NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

NCERT
SOLUTIONS
CLASS - 12th

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Class: 12th
Subject : Physics
Chapter : 4
Chapter Name : Moving Charges And Magnetism

Q4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40
A. What is the magnitude of the magnetic eld B at the centre of the coil?

Answer. Number of turns on the circular coil, n = 100 Radius of each turn, r = 8.0 cm = 0.08 m
Current owing in the coil, I = 0.4 A Magnitude of the magnetic eld at the centre of the coil is
given by the relation,
μ0 2πnI
|B| =
4π r

Where,
μ = = Permeability of free space
0
−7
4π×10 2π×100×0.4
|B| = ×
4π 0.08
−4
= 3.14 × 10 T

Hence, the magnitude of the magnetic eld is 3.14 × 10 −4
T

Page : 169 , Block Name : Exercise

Q4.2 A long straight wire carries a current of 35 A. What is the magnitude of the eld B at a point
20 cm from the wire?

Answer. Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m Magnitude of the magnetic eld at this point is
given as:
μ0 2I
=
4π r

Where,
μ = Permeability of free space = = 4n × 10
−7 −1
0 TmA
−7
4π×10 ×2×35
B =
4π×0.2
−5
= 3.5 × 10 T

Hence, the magnitude of the magnetic eld at a point 20 cm from the wire is 3.5 × 10 −5
T

Page : 169 , Block Name : Exercise

Q4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south
direction. Give the magnitude and direction of B at a point 2.5 m east of the wire?

Answer. Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
Magnitude of the distance of the point from the wire, r = 2.5 m.
μ0 2I
Magnitude of the magnetic eld at that point is given by the relation, B =
4πr

Where,
μ = Permeability of free space = 4π × 10
−7 −1
0 TmA
−7
4π×10 ×2×50
B =
4π×2.5
−6
= 4 × 10 T

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current
in the wire is vertically downward. Hence, according to the Maxwell's right hand thumb rule, the
direction of the magnetic eld at the given point is vertically upward.

Page : 169 , Block Name : Exercise

Q4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is
the magnitude and direction of the magnetic eld due to the current 1.5 m below the line ?

Answer. Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m Hence, magnetic eld at that point is
given by the relation,
μ0 2I
B =
4πr

Where,
=Permeability of free space = 4π × 10 −7 −1
μ0 TmA
−7
4π×10 ×2×90 −5
B = = 1.2 × 10 T
4π×1.5

The current is owing from East to West. The point is below the power line. Hence, according to
Maxwell's right hand thumb rule, the direction of the magnetic eld is towards the South.

Page : 169 , Block Name : Exercise

Q4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A
and making an angle of 30º with the direction of a uniform magnetic eld of 0.15 T?

Answer. Current in the wire, I = 8 A
Magnitude of the uniform magnetic eld, B = 0.15 T
Angle between the wire and magnetic eld, θ = 30 ∘

Magnetic force per unit length on the wire is given as
f = BI sin θ
∘
= 0.15 × 8 × 1 × sin 30
−1
= 0.6Nm

Hence, the magnetic force per unit length on the wire is 0.6Nm −1

Page : 169 , Block Name : Exercise

Q4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis.
The magnetic eld inside the solenoid is given to be 0.27 T. What is the magnetic force on the
wire?

Answer. Length of the wire, I = 3 cm = 0.03 m
Current owing in the wire, I = 10 A
Magnetic eld, B = 0.27 T
Angle between the current and magnetic eld, θ = 90°
Magnetic force exerted on the wire is given as: F = BI/sin θ
= 0.27 x 10 x 0.03 sin 90°
−2
8.1 × 10 N

Hence, the magnetic force on the wire is 8.1 × 10 N .−2

The direction of the force can be obtained from Fleming's left hand rule.

Page : 169 , Block Name : Exercise

Q4.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same
direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer. Current owing in wire A, I = 8.0A A

Current owing in wire B, I = 5.0 B

A Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, I = 10 cm = 0.1 m
Force exerted on length / due to the magnetic eld is given as
μ0 2IA IB l
B =
4πr

Where,
μ = Permeability of free space = 4n × 10
−7 −1
0 TmA
−7
4π×10 ×2×8×5×0.1
B =
4π×0.04
−5
= 2 × 10 N

The magnitude of force is 2 × 10 N −5

This is an attractive force normal to A towards B because the direction of the currents in the wires
is the same.

Page : 169 , Block Name : Exercise

Q4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The
diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B
inside the solenoid near its centre.

Answer. Length of the solenoid, I = 80 cm = 0.8 m
There are ve layers of windings of 400 turns each on the solenoid.
Total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A

Magnitude of the magnetic eld inside the solenoid near its centre is given by the relation,
μ0 N I
B =
l

Where,
μ = Permeability of free space = 4Π × 10
−7 −1
0 TmA
−7
4π×10 ×2000×8
B =
0.8
−3 −2
= 8π × 10 = 2.512 × 10 T

Hence, the magnitude of the magnetic eld inside the solenoid near its centre is 2.512 × 10 −2
T

Page : 169 , Block Name : Exercise

Q4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is
suspended vertically and the normal to the plane of the coil makes an angle of 30º with the
direction of a uniform horizontal magnetic eld of magnitude 0.80 T. What is the magnitude of
torque experienced by the coil?

Answer. Length of a side of the square coil, 1 = 10 cm = 0.1 m
Current owing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic eld,θ = 30 ∘

Strength of magnetic eld, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic eld is given by the
relation, T = nBI A sin θ
Where,
A = Area of the square coil
2
⇒ / × I = 0.1 × 0.1 = 0.01m
∘
∴ T = 20 × 0.8 × 12 × 0.01 × sin 30

= 0.96Nm

Hence, the magnitude of the torque experienced by the coil is 0.96 Nm.

Page : 169 , Block Name : Exercise

Q4.10 Two moving coil meters, M and M have the following particulars:
1 2

R1 = 10Ω
−3 2
A1 = 3.6 × 10 m B1 = 0.25T

R2 = 14Ω N2 = 42
−3 2
A2 = 1.8 × 10 m B2 = 0.50T

(The spring constants are identical for the two meters).
Determine the ratio of
(a) current sensitivity and
(b) voltage sensitivity of M and M . 2 1

Answer. For moving coil meter M . 1

Resistance R = 10Ω 1

Number of turns N = 30 1

Area of cross section A = 3.6 × 10 1
−3
m
2

Magnetic eld Strength B = 0.25T 1

Spring Constant K = K 1

For moving coil meter M 2

Resistance R = 14Ω 2

Number of turns N = 42 2

Area of cross section A = 1.8 × 10 2
−3
m
2

Magnetic eld Strength B = 0.50T 2

Spring Constant K = K 2

(a) Current sensitivity of M is given as 1
N1 B1 A1
In =
K1

And current sensitivity of M is given as 2
N2 B2 A2
Vs1 =
K1

Vs1 N1 B1 A1 K1
=
Vsl K12 N1 B1 A1
−3
42×0.5×1.8×10 ×10×K
= −3
= 1
K×30×0.25×3.6×10

Hence ,the ratio of current sensitivity of M to M is 1.4 2 1

(b) Voltage sensitivity for M is given as : 2

N2 B2 A2
Vs2 =
K2 R2

And voltage sensitivity of is given as :

N1 B1 A1
Vs1 =
K1

Vs2 N2 B2 A2 K1 R1
=
Vs 1 K2 R2 N1 B1 A1

−3
42×0.5×1.8×10 ×10×K
= = 1
−3
K×14×30×0.25×3.6×10

Hence ,the ratio of current sensitivity of M to M is 1 2 1

Page : 169 , Block Name : Exercise

Q4.11 In a chamber, a uniform magnetic eld of 6.5G (1G = 10 is maintained. An electron
−4
T)

is shot into the eld with a speed of 4.8 × 10 ms normal to the eld. Explain why the path of
6 −1

the electron is a circle. Determine the radius of the circular orbit. e =
−19 −31
1.6 × 10 C, me = 9.1 × 10 kg)

Answer. Magnetic eld strength, B = 6.5G = 6.5 × 10 −4
T

Speed of the electron, v = 4.8 × 10 m/s
6

Charge on the electron, e = 1.6 × 10 C −19

Mass of the electron, m = 9.1 × 10 kg
−31
e

Angle between the shot electron and magnetic eld,θ = 90
∘

Magnetic force exerted on the electron in the magnetic eld is given as: F = evB sin θ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a
circular path of radius r.

Hence, centripetal force exerted on the electron,
2
mv
Fc =
r

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
Fc = F
2
mv
= evB sin θ
r
mv
r =
Be sin θ
−31 6
9.1×10 ×4.8×10
=
−4 −19 ∘
6.5×10 ×1.6×10 ×sin 90
−2
= 4.2 × 10 m = 4.2cm

Hence, the radius of the circular orbit of the electron is 4.2 cm.

Page : 169 , Block Name : Exercise

Q4.12 Obtain the frequency of revolution of the electron in its circular orbit. Does the answer
depend on the speed of the electron? Explain.

Answer. Magnetic eld strength, B = 6.5 × 10 −4
T

Charge of the electron, e = 1.6 × 10 C −19

Mass of the electron, m = 9.1 × 10 kg e
−31

Velocity of the electron, v = 4.8 × 10 m/s 6

Radius of the orbit, r = 4.2cm = 0.042m
Frequency of revolution of the electron = V
Angular frequency of the electron = ω = 2πv
Velocity of the electron is related to the angular frequency as:
v = rω

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence,
we can write:
2
mv
evB =
r
m m
eB = (rω) = (r2πv)
r r
Be
v =
2πm

This expression for frequency is independent of the speed of the electron. On substituting the
known values in this expression, we get the frequency as:
−4 −19
6.5×10 ×1.6×10
v = −31
2×3.14×9.1×10
6
= 18.2 × 10 Hz

≈ 18MHz

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the
electron.

Page : 169 , Block Name : Exercise

Q4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended
vertically in a uniform horizontal magnetic eld of magnitude 1.0 T. The eld lines make an angle
of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be
applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some
irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer. (a) Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil = πr = π(0.08) = 0.0201m
2 2 2

Current owing in the coil, I = 6.0 A
Magnetic eld strength, B = 1T
Angle between the eld lines and normal with the coil surface,θ = 60 ∘

The coil experiences a torque in the magnetic eld. Hence, it turns. The counter torque applied to
prevent the coil from turning is given by the relation,
T = nI BA sin θ … (i)
∘
= 30 × 6 × 1 × 0.0201 × sin 60

= 3.133Nm

(b) It can be inferred from relation (i) that the magnitude of the applied torque is not dependent
on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if
the circular coil in the above case is replaced by a planar coil of same area.

Page : 169 , Block Name : Exercise

Q4.14 Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same
vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of
16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is
anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the
magnitude and direction of the net magnetic eld due to the coils at their centre.

Answer. Radius of coil X, r1 = 16 cm = 0.16 m
Radius of coil Y, r2 = 10 cm = 0.1 m
Number of turns of on coil X, n1 = 20
Number of turns of on coil Y, n2 = 25
Current in coil X, 11 = 16 A
Current in coil Y, 12 = 18 A
Magnetic eld due to coil X at their centre is given by the relation,
μ0 n1 I1
B1 =
2r1

where,
μ = Permeability of free space = 4π × 10
−7 −1
0 TmA

−7
4π×10 ×20×16
∴ B1 =
2×0.16

= 4π × 10
−4
T (towards East)

Magnetic eld due to coil Y at their centre is given by the relation,
μ0 n2 I2
B2 =
2r2
−7
4π×10 ×25×18
=
2×0.10

= 9π × 10
−4
T T (towards West)

Hence, net magnetic eld can be obtained as:

B = B2 − B1
−4 −4
= 9π × 10 − 4π × 10
−4
= 5π × 10 T

= 1.57 × 10
−3
T (towards West)

Page : 170 , Block Name : Additional Exercises

Q4.15 A magnetic eld of 100 G v is required which is uniform in a region of linear dimension
about 10 cm and area of cross-section about 10 m . The maximum current-carrying capacity of
−3 2

a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core
is at most 1000 turns m Suggest some appropriate design particulars of a solenoid for the
−1

required purpose. Assume the core is not ferromagnetic.

Answer. Magnetic eld strength, B = 100 G = 100 × 10 −4
T

Number of turns per unit length, n = 1000 turns m −1

Current owing in the coil, I = 15 A
Permeability of free space, μ = 4π × 10 TmA
−7 −1
0

Magnetic eld is given by the relation,
B = μ0 nI
B
∴ nI =
μ0
−4
100×10
= = 7957.74
−7
4π×10

≈ 8000Am

If the length of the coil is taken as 50cm ,radius 4cm ,number of turns 400 and current 10A ,then
these values are not unique for the given purpose . There is always a possibility of some
adjustments with limits

Page : 170 , Block Name : Additional Exercises

Q4.16 For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic
eld at a point on its axis at a distance x from its centre is given by,
2
μ0 I R N
B =
3

2(x2 +R2 ) 2

(a) Show that this reduces to the familiar result for eld at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying
equal currents in the same direction, and separated by a distance R. Show that the eld on the axis
around the mid-point between the coils is uniform over a distance that is small as compared to R,
and is given by,
μ0 BN I
B = 0.72 −
R
approximately.

[Such an arrangement to produce a nearly uniform magnetic eld over a small region is known as
Helmholtz coils.]

Answer. Radius of circular coil = R
Number of tums on the coil = N
Current in the coil = I
Magnetic eld at a point on its axis at distance x is given by the relation,
2
μ0 I R N
B =
3

2(x2 +R2 ) 2

Where
μ = Permeability Of free space
0

(a) If the magnetic eld at the centre of the coil is considered, then

2
μ0 I R N μ0 I N
∴ B = 3
=
2R 2R

This is the familiar result for magnetic eld at the centre of the coil.

(b) Radii of two parallel co-axial circular coils R
Number Of turns on each coil = N
Current in both coils = I
Distance between both the coils =R
Let us consider point Q at distance d from the centre.
Then, one coil is at a distance of + d from point Q.
R

2

Magnetic eld at point Q is given as:
2
μ0 N I R
B2 =
3
2
R 2
2
2[( −d) +R ]
2

Total magnetic eld,
B = B1 + B2
3 3

2 2 2 2 2
μ0 I R R R
2 2
= [{( − d) + R } + {( + d) + R } ]
2 2 2

3 3
2 2
− 2
−
μ0 I R 5R 2
2
5R 2
2

= [( + d − Rd) + ( + d + Rd) ]
2 4 4

3 3 3
2
μ0 /R 2 2 2 2 2 2
5R 4 d 4 d 4 d 4 d
= × ( ) [(1 + 2
− ) + (1 + 2
+ ) ]
2 4 5 R 5 R 5 R 5 R

For d << R neglecting the factor
2
d

R2

we get
3 3 3
2 2
μ0 I R 5R 2 4d 2 4d 2

= × ( ) × [(1 − ) + (1 + ) ]
2 4 5R 5R

3
2
μ0 I R N 4 2
6d 6d
≈ × ( ) [1 − + 1 + ]
3 5 5R 5R
2R

3

4 2 μ0 I N μ0 I N
B = ( ) = 0.72 ( )
5 R R

Hence, it is proved that the eld on the axis around the mid-point between the coils is uniform.

Page : 170 , Block Name : Additional Exercises

Q4.17 A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm,
around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the
magnetic eld
(a) outside the toroid,
(b) inside the core of the toroid, and
(c) in the empty space surrounded by the toroid.

Answer. Inner radius of the toroid ,r = 25cm = 0.25m 1

Outer radius of the toroid ,r = 26cm = 0.26m 2

Number of turns on the coil, N = 3500
Current in the coil, I = 11 A

(a) Magnetic eld outside a toroid is zero. It is non-zero only inside the core of a toroid.

(b) Magnetic eld inside the core of a toroid is given by the relation,
μ0 N I
B =
l

Where
μ = permeability of free space = 4π × 10
−7 −1
0 TmA

l= length of toroid =
r1 +r2
= 2π [ ]
2

= π(0.25 + 0.26)

= 0.51π
−7
4π×10 ×3500×11
∴ B =
0.51π
−2
≈ 3.0 × 10 T

(c) Magnetic eld in the empty space surrounded by the toroid is zero.

Page : 170 , Block Name : Additional Exercises

Q4.18 Answer the following questions:

(a) A magnetic eld that varies in magnitude from point to point but has a constant direction (east
to west) is set up in a chamber. A charged particle enters the chamber and travels unde ected
along a straight path with constant speed. What can you say about the initial velocity of the
particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic eld varying
from point to point both in magnitude and direction, and comes out of it following a complicated
trajectory. Would its nal speed equal the initial speed if it suffered no collisions with the
environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic eld in
north to south direction. Specify the direction in which a uniform magnetic eld should be set up
to prevent the electron from de ecting from its straight line path.

Answer. (a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic
eld. Hence, it travels along a straight path without suffering any de ection in the eld.

(b) Yes, the nal speed of the charged particle will be equal to its initial speed. This is because
magnetic force can change the direction of velocity, but not its magnitude.

(c) An electron travelling from West to East enters a chamber having a uniform electrostatic eld
in the North-South direction. This moving electron can remain unde ected if the electric force
acting on it is equal and opposite of magnetic eld. Magnetic force is directed towards the South.
According to Fleming's left hand rule, magnetic eld should be applied in a vertically downward
direction.

Page : 170 , Block Name : Additional Exercises

Q4.19 An electron emitted by a heated cathode and accelerated through a potential difference of
2.0 kV, enters a region with uniform magnetic eld of 0.15 T. Determine the trajectory of the
electron if the eld
(a) is transverse to its initial velocity,
(b) makes an angle of 30º with the initial velocity.

Answer. Magnetic eld strength, B=0.15 T
Charge on the electron, e = 1.6 × 10 C −19

Mass of the electron, m = 9.1 × 10 kg −31

Potential difference, V = 2.0kV = 2 × 10 V
3

Thus, kinetic energy of the electron =ev
1 2
⇒ eV = mv
2

2eV
v = √
m

where, V = velocity of the electron

(a) Magnetic force on the electron provides the required centripetal force of the electron Hence,
the electron traces a circular path of radius r
Magnetic force on the electron is given by the relation b ev
Centripetal force =
2
mv

r
2
mv
∴ Bev =
r
mv
r =
Be

From equations above we get
1

m 2eV 2

r = [ ]
Be m

1
−31 −19 3 2
9.1×10 2×1.6×10 ×2×10
= −19
× ( −31
)
0.15×1.6×10 9.1×10

−5
= 100.55 × 10
−3
= 1.01 × 10 m

= 1mm

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic Tel.

(b) When the eld makes an angle θ of 30° with initial velocity, the initial velocity will
Be v = v sin θ
1

From above we can write the expression for new radius as
mv1
r1 =
Be
mv sin θ
=
Be
1
−31 −19 3 2
9.1×10 2×1.6×10 ×2×10 ∘
−19
× [ −31
] × sin 30
0.15×1.6×10 9×10
1
−31 −19 3
9.1×10 2×1.6×10 ×2×10 2
∘
= × [ ] × sin 30
−19 −31
0.15×1.6×10 9×10

−3
= 0.5 × 10 m

= 0.5mm

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic eld direction

Page : 171 , Block Name : Additional Exercises

Q4.20 A magnetic eld set up using Helmholtz coils (described in Exercise ) is uniform in a small
region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic eld is
maintained in a direction normal to the common axis of the coils. A narrow beam of (single
species) charged particles all accelerated through 15 kV enters this region in a direction
perpendicular to both the axis of the coils and the electrostatic eld. If the beam remains
unde ected when the electrostatic eld is 9.0 × 10 Vm , make a simple guess as to what the −5 −1

beam contains. Why is the answer not unique?

Answer. Magnetic eld, B = 0.75 T
Accelerating voltage, V = 15kV = 15 × 10 V 3

Electrostatic eld, E = 9 × 10 Vm 5 −1

Mass of the electron = m
Charge of the electron = e
Velocity of the electron = V
Kinetic energy of the electron = eV
1 2
mv = eV
2

---1
2
e v
∴ =
m 2V

Since the particle remains unde ected by electric and magnetic elds, we can infer that the
electric eld is balancing the magnetic eld.

.eE = evB
v= E/B ...(2)
Putting equation (2) in equation (1), we get
2
E
( )
B 2
e 1 E
= =
m 2 V 2V B
2

2
3
(9.0×10 )
7
= 2
= 4.8 × 10 C/kg
2×15000×(0.75)

This value of speci c charge e/m is equal to the value of deuteron or deuterium ions. This is not a
unique answer. Other possible answers are He . L etc. ++ ++

Page : 171 , Block Name : Additional Exercises

Q4.21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two
vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic eld should be set up normal to the conductor in order that the tension in the
wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the
magnetic eld same as before? (Ignore the mass of the wires.) g=9.8 s −2

Answer. Length of the rod, I = 0.45 m
Mass suspended by the wires, m = 60g = 60 × 10 −3
kg

Acceleration due to gravity, g = 9.8m/s 2

Current in the rod owing through the wire, I = 5 A

(a) Magnetic eld (B) is equal and opposite to the weight of the wire i.e.,
BH = mg
mg
∴ B =
H
−3
60×10 ×9.8
= = 0.26T
5×0.45

A horizontal magnetic eld of 0.26 T normal to the length of the conductor should be set
−3
= 0.26 × 5 × 0.45 + (60 × 10 ) × 9.8

= 1.176N

(b) If the direction of the current is revered, then the force due to magnetic eld and the weight of
the wire acts in a vertically downward direction.
∴ Total tension in the wire = BIl + mg

−3
= 0.26 × 5 × 0.45 + (60 × 10 ) × 9.8

= 1.176N

Page : 171 , Block Name : Additional Exercises

Q4.22 The wires which connect the battery of an automobile to its starting motor carry a current
of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm
long and 1.5 cm apart? Is the force attractive or repulsive?

Answer. Current in both wires, I = 300 A
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, 1 = 70 cm = 0.7 m
Force between the two wires is given by the relation,
2
μ0 I
F =
2πr

Where,
Permeability of free space = 4π × 10 −7 −1
μ0 = TmA
−7 2
4π × 10 × (300)
∴ F =
2π × 0.015

= 1.2N/m

Since the direction of the current in the wires is opposite, a repulsive force exists between them.

Page : 171 , Block Name : Additional Exercises

Q4.23 A uniform magnetic eld of 1.5 T exists in a cylindrical region of radius 10.0 cm, its
direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to
south direction passes through this region. What is the magnitude and direction of the force on
the wire if,

(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance
of 6.0 cm?

Answer. Magnetic Field Strength ,B = 1.5T
Radius of cylindrical region ,r=10cm =0.1 m
Current in the wire passing through the cylindrical region, I= 7A

(a) If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical
region. Thus, I = 2r = 0.2 m
Angle between magnetic eld and current, θ = 90 ∘

Magnetic force acting on the wire is given by the relation,
F = BI / sin θ
∘
= 1.5 × 7 × 0.2 × sin 90

= 2.1N

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

(b) New length of the wire after turning it to the Northeast-Northwest direction can be given as: :
l
l1 =
sin θ

Angle between magnetic eld and current, θ = 45 ∘

Force on the wire, F = BI / sin θ 1

= BH

= 1.5 × 7 × 0.2

= 2.1N

Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle θ
because | sin θ is xed.

(c) The wire is lowered from the axis by distance, d = 6.0 cm
Let l be the new length of the wire.
2

2
l2
∴ ( ) = 4(d + r)
2

= 4(10 + 6) = 4(16)

∴ l2 = 8 × 2 = 16cm = 0.16m

Magnetic force exerted on the wire,
F2 = BI l2

= 1.5 × 7 × 0.16

= 1.68N

Hence, a force of 1.68 N acts in a vertically downward direction on the wire.

Page : 171 , Block Name : Additional Exercises

Q4.24 A uniform magnetic eld of 3000 G is established along the positive z-direction. A
rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop
in the different cases shown in Fig. What is the force on each case? Which case corresponds to
stable equilibrium?

Answer. Magnetic Field strength B = 3000G = 3000 × 10 −4
T = 0.3T

Length of the rectangular loop , b=5cm
Area of the loop,
2 −4 2
A = I × b = 10 × 5 = 50cm = 50 × 10 m

Current in the loop, I = 12 A
Now, taking the anti-clockwise direction of the current as positive and vise-versa:

→×B
(a) Torque,τ→ = I A →
From the given gure, it can be observed that A is normal to the y-z plane and B is directed along
the z-axis.
−4 ^ ^
∴ τ = 12 × (50 × 10 ) i × 0.3k
−2 ^
= −1.8 × 10 jNm

The torque is along the negative y-direction. The force on the loop is zero because the angle
between A and B is zero.

(b) This case is similar to case (a). Hence, the answer is the same as (a).

(c) Torque 1.8 × 10 Nm From the given gure, it can be observed that A is normal to the x-z
−2

plane and B is directed along the z-axis.
−4 ^ ^
∴ τ = −12 × (50 × 10 ) j × 0.3k
−2
= −1.8 × 10 iNm

The torque is 1.8 × 10 −2
Nm along the negative x direction and the force is zero.

(d) Magnitude of torque is given as:
|τ | = I AB
−4
= 12 × 50 × 10 × 0.3
−2
= 1.8 × 10 Nm

Torque is 1.8 × 10 −2
Nm at an angle of 240° with positive x direction. The force is zero.

−4 ^ ^
= (50 × 10 × 12) k × 0.3k

= 0

(e) Forque 1.8 × 10 −2
Nm
−4 ^ × 0.3k
^
= (50 × 10 × 12) k

= 0

Hence, the torque is zero. The force is also zero.

→×B
(f) Torque τ = I A →
−4 ^ ^
= (50 × 10 × 12) k × 0.3k

= 0

→ and B
Hence, the torque is zero. The force is also zero. In case (e), the direction of I A → is the same
and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its
equilibrium is stable.
→and B
Whereas, in case (f), the direction of I A → is opposite. The angle between them is 180°. If
disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.

Page : 171 , Block Name : Additional Exercises

Q4.25 A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic eld of 0.10 T
normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic eld?
(The coil is made of copper wire of cross-sectional area 10 m , and
−5 2

the free electron density in copper is given to be about
29 −3
10 m ⋅)

Answer. Number of turns on the circular coil, n = 20
Radius of the coil, r = 10 cm = 0.1 m
Magnetic eld strength, B = 0.10 T
Current in the coil, I = 5.0 A

(a) The total torque on the coil is zero because the eld is uniform.
(b) The total force on the coil is zero because the eld is uniform.
(c) Cross-sectional area of copper coil, A = 10 m −5 2

Number of free electrons per cubic meter in copper, N = 10 /m 29 3

Charge on the electron, e = 1.6 × 10 −19
C

Magnetic force, F = Bev d

Where, v = Drift velocity of electrons
d

I
=
N eA
BeI
F =
N eA
0.10 × 5.0
−25
= = 5 × 10 N
29 −5
10 × 10

Hence, the average force on each electron is 5 × 10 −25
N

Page : 172 , Block Name : Additional Exercises

Q4.26 A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0
cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the
wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two
leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in
the wire. What value of current (with appropriate sense of circulation) in the windings of the
solenoid can support the weight of the wire? g= = 9.8ms . −2

Answer. Length of the solenoid, L = 60 cm = 0.6 m
Radius of the solenoid, r = 4.0 cm = 0.04 m
It is given that there are 3 layers of windings of 300 turns each. ...
Total number of turns, n = 3 x 300 = 900
Length of the wire, I = 2 cm = 0.02 m
Mass of the wire, m = 2.5g = 2.5 × 10 kg −3

Current owing through the wire, i = 6 A
Acceleration due to gravity, g = 9.8m/s 2

μ0 nI
Magnetic eld produced inside the solenoid, where, B = L

μ0 = = Permeability of free space e = 4π × 10 TmA −7 −1

I = Current owing through the windings of the solenoid
Magnetic force is given by the relation,
F = Bil

μ0 nI
= il
L

Also, the force on the wire is equal to the weight of the wire.
μ0 nlil
∴ mg =
L

mgL
I =
μ0 nil

−3
2.5×10 ×9.8×0.6
= −3
= 108A
4π×10 ×900×0.02×6

Hence, the current owing through the solenoid is 108 A.

Page : 172 , Block Name : Additional Exercises

Q4.27 A galvanometer coil has a resistance of 12 Ω and the metre shows full scale de ection for a
current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Answer. Resistance of the galvanometer coil, G = 12 Ω
Current for which the galvanometer shows full scale de ection, I 8
−3
= 3mA = 3 × 10 A

Current for which there is full scale de ection,
V
R = − G
I8
Range of the voltmeter is
18
= − 12 = 6000 − 12 = 5988Ω
−3
3 × 10

which needs to be converted to 18 V.
V=18 v
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a
voltmeter. This resistance is given as:
V
R = − G
Ig

18
= − 12 = 6000 − 12 = 5988Ω
−3
3 × 10

Hence, a resistor of resistance is to be connected in series with the galvanometer.

Page : 172 , Block Name : Additional Exercises

Q4.28 A galvanometer coil has a resistance of 15 Ω and the metre shows full scale de ection for a
current Of 4 mA. How will you convert the metre into an ammeter Of range O to 6 A?

Answer. Resistance of the galvanometer coil,G = 15Ω
Current for which the galvanometer shows full scale de ection,
−3
I8 = 4mA = 4 × 10 A

Range of the ammeter is O, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it
into an ammeter. The value of S is given as:
Ix G
S =
I − Is
−3
4 × 10 × 15
=
−3
6 − 4 × 10
−2
6 × 10 0.06
S = =
6 − 0.004 5.996

≈ 0.01Ω = 10mΩ

Hence, a 10
mΩ shunt resistor is to be connected in parallel with the galvanometer.

Page : 172 , Block Name : Additional Exercises