Question

A uniform magnetic field of 3000 G is established along the positivez-direction. A rectangular loop of sides 10 cm and 5 cm carries acurrent of 12 A. What is the torque on the loop in the different casesshown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?2018

Answer 1

Given: The magnitude of the uniform magnetic field is 3000 G, The direction of magnetic field is established along positive zaxis, the length of the rectangular loop is 10 cm, the width of the rectangular loop is 5 cm and the magnitude of the current flowing through the loop is 12 A.

a)

Suppose the anticlockwise current is positive and vice-versa.

The formula for the torque is given by,

τ → =I A → × B → (1)

Where, the magnitude of the magnetic field is B.

Let l be the length and b be the width of the loop.

The area of the loop is,

A=10×5 =50  cm 2 =50× 10 −4   m 2

The angle between the area and the magnetic field is 90°.

By substituting the values in equation (1), we get

τ → =12×( 50× 10 −4 ) i ^ ×0.3 k ^ =−1.8× 10 −2 j ^  Nm

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

Thus, the net torque is −1.8× 10 −2 j ^  Nm and the net force is zero.

b)

By substituting the values in equation (1), we get

τ → =12×( 50× 10 −4 ) i ^ ×0.3 k ^ =−1.8× 10 −2 j ^  Nm

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

Thus, the net torque is −1.8× 10 −2 j ^  Nm and the net force is zero.

c)

The direction of magnetic field is along z axis and the area of the loop A is normal to the x-z plane.

By substituting the values in equation (1), we get

τ → =12×( 50× 10 −4 ) j ^ ×0.3 k ^ =−1.8× 10 −2 i ^  Nm

The direction of the torque is in negative x-direction. The force on the loop is zero because the angle between A and B is zero.

Thus, the net torque is −1.8× 10 −2 j ^  Nm and the net force is zero.

d)

By substituting the values in equation (1), we get

τ → =12×( 50× 10 −4 ) j ^ ×0.3 k ^ =−1.8× 10 −2 i ^  Nm

The direction of the torque is 240° with positive x direction. The force on the loop is zero because the angle between A and B is zero.

e)

The direction of magnetic field is along z axis and the area of the loop A is normal to the x-y plane.

By substituting the values in equation (1), we get

τ → =12×( 50× 10 −4 ) k ^ ×0.3 k ^ =0 Nm

The net torque on the loop is zero. The force on the loop is zero because the angle between A and B is zero.

f)

The direction of current is clockwise.

The direction of magnetic field is along z axis and the area of the loop A is normal to the x-y plane.

By substituting the values in equation (1), we get

τ → =12×( 50× 10 −4 ) k ^ ×0.3 k ^ =0 Nm

The net torque on the loop is zero. The force on the loop is zero because the angle between A and B is zero.

In case (e), the direction of A → and B → is same and the angle between them is zero. If they displaced they come back to equilibrium. Thus, the equilibrium is stable.

In case (f), the direction of A → and B → is opposite and the angle between them is 180°. If they displaced they does not come back to equilibrium. Thus, the equilibrium is unstable.