# Physics Class 12 NCERT Solutions: Chapter 4 Moving Charges and Magnetism Part 9 (For CBSE, ICSE, IAS, NET, NRA 2022)

Get top class preparation for NSO Class-12 right from your home: fully solved questions with step-by-step explanation- practice your way to success.

Q: 24. A uniform magnetic field of is established along the positive z-direction. A rectangular loop of sides and carries a current of . What is the torque on the loop in the different cases shown in Figure? What is the force on each case? Which case corresponds to stable equilibrium?

Answer:

Magnetic field strength,

Length of the rectangular loop,

Width of the rectangular loop,

Area of the loop,

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 plane and B is directed along the z-axis.

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

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 along the negative x direction and the force is zero.

(d) Magnitude of torque is given as:

Torque is at an angle of 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 . If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.

Q: 25. A circular coil of 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 , and the free electron density in copper is given to be about .)

Answer:

Number of turns on the circular coil,

Radius of the coil,

Magnetic field strength,

Current in the coil,

(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,

Number of free electrons per cubic meter in copper, N = 1029/m3

Charge on the electron,

Magnetic force,

Where,

Drift velocity of electrons

Hence, the average force on each electron is .