Solution: As the field due to an arc at the centre is given by
B=
0 I
4 r
So, BO =
0 I1
 I 2  
 0 2
4 r
4
r
But as, (VA – VB) = I1R1
= I2R2
i.e., I1
R1
R2
= I1
L1
L2
(as R \ L)
or, I2  I1

2  
(as L = rф)
So, BO =
0 I1
 I
 0 1
4 r
4 r
0
i.e., the field at the center of the coil is zero and is independent of Ө.
Problem. A long wire is bent into the shape without cross contact at P.
Determine the magnitude and direction of field at C. Now the circular
loop is rotated without distortion about its diameter perpendicular to
the straight portion of the wire. Determine the field at C if the point A
of the loop goes into the paper on rotation.
Solution: The field at point C will be the resultant of the fields due to circular loop
and wire. So in situation (A) as the field due to both, the loop and the
wire at C will be out of the page,
B R = BW + B C
=
0 2I
4 R
i.e., BR 

0 2I
4 R
0 2I
1   out of the page
4 R
However, in situation (B), the field due to the coil will be along its axis
(which is parallel to the straight wire) while due to straight wire out of
the page. So BC and BW in the this situation will be at right angles to each
other so that
BR =
=
B
2
W
0 2I
4 R
 B2C

1  2
B 
and,   tan1  W 
 BC 
 1
= tan1  
 
18
FORCE ON A CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD
When a current carrying conductor is placed in a magnetic field, the
conductor experiences a force in a direction perpendicular to both the direction
of magnetic field and the direction of current flowing in the conductor. This force
is also called Lorentz force.
F
i
B
i
The direction of this force can be found out either by Fleming’s left hand rule
or by right hand palm rule.
The magnetic force is
F = ilB sin Ө

In vector form, F  i l  B
Where

B = intesity of magnetic field
i = current in the conductor
l = length of the conductor
and
Ө = angle between the length of conductor and direction of
magnetic field.
Case :
(i)
If Ө= 90° or sin Ө= 1 then F = ilB (maximum)
Therefore, force will be maximum when the conductor carrying current
is perpendicular to magnetic field.
(ii)
If Ө= 0° or sin Ө= 0,
Then F = ilB × 0 = 0
Thus, the force will be zero, when the current carrying conductor is
parallel to the field.
Example 12. A straight current carrying conductor is placed in such a way that the
current in the conductor flows in the direction out of the plane of the paper. The
conductor is placed between two poles of two magnets, as shown. The conductor
will experience a force in the direction towards.
P
R
S
S
N
Q
(A) P
(B)
Q
(C)
R
(D) S
Sol. The direction of magnetic field on the conductor is along SR.
But F  i l  B
F  i l kˆ  (–B ˆi)
F  i l B ˆj  i l B(ˆj)
Hence, the direction of magnetic force on the wire is towards Q.
Hence option (B) is correct
Example 13. In the figure shown a semicircular wire loop is placed in uniform
magnetic field B = 1.0 T. The plane of the loop is perpendicular to the magnetic
field. Current i = 2A flows in the loop in the direction shown. Find the magnitude
of the magnetic force in both the cases (a) and (b). The radius of the loop is 1.0m.
×
×
×
×
×
×
i = 2A
×
×
1m
×
×
×
×
×
×
(a)
×
×
×
×
×
i = 2A
×
×
×
×
×
×
×
×
×
×
×
×
Sol. Refer figure (a) :
(b)
×
It forms a closed loop and the current completes the loop. Therefore, net
force on the loop in uniform field should be zero.
Refer figure (b) :
In this case although it forms a closed loop, but current does not complete
the loop. Hence, net force is not zero.
×
× C ×
×B
×
×
×
×
×
A
×
×
×
×
×
×
D
×
FACD  FAD
Floop  FACD  FAD  2FAD
|Floop | 2|FAD |
|Floop | 2ilB sin (l = 2r = 2.0 m)
= (2) (2) (2) (1) sin 90º = 8 N
Example 14. An arc of a circular loop of radius ‘R’ is kept in the horizontal plane
and a constant magnetic field ‘B’ is applied in the vertical direction as shown in
the figure. If the arc carries current ‘I’ then find the force in the arc.
×
×
×
×
×
×
×
×
×
×
×
I×
×
×
×
× × ×
B
× × ×
× ×
90º × ×
× × ×
Sol. As we know, magnetic force on a closed loop placed in uniform magnetic
field is zero.
(1)
R
/2
R
(2)
B1  B2  0
B1  B2  IlB(ˆj)
B1  Il B ˆj
B1  IB( 2R)ˆj
B1  2 IBR
Example. A conducting wire bent in the form of a parabola y2 = 2x carries a
current i = 2 A wire is placed in a uniform magnetic field B  4kˆ Tesla. The
magnetic force on the wire is (in newton)
y(m)
A
2
B
(A)
16 ˆi
(B)
32 ˆi
(C)
32 ˆi
x(m)
16 ˆi
(D)
Sol. The net magnetic force on closed loop is zero.
A
B
Force on parabola + force on straight wire AB = 0
Force on the parabola = – force on straight wire AB

  Iˆj  B


 
 2 4 ˆj  4kˆ
F  32 ˆi
Hence option (C) is correct.
Example. A conductor of length ‘l’ and mass ‘m’ is placed along the east-west line
on a table. Suddenly a certain amount of charge is passed through it and it is
found to jump to a height ‘h’. The earth’s magnetic induction is B. The charge
passed through the conductor is: (B is horizontal)
(A)
1
Bmgh
(B)
2gh
Blh
(C)
gh
Blh
(D)
m 2gh
Bl
Sol. The magnetic force is
F = I lB
or F 
dq
lB
dt
or dq 
Fdt
lB
or q 
Fdt mv 0

lB
lB
But 02 = v02 – 2gh
v 0  2gh
q 
m 2gh
lB
Hence option (D) is correct
Example. A U-shaped wire of mass m and length l is immersed with its two ends
in mercury (see figure). The wire is in a homogeneous field of magnetic induction

B. If a charge, that is, a current pulse q  idt , is sent through the wire, the wire
will jump up.
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
l
×
×
×
B
i
m
Hg
Calculate, from the height ‘h’ that the wire reaches, the size of the charge or
current pulse, assuming that the time of the current pulse is very small in
comparison with the time of flight. Make use of the fact that impulse of force

equals Fdt , which equals mv. Evaluate ‘q’ for B = 0.1 Wb/m2, m = 10 gm, l = 20
cm & h = 3 meters. [g = 10 m/s2]
Sol. I l B = F
Fdt = mv
or I l Bdt = mv
But
1 2
mv  mgh
2
v  2gh
Idt 
mv m

2gh
lB lB
m
10  10 3
q
2gh 
2  10  3  15 coulmb.
lB
20  102  0.1
Example.
A metal ring of radius r = 0.5 m with its plane normal to a uniform magnetic
field B of induction 0.2 T carries a current I = 100 A. The tension in newtons
developed in the ring is:
(A)
100
(B)
50
(C)
25
(D)
10
Sol.
l of the loop.
F
Tcos /2
/2
/2
Tcos /2
T
T
+
B
For equilibrium,
F  2T
sin 

 2T  T
2
2
or I∆lB = TӨ
or IrӨB = TӨ
T = IrB = 100 × 0.5 × 0.2 = 10 N
Hence option (D) is correct
Example:
Crying a current of magnitude 2.5 ampere may be placed so the
induction at R is zero.
Solution:
(a)
As force on a charged particle in a magnetic is given by

F  q v B

The field due to both current-carrying wires will be along ???
y-axis. Now as the particle is moving along positive ???? the
angle between the direction of motion of charged particle
 
  rad.
2
So, F = qvB sin 90
i.e., B =
F
qv
Substituting the given data,
B=
3.2  1020
1.6  1019  4  105
= 5 × 10–7 T
Now, as this field is produced by wires P and Q and point R
on the same side of both,
0  2IP 2IQ 



4  dP
dQ 
= 5 × 10–7 T
i.e.,
2  2.5 2I

5
5
2
i.e., I = 4A Ans.
(b)
If the distance of point R from third current-carrying wire is
x, then BR = 0 implies
0 2  2.5
 5  107  0
4
x
Now there two possibilities:
(i)
If the current in third wire is into the page, I = + 2. 5 A,
x = – 1, i.e., the point R must be to the left of the wire, i.e.,
the wire must be to the right of R, i.e., at A with RA = 1 m.
(ii)
If the current in third wire is out of the page, I = – 2.5 A, so x
= +1, i.e., the point R must be to the right of the wire, i.e.,
the wire must be to the left of R, i.e., at B with RB = 1 m.
Problem:
A very long thin strip of metal of width 'b' carries a current I along its
length. Find the magnitude of magnetic field in the plane of the strip
at a distance 'a' from the edge nearest to the point.
Solution:
Assuming the strip to be made up of a large elements parallel to its
length, consider an element of length dx at a distance x from the point
P. Treating the a long element as a long current-carrying wire,
dB =
0 2I'
4 x
Now as the current in the strip of width b is I, so the current in the
element of width dx will be
I' =
I
dx
b
and hence,
dB =
0 2I dx
4 b x
So, B =
=
Note:
0 2I ab dx
4 b a
x
0 2I
b

loge 1   Ans.
2 b
a

If the point is distant from the strip, i.e., a >> b,
b  b b2

loge 1     2  ...
a  a 2a

So, B =
=
b
a
0 2I b

2 b a
0 2I
4 a
i.e., for a distant point the strip behaves as a current carrying wire
which justifies our result.
Problem:
In the Bohr model of hydrogen atom, the electron circulates around
the nucleus in the path of radius 5.1 × 10–11 m at a frequency of 6.8 ×
1015 rev/s. Calculate the magnetic induction B at the centre of the
orbit. What is the equivalent dipole moment?
Solution:
The revolving electron is equivalent to a circular current
I = ef
= 1.6 × 10–19 × 6.8 × 1015
= 1.088 mA
So the field at the centre of the path
B=
0 2I
4 R
= 10
7
2    1.088  103

5.1  1011
= 13.4 Wb/m2
Further, as a current-carrying loop is equivalent to a magnetic dipole of
moment
M = NIS
= R2I
[as N = 1]
–11 2
) × 1.088 × 10–3
i.e., M = 8.88 × 10–24 A-m2 Ans.
Problem:
A charge of 1 coulomb is placed at one end of a non-conducting rod of
length 0.6 m. The rod is rotated in a vertical plane about a horizontal
axis passing through the other end of the rod with angular frequency
104
the magnetic field at a point on the axis of rotation
at a distance of 0.8 m from the centre of the path.
Now half of the charge is removed from one end and placed on the
other end. The rod is rotated in a vertical plane about a horizontal
axis passing through the mid point of the rod with the same angular
frequency. Calculate the magnetic field at a point on the axis at a
distance of 0.4 m from the centre of the rod.
Solution:
As revolving charge q is equivalent to a curret
I = qf
= q

2
104 
=1
2
= 5 × 103 A
and field at a distance x from the current-carrying coil is given by
0 2NIR 2
B=
4 R 2  x2 3 / 2

So, B = 10

7

2  1  5  103  0.6
0.62  0.82 


2
3 /2
= 1.13 × 10–3 T
Now, if half of die charge is placed at the other end and the rod is
rotated at the same frequency,
 q
 q
I' =   f    f
 2
 2
= qf
= I = 5 × 103 A
However, according to the given problem in the situation R' = 0.3 m
and x' = 0.4 m,
So, B' = 10
7
2  1  5  103  0.3
0.32  0.42 


2
3 /2
= 2.26 × 10–3 T Ans.
Problem.
Two circular coils each of 100 turns are held such that one lies in the
vertical plane and the other in the horizontal plane with their centres
coinciding. The radii of the vertical and horizontal coils are
respectively 20 cm and 30 cm. If the directions of the currents in them
are such that the earth's magnetic field at the centre of the coils is
exactly neutralized, calculate the currents in each coil. [H = 27.8 A/m;
angle of dip = 30°]
Solution:
As the field due to a current-carrying coil is along its axis, the vertical
coil will produce horizontal field and horizontal coil vertical, i.e.,
0 2NVIV
 BH
4 R V
and
0 2NHIH
 BV
4 RH
But as tan  
BV
,
BH
BV = BH tan  
BH
3

  
as


  

6 

and, 1
A
Wb
 4  107 2
m
m
So, 10
7
2  100  IV
0.2
= 4  107  27.8
i.e., IV =
0.2  2  27.8
100
= 0.1112 A Ans.
and, 10
7
2  100  IH
0.3
= 4  107
i.e., IH =
27.8
3
2  3  27.8
1000
= 0.0963 A Ans.
Problem:
A circular coil of radius 0.157 m has 50 turns. It is placed such that its
axis is in the magnetic meridian. A dip needle is supported at the
centre of the coil with its axis of rotation horizontal and in the plane
of the coil. The angle of dip is 30° when a current flows through the
coil. The angle of dip becomes 60° on reversing the current. Find the
current in the coil assuming that the magnetic field due to the coil is
smaller than the horizontal component of earth's magnetic field [BH =
3 × 10–5 T].
Solution:
As angle of dip,
tan
 BV 
 B 
H
and in the setting specified in the problem,
BH
B H ± BC
with BH > BC
So that, tan 30° =
and tan 60° =
i.e.,
=
BV
BH  BC 
BV
BH  BC 
BH  BC
tan60

BH  BC
tan30
3
3
 1 


3
So, BC =
=
1
BH
2
1
 3  105
2
= 1.5 × 10–5 T
Now as field at the centre of a current-carrying coil is given by
B=
0 2NI
4 R
so 10
7
i.e., I =
2NI
 1.5  105
R
0.157  150
2  3.14  50
= 0.075 A Ans.
Problem:
Calculate the field at the centre of a semi- circular wire of radius R in
situations, if the straight wire is of infinite length.
Solution:
Keeping in mind that the field due to a straight wire of infinite length
for a point at a distance d from one of its ends is zero if the point is
 0I 
if the point is on a line perpendicular to its
4d
along its length and 

 0I 
and here
4R 
length while at the centre of a semicircular coil is 

BR  Ba  Bb  Bc
(A)
BR  0 
=
(B)
(C)
0 I
   into the page
4 R
BR 
=
0 I
4 R

0 I
4 R

0 I
4 R
0 I
   2 out of the page
4 R
BR 
=
0 I
 0
4 R
0 I
4 R

0 I
 I
 0
4 R
4 R
0 I
   2 into the page
4 R
Problem.
The wire loop PQRSP formed by joining two semi-circular wires of
radii R1 and R2 carries a current I. What is the magnetic induction at
the centre O and magnetic moment of the loop in cases (A) and (B)?
Solution:
As the point O is along the length of the straight wires, so the field at O
due to them will be zero and hence
(A)
B
0  I
I


4  R2
R1
i.e., B 



1
0
1
out of the page
I 

4  R1 R 2 
and, M  NIS
1
1
R 22   R12
2
2
= 1 I
i.e., M 
(B)



1
I R 22  R12  into the page
2
Following as in case (A), in this situation,
B
1
0
1
into the page
I 

4  R1 R 2 
and, M 
1
I R 22  R12  into the page
2
Problem.
A battery is connected between two points A and B on the
circumference of a uniform conducting ring of radius r and resistance
R. One of the arc AB of the ring subtends an angle at the centre.
What is the value of the magnetic field at the centre due to the
current in the ring?
Solution:
As the field due to an arc at the centre is given by
B=
0 I
4 r
So, BO =
0 I1 0 I2 2  

4 r
4
r
But as, (VA – VB) = I1R1
= I2R2
i.e., I2 = I1
= I1
R1
R2
L1
L2
as R  L 
or, I2 = I1

2  
as L  r
so, BO =
i.e.,
0 I1
 I
 0 1
4 r
4 r
0
Note:
In the situation, the current in the two semicircles will divide equally
and will produce equal fields at the centre in opposite directions
resulting in zero net field which verifies the correctness of the above
result, i.e., the field at the centre is zer
Problem:
A long wire is bent into the shape without cross contact at P.
Determine the magnitude and direction of field at C. Now the circular
loop is rotated without distortion about its diameter perpendicular to
the straight portion of the wire. Determine the field at C if the point A
of the loop goes into the paper on rotation.
Solution:
The field at point C will be the resultant of the fields due to circular
loop and wire. So in situation (A) as the field due to both, the loop and
the wire at C will be out of the page,
B R = R W + BC
=
0 2I
4 R
i.e., BR 

0 2I
4 R
0 2I
1   out of the page
4 R
However, in situation (B), the field due to the coil will be along its axis
(which is parallel to the straight wire) while due to straight wire out of
the page. So BC and BW and in this situation will be at right-angles to
each other so that
BR =
=
B
2
W
 B2C

0 2I
1  2
4 R
B 
tan1  W 
 BC 
 1
= tan1  
 
18
Problem:
A solenoid of length 0.4 m and diameter 0.6 m consists of a single
layer of 1000 turns of fine wire carrying a current of 5 × 10–3 ampere.
Calculate the magnetic field on the axis at the middle and at the end
of the solenoid.
Solution:
In case of a solenoid the field at a point on the axis as shown in Fig.
13.34 (A) is given by
B=
0
2nI  sin   sin 
4
Here, n 
N
L
=
1000
0.4
= 2.5 × 103 turns/m
So, B = 10–7 ×
× 103 × 5 × 10–3
i.e., B = 2.5 × 10–6 [sin + sin ]
So (a) when the point is at the middle on the axis =
L
L2  4r2
=
=
0.4
0.4
 4 0.3
2
2
4
7.2
–6
= 8.7 × 10–6 T Ans.
L
L2  r2
=
0.4
0.4
2
 0.3
2
×2×
4
7.2
=
4
5
10–6 ×
4
5
= 6.28 × 10–6 T Ans.
Note:
If the solenoid had been of infinite length for case (a),
=5
–6
T while for case (b), = 0 and =
=

, so B = 2.5
2

, so B
2
× 10–6
T.
13.2. Force on a Current-Carrying Conductor in a Magnetic Field
Through experiments Ampere established that when a current
element* Id L is placed in a magnetic field B , it experiences a force
dF  IdL  B
…(1)
From this expression it is clear that:
(a)
The magnitude of force is
…(2)
where is the angle between the vectors I d L and, B. So:
(i)
Force on a current element will be minimum (= 0) when sin
= min = 0, i.e., = 0° or 180°, i.e., a current element in a
magnetic field does not experience any force if the current in
it is collinear with the field.
(ii)
The force on the current element will be maximum (= BI dL)
when sin = max = 1, i.e., = 909, i.e., force on a current
element in a magnetic field is maximum (= BI dL) when it is
perpendicular to the field.
*Current element is defined as a vector having magnitude
equal to the product of current with a small part of length of
the conductor and the direction in which the current is
flowing in that part of the conductor.
(b)
The direction of force is always perpendicular to the plane
containing Id L and B is same as that of cross-product of


two vectors A  B with A  IdL.
The direction of force when current element Id L and B are
perpendicular to each other can also be determined by
applying either of the following rules:
(i)
Fleming's Left-hand Rule:
Stretch the forefinger, central finger and thumb of left hand
mutually perpendicular. Then if the forefinger points in the
direction of field ( B ), the central finger in the direction of
current I, the thumb will point in the direction of force (or
motion).
(ii)
Right-hand Palm Rule:
Stretch the fingers and thumb of right hand at right angles to
each other. Then if the fingers point in the direction of field
B and thumb in the direction of current I, the normal to
palm will point in the direction of force (or motion).
Regarding the force on a current-carrying conductor in a
magnetic field it is worth mentioning that:
(1)
As the force BI dL sin is not a function of position r, the
magnetic force on a current element is non-central [a central
force is of the form F = Kf (r) nr ]
(2)
The force d F is always perpendicular to both B and Id L
though B and Id L may or may not be perpendicular to each
other.
(3)
In case of current-carrying conductor in a magnetic field if
the field is uniform, i.e., B = constt.,
*In this situation, d F and  may or may not be zero.
F   IdL  B
= I  dL   B


and as for a conductor
 dL represents the vector sum of all
the length elements from initial to final point, which in
accordance with the law of vector addition is equal to the
length vector L ' joining initial to final point, so a currentcarrying conductor of any arbitrary shape in a uniform field
experiences a force
F  I  dL   B
= IL '  B
...(3)
where L ' is the length vector joining initial and final points
of the conductor.
(4)
If the current-carrying conductor in the form of a loop of any
arbitrary shape is placed in a uniform field,
F
 IdL  B
= I  dL   B


and as for a closed loop, the vector sum of d L always zero.
So, F = 0*
[as
 dL = 0]
…(4)
i.e., the net magnetic force* on a current loop in a uniform
magnetic field is always zero.
Here it must be kept in mind that in this situation different
parts of the loop may experience elemental force due to
which the loop may be under tension or may experience a
torque.
(5)
If a current-carrying conductor is situated in a non-uniform
field, its different elements will experience different forces;
so in this situation,
FR  0 but 
may or may not be zero ...(5)
If the conductor is free to move, it translates with or without
rotation.
(6)
The net force on a current-carrying conductor due to its own
field is zero; so if there are two long parallel current-carrying
wires 1 and 2, wire-1 will be in the field of wire-2 and viceversa. So force on dL2 length of wire-2 due to field of wire-1,
dF2 = I2
dL2dB1 =
0 2I1I2
dL 2
4 d
0 2I1 

as
B

1

4 d 

i.e.,
dF2
 2I I
 0 12
dL2
4 d
Same will be true for wire-1 in the field of wire-2. The
direction of force in accordance with Fleming's left hand rule.
So force per unit length in case of two parallel currentcarrying wires separated by a distance d is given by:
dF 0 2I1I2

dL 4 d
and the force between the wires is attractive if the current in
them is in the same direction, otherwise repulsive [this is
opposite to that of what happens between two charges].
Note:
Through this concept the SI unit of current, ampere, is
defined as the current which when passed through each of
two parallel infinitely long straight conductors placed in free
space at a distance of 1 m from each other, produces
between them a force of 2 × 10–7 newton for one metre of
their length.
(7)
In case of a current-carrying conductor in a magnet field if
the conductor experiences a force and is free to move, work
will be done and hence its kinetic energy or speed will
change, i.e.,
W = KE

with W = F.ds
Question.
What is the force experienced by a semi-circular wire of radius R
when it is carrying a current I and is placed in a uniform magnetic
field of induction B?
Answer:
As force on a current element Id L in a field B is given by:
dF  IdL  B
the force on a current-carrying conductor in a uniform field will be
F   IdL  B
= I dL   B


[as I and B = constt.]
and if n is a unit vector in the direction PQ,
 dL  PQ
= 2R n

So, F = 2RI n  B
(A)

As in this situation n and B are parallel,
= 0°
F = 2RBI sin 0
= 0 Ans.
(B)
As in this situation also B is perpendicular to n
F = 2RBI sin 90
= 2RBI out of the page Ans.
(C)
As in this situation also B is perpendicular to n,
F = 2RBI
and directed vertically down. Ans.
Question VI:
A wire ABCDEF (with each side of length L) bent and carrying a
current I is placed in a uniform magnetic induction B parallel to
positive y-direction. What is the force experienced by the wire?
Answer:
If we join A to F by a straight conductor it will become a closed loop
and as in case of closed loop in a uniform magnetic field,
F =0
So if FW is the force on the given network of wires and F ' on the wire
AF,
FW  F '  0,
i.e., FW   F '
But as F ' = BIL sin 90° along negative z-axis, so
FW = BIL along z-axis Ans.
Question.
Two infinitely long parallel wires carry equal current in the same
direction.
(a)
What is the direction of the magnetic field due to one of
the wires at any point along the other wire?
(b)
What is the direction of force on one wire due to the other?
(c)
By what factor does this force change if the current each
wire is doubled?
(d)
What is the direction of the magnetic field at a point
midway between the two wires?
Answer:
(a)
In accordance with right hand screw rule, the field due to I1
at the position of I2 is perpendicular to the plane of the page
and into it.
(b)
As similar currents attract each other, the force on I2 will be
perpendicular to it and towards I1 .
(c)
As force per unit length between are parallel currentcarrying conductors is given by
dF 0 2I1I2

dl
4 d
so if I1 and I2 both are doubled, the force will increase by a
factor of 4.
(d)
For a point between the wires the field due to I1 will be into
the page while due to I2 out of the page. So at a point
midway between the wires for I1 = I2 the two fields will cancel
each other and so there is no question of direction.
Question.
Two parallel wires carrying current in the same direction attract each
other while two similar charges moving parallel to each other in the
same direction repel each other. Can you resolve this apparant
contradiction?
Answer:
In case of two similar current-carrying wires only a magnetic force
Fm =
0 2I1I2
4 r
…(1)
acts between the wires as the wires are electrically neutral. However,
in case of motion of two similar charged particles moving parallel to
each other in the same direction two forces, viz., magnetic and electric,
act between them. The magnetic force is attractive and is given by Eq.
(1) while the electric force is repulsive and is given by Coulomb's law,
i.e.,
Fe =
1 q1q2
40 r2
So the net force between them will be
F  Fe  Fm
However, as the electric force is much stronger than the magnetic, so
the net interaction between two similar charges moving parallel to
each other will be repulsive inspite of the attractive magnetic force
between them.*
Problem.
A straight wire of length 30 cm and mass 60 mg lies in a direction 30°
east of north. The earth's magnetic field at this site is horizontal and
has a magnitude of 0.8 G. What current must be passed through the
wire so that it may float in air? [g = 10 m/s2]
Solution:
A current I is passed through the wire from end A towards B it will experience a
force BIL sin vertically up and hence will float if
BIL sin = mg
i.e., I =
=
mg
BL sin 
60  106  10
0.8  10
4
 30  10
2
 1
 
 2
= 50 A Ans.
(A)
The Discovery and the Laws
Oersted in 1820 by observing the deflection of a compass needle
when it is brought near a current carrying conductor discovered
that a current carrying conductor produces a magnetic field in the
space surrounding it.
The field is described by the vector B called magnetic induction, flux
density or simply field – vector with units Wb/m2 or tesla.
BIOT-SAVART LAW
B
0 IdL  r
4  r3
…(1)
or Ampere's circuital law
 B.dL   I
…(2)
0
Biot-Savart law is an inverse-square law and is the magnetic analogue of
Coulomb's law E 
Gauss's law
1
r
while Ampere's law is analogous to
dq
40 
r3
 E.ds 
1
q .
0
The direction of B at a point in case of linear and circular current
carrying conductors is specified as follows :
(a)
For linear currents :
If the wire is grasped in the palm of the right hand with the stretched
thumb pointing in the direction of the current, the fingers curl in the
direction of the field (i.e., the lines of force).
(b)
For circular currents :
If the direction of the current coincides with the direction of the curl of
the fingers of the right hand, the stretched thumb points in the direction
of the field perpendicular to the plane of the paper is represented by (×)
if into the page and by (.) if out of the page.
Problem 28 (b). A circular loop of radius R is bent along a diameter and given a
shape as shown in the figure (A). One of the semicircles (KNM) lies in
the x-z plane and the other one (KLM) in the y-z plane with their
centres at the origin. Current I is flowing through each of the
semicircles.
(a)
A particle of charge q is released at the origin with a velocity   0i.
Find the instantaneous force F on the particle. Assume that space is
gravity free.
(b)
If an external uniform magnetic field B J is applied, determine the forces
F1 and F2 on the semicircles KLM and KNM due to this field and the
net force F on the loop.
Solution :
(a)
Magnetic field due to semicircular current loop.
B
0 I
.
4 R
and direction perpendicular to plane of loop outward if current is
anticlockwise and towards the centre if current is clockwise if looking
from that side.
So, due to KNM the direction of B will be in +y-direction as looking from
the y-axis the current is anticlockwise. Similarly due to KLM, the direction
of B is along –x axis. So, due to both segments,
B  BKNM  BKLM
 
0I
I
j  0 i
4R
4R
The force on the moving charged particle in a magnetic field,

F  q v B


  

q0I
0i  j  i
4R

q0I
 k
4R 0
 
i.e., along (–z) direction Ans.
(b)
The force on a current loop in external magnetic field is zero, so force F1
on segment KLM is the same as the force on segment KM carrying the
current in same direction.
 
FKLM  I dI  B


 I 2Rk  jB  2IRBi


and FKNM  I 2Rk  jB  2IRBi
So, net force on the loop = 4IRB i
i.e., 4IRB Newton in + x-direction.
Problem 29 (a). A current of 10 A flows around a closed path in a circuit which is
in the horizontal plane as shown in the figure (C). the circuit consists of
eight alternating arcs of radii r1  0.08m and r2  0.12m. Each
are subtends the same angle at the centre.
(a)
Find a magnetic field produced by this circuit at the centre.
(b)
An infinitely long straight wire carrying a current of 10A is passing
through the centre of the above circuit vertically with the direction of
the current being into the plane of the circuit. What is the force acting
on the wire at the centre due to the current in the circuit? What is the
force acting on the arc AC and the straight segment CD due to the
current at the centre?
Solution : Contribution to magnetic field due to straight wire sections converging
towards centre is zero. wire sections along arcs contribute to resultant
magnetic field. Magnetic field due to an arc subtending an angle  at
centre is given by
B
0i


2R 2
Magnetic field due to the arcs of radius r1  4 
[as each arc subtends
0i 1 0i
 
2r1 8 4r1

angle at centre and magnetic field due to each
4
arc is in same direction]
Similarly magnetic field due to the arcs of radius r2 
Hence resultant magnetic field 
0i
4r2
0i  1 1 
  .
4  r1 r2 
On substituting numerical values we get
B  6.54  105 T. Ans.
(b)
(i) Resultant magnetic field at the centre points upwards normal to plane
of paper. The current in the wire at centre is antiparallel to the magnetic
field; hence force on wire is zero.
(ii)
Magnetic field due to a long wire circulates around it hence magnetic
field is tangential to arc AC so it will not experience force.
(iii)
Section CD lies in a variable magnetic field force. So force experienced by
a small element dx at a distance x from long wire at centre =
Resultant force on CD 

r2
r1
0i
i dx
2x
0i
i dx
2x
0i2
r

loge 2
2
r1
On substituting numerical values, we get
F  8.11  106 N downwards
Current Loop
(A)
Magnetic Moment of a Current loop
According to magnetic effects of current, in case of a current-carrying
coil for a distance axial point,
 0 2NIR2
B
n
4  R2  x2 3/2
 0 2NIR2
n
4 x3
(with x >> R)
If we compare this result with the field due to a small bar magnet for a
distant axial point, i.e.,
B
 0 2M
4 x3
we find that a current-carrying coil for a distant point behaves as a
magnetic dipole of moment
M  lS
with S  r2Nn
…(1)
So the magnetic moment of a current carrying coil is defined as the
product of current in the coil with the area of coil in vector form.
Regarding magnetic moment of a current loop it is worth noting that:
(1)
Magnetic moment of a current loop is a vector perpendicular to the
plane of the loop as shown in Fig. 13.57 with dimension [L2A] and unit Am2.
(2)
It depends on the current in the loop and its area, but is independent of
the shape of the loop, i.e., circular or rectangular etc.
(3)
In case of a charged particle having charge q and moving in circle of
radius R with velocity v as
I  qf  q
v
2R
and S   R2 n
1
M  IS  qvRn
2
…(2)
But an angular momentum of the charged particle in this situation is
L  mvRn, so
M
q
L
2m
…(3)
i.e., the magnetic moment of a charged particle moving in a circle is
 q 
  times of its angular momentum and is directed opposite of L if
2m
the charge is negative.
(B)
Torque on a Current Loop
When a current-carrying coil is placed in a uniform magnetic field the net
force on it is always zero. However, as its different parts experience
forces in different directions so the loop may experience a torque (or
couple) depending on the orientation of the loop and the axis of
rotation. For this, consider a rectangular coil in a uniform field B which is
free to rotate about a vertical axis PQ and normal to the plane of the coil
The arm AB and CD will experience forces B(NI)b vertically up and down
respectively. These two forces together will give zero net force and zero
torque (as are collinear with axis of rotation), so will have no effect on
the motion of the coil.
Now the forces on the arms AC and BD will be BINL in the direction out
of the page and into the page respectively, resulting in zero net force,
but an anticlockwise couple of value.
with S = NLb
….(1)
Now treating the current-carrying coil as a dipole of moment M  IS,
Eqn. (1) can be written in vector from as
  M B
[with M  IS  NiSn ]
…(2)
This is the required result and from this it is clear that:
(1)
Torque will be minimum (= 0) when s
the plane of the coil is perpendicular to magnetic field i.e., normal to the
coil is collinear with the field.
(2)
the plane of the coil is parallel to the field i.e., normal to the coil is
perpendicular to the field.
(3)
By analogy with electric or magnetic dipole in a field, in case of a currentcarrying coil in a field,
U   M.B
with F = 
dU
dr
and W = MB(1 –
The values of U and W for different orientations of the coil in the field.

  0 M is parallel to B

  0  min
W  0  min
U   MB  min
Stable equilibrium (A)

  90 M is  to B

  MB  max
W  MB
U 0
No equilibrium (B)

  180 M is antiparallel to B

  0  min
W  2MB  max
U  MB  max
Unstable equilibrium (C)
(4)
Instruments such as electric motor, moving coil galvanometer and
tangent galvanometers etc. are based on the fact the a current-carrying
coil in a uniform magnetic field experiences a torque (or couple).
(C)
Galvanometers
(a)
Moving coil galvanometer:
In case of moving coil galvanometer [Fig. 13.60(A)] the deflecting torque
due to current in the coil BINS* is balanced by the restoring couple dur
to elasticity of spring supporting the coil. So if c is the restoring couple
per unit twist and  is the deflection (i.e., rotation) of the coil.
BINS  c
i.e., I  KG
with KG 
c
 Galvanometer constant
NSB
i.e., in case of moving coil galvanometer, the current passing through the
coil is directly proportional to its deflection.
(b)
Tangent galvanometer :
In case of tangent galvanometer a magnetic compass needle is placed
horizontally at the centre of a vertical fixed current-carrying coil whose
plane is in the magnetic meridian. So if the needle in equilibrium
subtends an angle  with the earth's magnetic field,
M  BH  M  BC
or, MBH sin   MBC sin 90  
i.e., BC  BH tan 
or,
0 2NI
 BH tan 
4 R
0 2NI 

asBC  4 R 


i.e., I = K tan 
with K 
4 RBH
.
 Reduction factor of the tangent
0 2N
galvanometer, i.e., in case of a tangent galvanometer when the plane of
coil is in magnetic meridian, current in the coil is directly proportional to
the tangent of deflection of magnetic needle.
NOTE :
A part from these, we also have a third type of galvanometer called 'hotwire galvanometer' which is based on the heating effects of current; so
in it the deflection  is directly proportional to the heating effect, i.e.,
I2R   or I  K 
As this galvanometer is based on he heating effects of current, it works
both in ac and dc.
Question-IX: For a given length L of a wire carrying a current I, how many
circular turns would produce the maximum magnetic moment and of
what value?
Answer : The magnetic moment of a loop having area S and current I is given by
M = IS
and for a circular coil having N turns,
S  R2N
so, M  R2NI
…(1)
Now as the length of the wire from which the loop is formed is given to
be L, so
L  2RN
i.e., R  L / 2N
…(2)
So substituting the value of R from Eqn. (2) in (1),
L2
M  NI 
42N2
i.e., M 
IL2
4N
…(3)
From Eqn. (3) it is clear that M will be maximum when N = min = 1, i.e.,
the coil has only one turn and
Mmax

1 2
IL
4
Question-X: Under what condition is the reading of a tangent galvanometer
most accurate?
Answer: In case of a tangent galvanometer, as
I  K tan ,
i.e., dI  K sec2  d
So,
dI
d
2d


I
sin  cos  sin2
Hence the error in the measurement will be least when
sin2  max  1,
i.e., 2  90.
This in turn implies that the reading of a tangent galvanometer will be
most accurate when the deflection in it is 45.
Problem 29(b). The earth has a magnetic dipole moment of
7.8  1022 A  m2. What current would have to e set up in a single
turn of wire going around the earth at its magnetic equator if we wish
to set up such a dipole through magnetic effects of current? Given the
radius of earth to be 6.4  103 km.
Solution : By definition,
M  IS  IR2
M
7.8  1022
So, I 

R 2
  6.4  106

 6.06  108 A Ans.

2
Problem 30. A charge Q is spread uniformly over an insulated ring of radius R.
What is the magnetic moment of the ring if it is rotated with an angular
velocity  with respect to the normal axis?
Solution : As the charge on element dL of the ring will be
dq 
Q
dL
2R
so the current due to circular motion of this charge,
dI  dq  f 
Q

dL 
2R
2
as   2f 
and hence magnetic moment due to current dI,
dM   dI S 
Q dL
 R2
2
4 R
as S  R2 
so, M 
QR
QR
dL 
2R

4
4
as dL  2R 
 

i.e., M 
1
QR 2 Ans.
2
Problem 31. A coil in the shape of an equilateral triangle of side 0.02 m is
suspended from a vertex such that it is hanging in a vertical plane
between the pole pieces of a permanent magnet producing a
horizontal magnetic field of 5  102 T. Find the couple acting on the
coil when a current of 0.1 ampere is passed through it and the
magnetic field is parallel to its plane.
Solution: As the coil is in the form of an equilateral triangle, its area,
S

1
L  L sin60
2
1
3
2
 0.02 
2
2
 3  104 m2
And so its magnetic moment
M  IS  0.1  3  104
 3  105 A  m2
…(1)
Now the couple on a current-carrying coil in a magnetic field is given by
  MB
i.e.,   MB sin 
and when the plane of the coil is parallel to the magnetic field, the angle
between M (i.e., normal to the area of coil) and B will be 90° and hence
  MB sin 90  MB
i.e.,  


3  105  5  102
 5 3  107N  m Ans.
Problem 32. A solenoid of length 0.4 m a having 500 turns of wire carries a
current of 3 ampere. A thin coil having 10 turns of wire and of radius
0.01 m carries a current of 0.4 ampere. Calculate the torque required to
hold the coil in the middle of the solenoid with its axis perpendicular to
the axis of the solenoid.
Solution : The field of the solenoid (for an internal point not near its ends) will be
B
0
  N 
4nI  0 4 I

4
4  L 
N

 asn  L 


i.e., B  107 4  
500

 3
0.4

 1.5  103 T
And the magnetic moment of the coil
M  IS  INR2
i.e., M  0.4  10    0.01
2
i.e., M  4  104 A  m2
So the torque on the coil
  M  B  MB sin 
Now as the axis of coil, i.e., direction of M is perpendicular to the axis of
the solenoid, i.e., direction of B,   90 and hence

 

  4  104  1.5  103 sin90
6  106N  m Ans.
13.4 (A) Force on a Charge in a Magnetic Field
A current element I dL in a magnetic field B experiences a force
dF  IdL  B
…(1)
Now if the current element I dL is due to motion of charged particles,
each having a charge q moving with velocity v through a cross-section S,
I dL  nS dL qv  nd qv
with d  S dL 
so Eqn. (1) reduce to

dF  nd q v  B

And as n d represents the total number of charged particles in volume
d (n being the number of charged particles per unit volume), the force
on a charged particle will be
F

1 dF
 q v B
n d

…(2)
This is the required result and from this it is clear that:
(1)
The force F is always perpendicular to both the velocity v and the field
B in accordance with RHSR, though v and B themselves may or may
not be perpendicular to each other.
(2)
If the charged particle is at rest (i.e., placed or situated) in a magnetic
field, v  0, so no force will act on it, i.e., a charged particle at rest in a
steady magnetic field does not experience any force.
(3)
If the motion of the charged particle is collinear with the field, i.e.,
  0 or 180,
F  qB sin   0
[as sin 0 = sin 180 = 0]
i.e., a moving charged particle does not experience any force in a
magnetic field if its motion is parallel or antiparallel to the field.
(4)
As the magnitude of the force experienced by a charged particle in a
magnetic field.
F  qB sin  ,
the force will be maximum   qB when
sin   max  1,i.e.,   90,
i.e., the particle is moving perpendicular to the field. In this situation all
the three vectors F, v andB
are mutually perpendicular to each other
(5)
In case of motion of a charged particle in a magnetic field, as force (if it
acts) is always perpendicular to the motion, i.e., displacement, so the
work done,
W   F.ds   Fds cos 90  0 as   90 , i.e., work done in
motion of a charged particle in a magnetic field is always zero. And as by


work-energy theorem W   KE, the kinetic energy  
1

m2  and

2
hence speed  remains constant. However, in this situation the force
changes the direction of motion, i.e., the velocity v of the charged
particle.
(6)
As in case of motion of a charged particle in a magnetic field


F  q v  B so magnetic induction B can be defined* as a vector
having the direction in which a moving charged particle doesnot
experience any force in the field and magnitude equal to the ratio of the
magnitude of maximum force to the product of magnitude of charge
with velocity, i.e. B = fmax/qv.
-particle enter a region of
constant magnetic field with equal velocities. The magnetic field is along the
inward normal to the plane of the paper. The tracks of the particles. Relate the
tracks to the particles.
Answer: As force on a charged particle in a magnetic field is given by


F  q v  B so:
(a)
As for neutron q = 0, F = 0, hence it will pass undeflected, i.e., track C
corresponds to neutron.
(b)
Now if the particle is negatively charged, i.e., electron, in accordance
with RHSR, it will experience a force to the right; so track D corresponds
to electron.
(c)
Further if the charged particle is positive, in accordance with RHBR, it will
experience a force to the left; so both tracks A and B correspond to
-particles). Now in case of
motion of charged particle perpendicular to the magnetic field the path
is a circle with radius
r
mv
qB
i.e., r 
m
q
 m
 4m
and as    
 q    2e 
 m
m
,
while   
 q p
e
 m
 m
i.e.,      , so
 q    q p
r  rp
-particle while A to proton.
Question X: What is the ratio of radii of paths when an electron and a proton
enter at right angles to a uniform field with same (a) velocity (b)
 mp 
 1840.
momentum and (c) kinetic energy? Given that 
 me 
Answer: In case of circular motion of a charged particle in a uniform magnetic
field.
r
(a)
mv
p


qB
qB
2mK
qB
As in this situation v, B an q are same for the two particles, so
rp
re

mp
me
 1840
i.e., rp  1840 re Ans.
(b)
As in this situation p, B and q are same for the two particles, so
rp
re

1
1
1
i.e., rp = re Ans.
(c)
As in this situation kinetic energy K, B and q are same for the two
particle, so,
rp
re

mp
me
 1840
i.e., rp  43 re Ans.
Question XII: A beam of electrons passes undeflected through a region. What
possible conclusions could be drawn regarding the existence of an
electric and a magnetic fields?
Answer: If an electron passes undeflected through a region, the following are the
probabilities of existence of an electric or a magnetic or both fields:
(a)
No electric field is present and the particle is moving parallel to the
magnetic field. In this situation E  0 but B  0.
(c)
Both the fields are present and collinear and the particle is moving
parallel to them. (This situation is equivalent to (a) + (b))
In this situation E  0 and B  0.
(d)
Both the fields are present and perpendicular to each other and the
particle is moving perpendicular to both of them with Fe = Fm. In this
situation also E  0 and B  0.
NOTE:
If the velocity of a particle remains unchanged in passing through a
certain region, apart from E  0 and B  0 only situations (a), i.e.,
E  0 and B  0 and (d), i.e., E  0 and B  0 are possible, as in
other situations the speed and hence the velocity will not remain
constant.
Question XIII. If a charged particle is deflected either by an electric or a
magnetic field, how can we ascertain the nature of the field?
Answer: By observing the trajectory and measuring the kinetic energy of the
charged particle as in a magnetic field the trajectory is a circle in a plane
perpendicular to the field and the KE (and hence the speed) remains
constant while in an electric field the trajectory is a parabola in a plane
parallel to the field and the KE (and hence the speed) of the particle
changes.
Question XIV. A metallic block carrying current I is subjected to a uniform
magnetic induction B . What is the force F experienced by the charge
moving with speed v and as a result of motion of this charge which face
will acquire lower potential?
Answer: As the block is of metal, the charge carriers are electrons; so for current
along positive x-axis, the electrons are moving along negative x-axis, i.e.
v   v i.
and as the magnetic field is along the y-axis,

i.e., B  B j, so F  q v  B

for this case yields
F   e  v i  B j
i.e., F  evBk
as i  j  k  Ans.


As force on electrons is towards the face ABCD, the electrons will
accumulate on it and hence it will acquire lower potential*.
*If the charge carriers had been positive, the face ABCD will acquire
higher potential.
NOTE:
E.H. Hall in 1879 observed that when a current passes through a slab of
material in the presence of a transverse magnetic field, a small potential
difference is established in a direction perpendicular to both, the current
flow and the magnetic field. This effect is called Hall effect and the
voltage thus developed is called Hall voltage. Theory shows that
VH  RH
IB
d
with RH 
1
ne
Question XV. Can a magnetic field set a resting electron into motion?
Answer: Yes; e.g., if a bar magnet is moved rapidly past a stationary charge, the
charge will set into motion. this is because in this situation relative to the
magnet, the charge will be in motion (in a direction opposite to that of
the magnet) and hence it will experience a force perpendicular to the
plane containing the field and the motion in accordance with RHSR.
NOTE:
Actually in this situation the field is not steady, and a time-varying
magnetic field produces an electric field due to which the charge
experiences a force and sets in motion.
Problem 41. A uniform magnetic field with a slit system is to be used as a
momentum filter for high energy charged particles. With a field of B
-particle each of energy 5.3
MeV. The magnetic field is increased to 2.3B tesla and deuterons are
passed into the filter. What is that energy of each deuteron
transmitted by the filter?
Solution: In case of circular motion of a charged particle in a magnetic field, as
r=
mv

qB
2mK
qB
r2q2B2
,
i.e., as K =
2m
so according to the given problem,
r2 2e B2
K 
2  4m
2
r2  e 2.3B
and, KD =
2 2m
2
2
K
2.3  4
i.e., D 
K
22  2
2
or KD = 5.3 
5.29
2
= 14.02 MeV Ans.
Problem 42. A cyclotron is operating with a flux density of 3 Wb/m2. The ion
which enters the field is a proton having mass 1.67 × 10–27 kg. If the
maximum radius of the orbit of the particle is 0.5 m, find (a) the
maximum velocity of the proton, (b) the kinetic energy of the particle,
and (c) the period for a half cycle.
Solution:
(a)
As in case of motion of a charged particle in a magnetic field,
r=
mv
qB
i.e., v =
qBr
m
So, vmax =
qBrmax
m
So, vmax =
1.6  1019  3  0.5
1.67  1027
= 1.43 × 108 m/s Ans.
(b)
1
mv2
2
KE =
=
1
 1.67  1027  1.43  108
2


2
i.e., KE = 1.71 × 10–11 J
1.71  1011
 108 MeV Ans.
=
1.6  1019
(c)
In case of circular motion, as
T=
=
2 r
v
2  0.5
 2.18  108 s,
8
1.43  10
so time for half cycle,
t
=
1
T
2
1
2.18  108  1.09  108 s Ans.
2


Problem 43. A beam of charged particles, having kinetic energy 103 eV and
containing masses 8 × 10–27 kg and 1.6 × 10–26 kg, emerges from the end
of an accelerator tube. There is a plate at a distance 10 –2 m from the
end of the tube and placed perpendicular to the beam. Calculate the
magnitude of the smallest magnetic field which can prevent the beam
from striking the plate.
Solution: The motion of a charged particle in a transverse field is a circle. So with
reference to the particle will not hit the plate if
r<d
i.e.,
mv
d
qB

mv 
as
r


qB 

or,
2mK
d
qB
as p  2mK 


or, B >
2mK
qd
 2mK 
 will be maximum when m is maximum, so
 qd 
Now as 
B>
2mmaxK
qd
2  1.6  10 10
26
i.e., B >
or B >
3
 1.6  1019

q  102
2
 1.6  1019
q
Now as in this problem charge on the particles is not specified, assuming
q = ne with n = 1, 2, ….,
B>
2
n
so, Bmin 
2
T
n
 q
with n =    1, 2, 3.... Ans.
 e
-particle is accelerated by a potential difference of 104 V. Find
the change in its direction of motion if it enters normally in a region of
thickness 0.1 m having transverse magnetic induction of 0.1 T [given
-particle = 6.4 × 10–27 kg].
Solution:
-particle is accelerated by 104 volt, its kinetic energy will be
K = (2e) (104 V) = 2 × 104 eV.
Now the path of a charged particle when it enters a magnetic field at
right angles is a circle with radius
r=
mv

qB
2mK
qB
So here,
2  6.4  10
r=
27
 2  104  1.6  1019

1/2
2  1.6  1019  0.1
= 0.2 m
Now as in case of a circle angle between tangents at two points will be
equal to the angle between normals at those points. As in a circle
tangent is normal to radius at every point, the change in direction of the
particle as it passes the field,
 d
  sin1  
r
 0.1
 30 Ans.
0.2 
= sin1 

Problem 45. The region between x = 0 and x = L is field with uniform, steady
magnetic field B0k. A particle of mass m, positive charge q and
velocity v0 i travels along x-axis and enters the region of the magnetic
field. Neglect the gravity throughout the question.
(a)
Find the value of L if the particle emerges from the region of magnetic
field with its final velocity at an angle 30° to its initial velocity.
(b)
Find the final velocity of the particle and the time spent by it in the
magnetic field, if the magnetic field now extends up to 2.1 L.
Solution:
(a)
sin 30 =
L
r
where r =
mv
, radius of circular path followed by charged particle in
qB
transverse magnetic field.
=
(b)
mv0
2qB
The length of magnetic field
= 2.1 L
=
2.1 mv 2.1

r
2qB
2
which is greater than r. So, the particle will complete the semicircle in
the magnetic field as shown in figure and will emerge parallel to x-axis.
Thus,
Velocity on emerging from magnetic field =  v0i
Time spent in magnetic field =
r
v0
=
mv0
qB0v0
=
m
Ans.
qB0
Problem 46. A particle of mass m = 1.6 × 10–27 kg and charge q = 1.6 × 10–19
coulomb enters a region of uniform magnetic field of strength 1 tesla
along the direction. The speed of the particle is 107 m/s. (a) The
magnetic field is directed along the inward normal to the plane of the
paper. The particle leaves the region of the field at point F. Find the
distance EF and . (b) If the direction of the magnetic field is along the
outward normal t the plane of the paper, find the time spent by the
particle in the region of the magnetic field after entering it at E.
Solution:
(a)
When a charged particle moves perpendicular to a magnetic field the
path is a circle with magnetic force qvB directed radially towards the
centre of the path. So if we draw normals at E and F they will meet at O
which is the centre of the circle.
  2  1  45 Ans.
 as 
1
 2  
And as here
r=
mv
qB
1.6  1027  107
 0.1 m
=
1.6  1019  1
So, EF = (2 × 0.1 × sin 45) m
= 10 2 cm Ans.
(b)
Now if the direction of the field is out of the page the force will act as
shown in Fig. 13.87 (B) and so now O will be the centre and the angle

  
 3
subtended by the path at O will be 2        . Now as in a
 2   2

magnetic field the speed of a particle remains constant, so if t is the time
taken by the particle in the field,
 r 

v
i.e., t = 

 3
0.1    
 2
So, t =
107
-8
= 4.71 × 10–8 s Ans.
Problem 47. A beam of protons with a velocity 4 × 105 m/s enters a uniform
magnetic field of 0.3 tesla at an angle of 60° to the magnetic field. Find
the radius of the helical path taken by the proton beam. Also find the
pitch of the helix (which is the distance travelled by a proton in the
beam parallel to the magnetic field during one period of rotation).
Mass of the proton = 1.67 × 10–27 kg.
Solution: When a charged particle is projected at an angle  to a magnetic field,
while
radius.
r=
m  v sin 
qB
 3
1.67  1027  4  105  

 2 
=
1.6  1019  0.3
=
2
 102 m
3
= 1.2 cm Ans.
And as
T=
2r
v sin 
=
2  1.2  102
 3
4  105  

 2 
= 2.175 × 1010–7 s
So pitch
5
= 4  10 
1
 2.175  107
2
i.e., p = 4.35 × 10–2 m
= 4.35 cm Ans.
Problem 48. An electron gun G emits electrons of energy 2 keV travelling in the
positive x-direction. The electrons are required to hit the spot S where
GS = 0.1 m and line GS makes an angle of 60° with the x-axis as shown
in Fig. 13.89. A uniform magnetic field B parallel to GS exists in the
region outside the electron gun. Find the minimum value of B needed
to make the electron hit S.
Solution: Resolving the velocity along the perpendicular to the magnetic field B
we find that the particle will travel al
the same time it will move in a circle, i.e., the path of the particle will be
helical with
r=
mv sin 
qB
So, T =
2r
v sin 
 m
 qB 
= 2 
and p = v cos   T
=
2m
 v cos 
qB
So the particle will hit S if
GS = np
=n
2m
v cos 
qB
where n = 1, 2, 3, …
i.e., B = n
2m
v cos 
q GS
And for B to be minimum, n = min = 1
i.e., Bmin =
=
2 cos 
mv
q GS
2 cos 
2mK

q
GS
as mv  p  2mK 


 1
2   
 2
i.e., Bmin =
0.1
2  9.1  1031  2  103  1.6  1019
1.6  1019
Bmin =
10  8  3
 104
16
= 15  104
= 4.71  103 T Ans.
Problem 49. An electron beam passes through a magnetic field of 2 × 10 –3
Wb/m2 and an electric field of 3.4 × 104 V/m, both acting
simultaneously. If the path of electrons remains undeflected, calculate
the speed of the electrons. If the electric field is removed, what will be
the radius of the electron path [me = 9.1 × 10–31 kg]?
Solution: An electron in an electric field experiences a force eE opposite to the
field; so it will pass undeflected only if the magnetic field exerts a force
which is equal in magnitude and opposite in direction of the that exerted
by the electric field, i.e., E, B and v are mutually perpendicular
In this situation,
F e = Fm,
i.e., eE = evB
E 3.4  104

so, v =
B
2  103
= 1.7 × 107 m/s Ans.
Now when the electric field is switched off, the particle in the magnetic
field will describe a circle with radius
r=
=
mv
qB
9.1  1031  1.7  107
1.6  1019  2  103
= 4.83 × 10–2 m
= 4.83 cm Ans.
Problem 50. A particle of mass 1 × 10–26 kg and charge + 1.6 × 10–19 coulomb
travelling with a velocity 1.26 × 106 m/s in +x direction enters a region
in which a uniform electric field E and a uniform magnetic field B are
present such that Ex = Ey = 0, Ez = – 102.4 kV/m and Bx = Bz = 0, By = 8 ×
10–2 Wb/m2. The particle enters this region at the origin at time t = 0.
(a) Determine the location (x, y and z co-ordinates) of the particle at t =
5 × 10–6 s. (b) If the electric field is switched off at this instant with the
magnetic field still present, what will be the position o the particle at t
= 7.45 × 10–6 s?
Solution:
(a)
According to the given problem the electric field is along negative z-axis
while the magnetic field along +y-axis and the particle having positive
charge is moving along the x-axis.,
So the electric force,
Fe  qE
 
= 1.6  1019  102.4  103  k
while the magnetic force

FB  q v  B


= 1.6 × 10–19 × 1.28 × 106 × 8 × 10–2 k

i.e., FB  1.6  1019  102.4  103 k
So that,
F  Fe  Fm  0
and as resultant force is zero,
a=
F
0
m
And hence the particle will move along +x-axis with constant velocity
v  1.28  106 i m / s. So the distance travelled by the particle in
time 5 × 10–6 s,
x0 = vt
= (1.28 × 106) × (5 × 10–6)
= 6.40 m
and hence the position of the particle at t = 5 × 10–6 s will be [6.4, 0, 0]
m. Ans.
(b)
Now when the electric field is switched off, the particle will describe a
circle in x-z plane with radius
r=
mv
qB
1  1026  1.28  106
 1m
=
1.6  1019  8  102
With reference to Fig. 13.91, the position of the particle
x = x0
y=0
and z = r(1 –
 ut 
 , so
r 
i.e.,   

1.28  106 7.45  5  106

1
= 1.28 × 2.45 = 3.136
 rad
= 6.4 m
and z = 1(1 –
So the co-ordinates of the point are (6.4, 0, 2m) Ans.
Problem 51. There is a constant homogeneous electric field of 100 V m–1 within
the region x = 0 and x = 0.167 m pointing in the positive x-direction.
There is a constant homogeneous magnetic field B within the region x =
0.167 m and x = 0.334 m pointing in the z-direction. A proton at rest at
the origin (x = 0, y = 0) is released in a positive x-direction. Find the
minimum strength of the magnetic field B, so that the proton is
detected back at x = 0, y = 0.167 m (mass of the proton = 1.67 × 10 –27
kg).
Solution: The situation described in the problem.
As electric field is along x-axis, so proton will be accelerated by the
electric field and will enter the magnetic field at A (i.e., x = 0.167, y = 0)
with velocity v along x-axis such that
1
mv2  W  Fd  qEd
2
1/2
 2qEd 
i.e., v = 

 m 
1/2
 2  1.6  1019  100  0.167 
=

1.67  1027


= 4 2  104
m
2
Now as proton is moving perpendicular to magnetic field so it will
describe a circular path in the magnetic field with radius r such that
r=
mv
qB
And as it comes back at C[x = 0, y = 0.167 m] its path in the magnetic
field will be semicircle such that
y = 2r =
2mv
qB
i.e., B =
2mv
qy
2  1.67  1027  4 2  104
i.e., B =
1.6  1019  0.167
=
1
 102  7.07 mT Ans.
2
NOTE:
To solve the problem properly here we have assumed that B is along
negative z-axis (and not along z-axis as given in the problem). If B is along
z-axis the proton will never reach the specified position of the detector.
Problem 52. A particle of mass m and charge q is moving in a region where
uniform, constant electric and magnetic fields E and B are present.
E and B are parallel to each other. At time t = 0 the velocity v0 of
the particle is perpendicular to E. (Assume that, its speed is always <<
c, the speed of light in vacuum.) Find the velocity v of the particle at
time t. You must express your answer in terms of t, q, m, the vectors
v0 , E and B, are their magnitudes v0. E and B.
Solution: As force due to parallel electric and magnetic fields on a charged
particle moving perpendicular to the fields will be at right angles to each
other [electric force being along the direction of E while magnetic
perpendicular to the plane containing v and B ] so magnetic force will
not affect the motion of charged particle in the direction of electric field
and vice-versa. So the problem is equivalent to superposition of two
independent motions.
So for motion of particle under electric field along,
ay 
i.e.,
qE
,
m
dvy
dt

qE
m
F


as
a

and F  qE

m


vy
or

0
t
dvy 
qE
0 m dt,
i.e., vy =
qE
t
m
…(1)
While at the same time the charged particle under the action of
magnetic field will describe a circle in the x-z plane with
r=
mv0
,
qB
i.e.,  
v0 qB

r
m
So the angular position of the particle at time t in the x-z plane will be
given by
  t 
qB
t
m
and hence in accordance with,
vx = v0
 qB 
t ….(2)
 m 
= v0 cos t  v0 cos 
and vz = v0
 qB 
t
 m 
= v0 sin t  v0 sin 
…(3)
So in the light of equations (1), (2) and (3)
vx = i v x  j v y  k v z
 qB 
 qE 
 qB 
t  j 
t  k v0 sin 
t
m 
m 
 m 
= i v0 cos 
But as here
i 
v0
,
v0
j
E B

E B
and k 
v0  B
v0B
 v0 
 qB 
t 
 v0 cos 
v
m


 0
So, V  
 E  qE
t
 
E
m
 
v  B
 qB 
 
t  Ans.
 v0 sin 
v
B
m


 0 
Problem 53. A uniform, constant magnetic field B is directed at an angle of 45°
to the x-axis in thx xy-plane. PQRS is a rigid, square wire frame carrying
a steady current I0, with its centre at the origin O. At time t = 0, the
frame is at rest in the position shown in the figure, with its sides
parallel to the x and y axes. Each side of the frame is of mass M and
length L.
(a)
What is the torque  about O acting on the frame due to the magnetic
field?
(b)
Find the angle by which the frame rotates under the action of this
torque in a short interval of time, t, and the axis about which this
rotation occurs. ( t is so short that any variation in the torque during
this interval may be neglected.)
Given: the moment of inertia of the frame about an axis through its
entre perpendicular to its plane is
4 2
ML .
3
Solution:
(a)
As magnetic field B is in x-y plane and subtends an angle of 45° with xaxis
Bx = B cos 45
 B 

2
=

and By = B cos 45
 B 

2
=

So in vector form
 B 
 B 
B  i
 j

 2
 2 
and M  I0Sk
= I0L2 k
So,   M  B
B 
 B
i
j
2
2 
= I0L2 k  

i.e.,  
I0L2B
 j  i Ans.

2 
i.e., torque has magnitude I0L2B and is directed along line QS from Q to S.
(b)
As by theorem of perpendicular axes, moment of inertia of the frame
about QS,
IQS =
=
14
2 2
2
ML

 3 ML
2  3

=
1
I
2 z

I
I0L2B  3 3 I0B

2ML2
2 M
from   0t 

1 2
t with 0  0 we have
2
1 2
t
2
=
1  3 I0B 
2
t 



2 2 M 
=
3 I0B
t2 Ans.
4 M
Problem 54. A wire loop carrying a current I is placed in the x-y plane (a) If a
particle with charge q and mass m is placed at the centre P and given a
velocity v along NP find its instantaneous acceleration. (b) If an
external uniform magnetic induction B  Bi is applied, find the force
and torque acting on the loop.
Solution:
(a)
As the case of current-carrying straight conductor and arc, the
magnitude of B is given by
BW =
0 I
sin   sin 
4 d
and BA =
0 I
4 r
So in accordance with RHSR,
B 
W
0
4

P
 
I
 2 sin 60  k
a cos 60
 
and BA
P


0 I  2 
 k
4 a  3 
and hence net B at P due to the given loop
B
 
0 2I 

3

k
4 a 
3 
…(1)
Now as force on a charged particle in a magnetic field is given by

F  q v B

So here
F  qvB sin 90 along PF
i.e., F 
0 2qvI 

3   along PF

4 a 
3
also so
a
F
m
= 107
2qvI 

3   along PF Ans.

ma 
3
(b)
As dF  IdL  B,
so F 
 I dL  B
As here I and B are constant,
F  I   dL   B  0
as

 dL  0 Ans.
Further as area of coil,
1
1

S    a2  .2a sin 60  a cos 60 k
2
3


= a2 
3

3
k
4 
So, M  IS

= Ia2 
3

3
k
4 
and hence   M  B

2
= Ia B 
3

3
 k i
4 


2
i.e.,   Ia B 
3


3
 j Nm
4 
as k  i  j Ans.