NCERT Solutions for Class 12 Science Physics Chapter 4 Moving Charges And Magnetism are provided here with simple step-by-step explanations. These solutions for Moving Charges And Magnetism are extremely popular among class 12 Science students for Physics Moving Charges And Magnetism Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 12 Science Physics Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 169:

#### Question 4.1:

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 field **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 flowing in the
coil, *I* = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given by the relation,

Where,

= Permeability of free space

= 4π
× 10^{–7} T m A^{–1}

Hence, the magnitude of
the magnetic field is 3.14 × 10^{–4} T.

#### Page No 169:

#### Question 4.2:

A
long straight wire carries a current of 35 A. What is the magnitude
of the field **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 field at this point is given as:

*B*

Where,

= Permeability of free space = 4π
× 10^{–7} T m A^{–1}

Hence, the magnitude of
the magnetic field at a point 20 cm from the wire is 3.5 × 10^{–5}
T.

#### Page No 169:

#### Question 4.3:

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.

Magnitude of the
magnetic field at that point is given by the relation, *B*

Where,

_{
=} Permeability of free space = 4π
× 10^{–7} T m A^{–1}

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 field at the given point is vertically upward.

#### Page No 169:

#### Question 4.4:

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 field 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 field at that point is given by the relation,

Where,

_{
=} Permeability of free space = 4π
× 10^{–7} T m A^{–1}

The current is flowing 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 field is towards the South.

#### Page No 169:

#### Question 4.5:

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 field of 0.15 T?

#### Answer:

Current in the wire, *I
*= 8 A

Magnitude of the
uniform magnetic field, *B* = 0.15 T

Angle between the wire
and magnetic field, *θ* = 30°.

Magnetic force per unit length on the wire is given as:

*f*
= *BI* sin*θ*

= 0.15 × 8 ×1 × sin30°

=
0.6 N m^{–1}

Hence, the magnetic
force per unit length on the wire is 0.6 N m^{–1}.

#### Page No 169:

#### Question 4.6:

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

#### Answer:

Length of the wire, *l*
= 3 cm = 0.03 m

Current flowing in the
wire, *I* = 10 A

Magnetic field, *B*
= 0.27 T

Angle between the
current and magnetic field, *θ* = 90°

Magnetic force exerted on the wire is given as:

*F*
= *BIl*sin*θ*

= 0.27 × 10 × 0.03 sin90°

=
8.1 × 10^{–2}
N

Hence, the magnetic
force on the wire is 8.1 × 10^{–2} N. The
direction of the force can be obtained from Fleming’s left hand
rule.

#### Page No 169:

#### Question 4.7:

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 flowing in wire
A, *I*_{A} = 8.0 A

Current flowing in wire
B, *I*_{B} = 5.0 A

Distance between the
two wires, *r* = 4.0 cm = 0.04 m

Length of a section of
wire A,* l* = 10 cm = 0.1 m

Force exerted on length
*l* due to the magnetic field is given as:

Where,

= Permeability of free space = 4π
× 10^{–7} T m A^{–1}

The magnitude of force
is 2 × 10^{–5} N. This is an attractive force
normal to A towards B because the direction of the currents in the
wires is the same.

#### Page No 169:

#### Question 4.8:

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,
*l* = 80 cm = 0.8 m

There are five layers of windings of 400 turns each on the solenoid.

Total
number of turns on the solenoid, *N* = 5 × 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 field inside the solenoid near its centre is given by the relation,

Where,

= Permeability of free space = 4π
× 10^{–7} T m A^{–1}

Hence, the magnitude of
the magnetic field inside the solenoid near its centre is 2.512 ×
10^{–2} T.

#### Page No 169:

#### Question 4.9:

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 field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

#### Answer:

Length of a side of the
square coil, *l* = 10 cm = 0.1 m

Current flowing in the
coil, *I* = 12 A

Number of turns on the
coil, *n* = 20

Angle made by the plane
of the coil with magnetic field, *θ* = 30°

Strength of magnetic
field, *B* = 0.80 T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

τ = *n* *BIA*
sin*θ*

Where,

*A* = Area of the
square coil

*l × l* = 0.1 × 0.1 = 0.01 m^{2}

∴ τ = 20 × 0.8 × 12 × 0.01 × sin30°

= 0.96 N m

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

#### Page No 169:

#### Question 4.10:

Two moving coil meters, M_{1} and M_{2} have the following particulars:

*R*_{1} = 10 Ω, *N*_{1} = 30,

*A*_{1} = 3.6 × 10^{–3} m^{2}*, **B*_{1} = 0.25 T

*R*_{2} = 14 Ω, *N*_{2} = 42,

*A*_{2} = 1.8 × 10^{–3} m^{2}, *B*_{2} = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}.

#### Answer:

For moving coil meter M_{1}:

Resistance, *R*_{1} = 10 Ω

Number of turns, *N*_{1} = 30

Area of cross-section, *A*_{1} = 3.6 × 10^{–3} m^{2}

Magnetic field strength, *B*_{1} = 0.25 T

Spring constant *K*_{1} = *K*

For moving coil meter M_{2}:

Resistance, *R*_{2} = 14 Ω

Number of turns, *N*_{2} = 42

Area of cross-section, *A*_{2} = 1.8 × 10^{–3} m^{2}

Magnetic field strength, *B*_{2} = 0.50 T

Spring constant, *K*_{2} = *K*

**(a)** Current sensitivity of M_{1} is given as:

And, current sensitivity of M_{2}_{ }is given as:

Ratio

Hence, the ratio of current sensitivity of M_{2} to M_{1} is 1.4.

**(b)** Voltage sensitivity for M_{2} is given as:

And, voltage sensitivity for M_{1}_{ }is given as:

${V}_{s1}=\frac{{N}_{1}{B}_{1}{A}_{1}}{{K}_{1}{R}_{1}}$

Ratio$\frac{{V}_{s2}}{{V}_{s1}}=\frac{{N}_{2}{B}_{2}{A}_{2}{K}_{1}{R}_{1}}{{K}_{2}{R}_{2}{N}_{1}{B}_{1}{A}_{1}}$

Hence, the ratio of voltage sensitivity of M_{2} to M_{1} is 1.

#### Page No 169:

#### Question 4.11:

In a chamber, a uniform
magnetic field of 6.5 G (1 G = 10^{–4} T) is
maintained. An electron is shot into the field with a speed of 4.8 ×
10^{6} m s^{–1} normal to the field. Explain
why the path of the electron is a circle. Determine the radius of the
circular orbit. (*e *= 1.6 × 10^{–19} C, *m*_{e}= 9.1×10^{–31} kg)

#### Answer:

Magnetic field
strength, *B* = 6.5 G = 6.5 × 10^{–4} T

Speed of the electron,
*v* = 4.8 × 10^{6} m/s

Charge on the electron,
*e* = 1.6 × 10^{–19} C

Mass of the electron,
*m*_{e} = 9.1 × 10^{–31} kg

Angle between the shot
electron and magnetic field, *θ* = 90°

Magnetic force exerted on the electron in the magnetic field 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,

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

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

#### Page No 169:

#### Question 4.12:

In Exercise 4.11 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 field
strength, *B* = 6.5 × 10^{−4} T

Charge of the electron,
*e* = 1.6 × 10^{−19} C

Mass of the electron,
*m*_{e} = 9.1 × 10^{−31} kg

Velocity of the
electron, *v* = 4.8 × 10^{6} m/s

Radius of the orbit, *r*
= 4.2 cm = 0.042 m

Frequency of revolution
of the electron = *ν*

Angular frequency of
the electron = *ω* = 2π*ν*

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:

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:

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

#### Page No 169:

#### Question 4.13:

**(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 field of magnitude 1.0 T. The field lines make an angle of
60º
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

Current
flowing in the coil, *I* = 6.0 A

Magnetic
field strength, *B* = 1 T

Angle between the field lines and normal with the coil surface,

*θ*
= 60°

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

τ
= *n* *IBA* sin*θ* … (*i*)

= 30 × 6 × 1 × 0.0201 × sin60°

= 3.133 N m

**(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 some irregular shape that encloses the same area.

#### Page No 170:

#### Question 4.14:

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 field due to the coils at their centre.

#### Answer:

Radius of coil X, *r*_{1}
= 16 cm = 0.16 m

Radius of coil Y,* r*_{2}
= 10 cm = 0.1 m

Number of turns of on
coil X, *n*_{1} = 20

Number of turns of on
coil Y, *n*_{2} = 25

Current in coil X, *I*_{1}
= 16 A

Current in coil Y, *I*_{2}
= 18 A

Magnetic field due to coil X at their centre is given by the relation,

Where,

= Permeability of free space =

Magnetic field due to coil Y at their centre is given by the relation,

Hence, net magnetic field can be obtained as:

#### Page No 170:

#### Question 4.15:

A
magnetic field of 100 G (1 G = 10^{−4}
T) is required which is uniform in a region of linear dimension about
10 cm and area of cross-section about 10^{−3}
m^{2}.
The maximum current-carrying capacity of 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^{−1}.
Suggest some appropriate design particulars of a solenoid for the
required purpose. Assume the core is not ferromagnetic

#### Answer:

Magnetic field
strength,* B* = 100 G = 100 ×
10^{−4} T

Number of turns per
unit length, *n* = 1000 turns m^{−1}

Current flowing in the
coil, *I* = 15 A

Permeability of free space, =

Magnetic field is given by the relation,

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

#### Page No 170:

#### Question 4.16:

For
a circular coil of radius *R
*and *N
*turns carrying current
*I*,
the magnitude of the magnetic field at a point on its axis at a
distance *x *from
its centre is given by,

**(a)**
Show that this reduces to the familiar result for field 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 field 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,

, approximately.

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

#### Answer:

Radius of circular coil
= *R*

Number of turns on the
coil = *N*

Current in the coil = *I*

Magnetic field at a
point on its axis at distance *x* is given by the relation,

Where,

= Permeability of free space

**(a) **If the
magnetic field at the centre of the coil is considered, then *x*
= 0.

This is the familiar result for magnetic field 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 from point Q.

Magnetic field at point Q is given as:

Also, the other coil is at a distance of from point Q.

Magnetic field due to this coil is given as:

Total magnetic field,

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

#### Page No 170:

#### Question 4.17:

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 field (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*_{1} = 25 cm = 0.25 m

Outer radius of the
toroid, *r*_{2} = 26 cm = 0.26 m

Number of turns on the
coil,* N* = 3500

Current in the coil, *I*
= 11 A

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

**(b) **Magnetic field inside the core of a toroid is given by the
relation,

*B*
=

Where,

= Permeability of free space =

*l*
= length of toroid

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

#### Page No 170:

#### Question 4.18:

Answer the following questions:

**(a)**
A magnetic field 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 undeflected 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 field varying from point to point both in magnitude and
direction, and comes out of it following a complicated trajectory.
Would its final 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 field in north to south direction. Specify the
direction in which a uniform magnetic field should be set up to
prevent the electron from deflecting from its straight line path.

#### Answer:

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

**(b) **Yes, the final 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 field in the North-South direction.
This moving electron can remain undeflected if the electric force
acting on it is equal and opposite of magnetic field. Magnetic force
is directed towards the South. According to Fleming’s left hand
rule, magnetic field should be applied in a vertically downward
direction.

#### Page No 171:

#### Question 4.19:

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

#### Answer:

Magnetic field
strength, *B* = 0.15 T

Charge on the electron,
*e* = 1.6 × 10^{−19
}C

Mass of the electron, *m*
= 9.1 × 10^{−31 }kg

Potential difference, *V*
= 2.0 kV = 2 × 10^{3}
V

Thus, kinetic energy of
the electron = *eV*

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

From equations (1) and (2), we get

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

**(b)** When the field makes an angle *θ*
of 30° with initial
velocity, the initial velocity will be,

From equation (2), we can write the expression for new radius as:

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

#### Page No 171:

#### Question 4.20:

A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field 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 field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^{5} V m^{−1}, make a simple guess as to what the beam contains. Why is the answer not unique?

#### Answer:

Magnetic field, *B* = 0.75 T

Accelerating voltage, *V* = 15 kV = 15 × 10^{3} V

Electrostatic field, *E* = 9 × 10^{5} V m^{−1}

Mass of the electron = *m*

Charge of the electron = *e*

Velocity of the electron = *v*

Kinetic energy of the electron = *eV*

Since the particle remains undeflected by electric and magnetic fields, we can infer that the force on the charged particle due to electric field is balancing the force on the charged particle due to magnetic field.

Putting equation (2) in equation (1), we get

This value of specific charge *e*/*m* is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He^{++,} Li^{++}, etc.

#### Page No 171:

#### Question 4.21:

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 field 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 field same as before?
(Ignore the mass of the wires.) g = 9.8 m s^{−2}.

#### Answer:

Length of the rod,* l*
= 0.45 m

Mass suspended by the
wires, *m* = 60 g = 60 ×
10^{−3} kg

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

Current in the rod
flowing through the wire, *I* = 5 A

**(a) **Magnetic field (*B*) is equal and opposite to the
weight of the wire i.e.,

A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming’s left hand rule gives an upward magnetic force.

**(b) **If the direction of the current is revered, then the force
due to magnetic field and the weight of the wire acts in a vertically
downward direction.

∴Total
tension in the wire = *BIl* + *m*g

#### Page No 171:

#### Question 4.22:

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, *l* = 70 cm = 0.7 m

Force between the two wires is given by the relation,

Where,

= Permeability of free space =

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

#### Page No 171:

#### Question 4.23:

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 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* = 1.5 T

Radius of the cylindrical region, *r* = 10 cm = 0.1 m

Current in the wire passing through the cylindrical region, *I* = 7 A

**(a) **If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region.

Thus, *l* = 2*r* = 0.2 m

Angle between magnetic field and current, *θ* = 90°

Magnetic force acting on the wire is given by the relation,

*F* = *BIl* sin *θ*

= 1.5 × 7 × 0.2 × sin 90°

= 2.1 N

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: :

Angle between magnetic field and current, *θ* = 45°

Force on the wire,

*F* = *BIl*_{1} sin *θ*

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

**(c) **The wire is lowered from the axis by distance, *d* = 6.0 cm

Thus the length of wire in magnetic field will be 16 cm as AB= *L *=2*x* =16 cm

Now the force,

*F *= *iLB *sin90° as the wire will be perpendicular to the magnetic field.

F= 7 × 0.16 × 1.5 =1.68 N

The direction will be given by right hand curl rule or screw rule i.e. vertically downwards.

#### Page No 171:

#### Question 4.24:

A uniform magnetic
field 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.
4.28? What is the force on each case? Which case corresponds to
stable equilibrium?

#### Answer:

Magnetic field
strength, *B* = 3000 G = 3000 ×
10^{−4} T = 0.3 T

Length of the
rectangular loop,* l* = 10 cm

Width of the
rectangular loop, *b* = 5 cm

Area of the loop,

*A* = *l* ×
*b* = 10 × 5 = 50 cm^{2
}= 50 × 10^{−4}
m^{2}

Current in the loop, *I*
= 12 A

Now, taking the anti-clockwise direction of the current as positive and vise-versa:

**(a) **Torque,

From
the given figure, it can be observed that *A *is normal to the
*y*-*z* plane and *B* is directed along the *z*-axis.

The
torque is
N m 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

From
the given figure, it can be observed that *A *is normal to the
*x*-*z* plane and *B* is directed along the *z*-axis.

The
torque is
N m along the negative *x *direction and the force is zero.

**(d)** Magnitude of torque is given as:

Torque
is
N m at an angle of 240° with
positive *x *direction. The force is zero.

**(e)** Torque

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

**(f)** Torque

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

In case (e), the direction of and is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.

Whereas, in case (f), the direction of and 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 No 172:

#### Question 4.25:

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field 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 field?

(The
coil is made of copper wire of cross-sectional area 10^{−5}
m^{2},
and the free electron density in copper is given to be about 10^{29}
m^{−3}.)

#### Answer:

Number of turns on the
circular coil,* n* = 20

Radius of the coil,*
r* = 10 cm = 0.1 m

Magnetic field
strength, *B* = 0.10 T

Current in the coil, *I*
= 5.0 A

**(a) **The total torque on the coil is zero because the field is
uniform.

**(b) **The total force on the coil is zero because the field is
uniform.

**(c) **Cross-sectional area of copper coil, *A* = 10^{−5}
m^{2}

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

Charge
on the electron, *e* = 1.6 ×
10^{−19} C

Magnetic
force, *F* = *Bev*_{d}

Where,

*v*_{d}_{
}= Drift velocity of electrons

Hence, the average force on each electron is

#### Page No 172:

#### Question 4.26:

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
*= 9.8 m s^{−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 ×
300 = 900

Length of the wire, *l*
= 2 cm = 0.02 m

Mass of the wire, *m*
= 2.5 g = 2.5 × 10^{−3}
kg

Current flowing through
the wire, *i* = 6 A

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

Magnetic field produced inside the solenoid,

Where,

= Permeability of free space =

*I *= Current
flowing through the windings of the solenoid

Magnetic force is given by the relation,

Also, the force on the wire is equal to the weight of the wire.

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

#### Page No 172:

#### Question 4.27:

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection 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 there is full scale deflection,
= 3 mA = 3 ×
10^{−3}
A

Range of the voltmeter is 0, 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:

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

#### Page No 172:

#### Question 4.28:

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

#### Answer:

Resistance
of the galvanometer coil, *G*
= 15 Ω

Current for which the galvanometer shows full scale deflection,

= 4 mA = 4 ×
10^{−3}
A

Range of the ammeter is 0, 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:

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

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