Question

A magnet of magnetic dipole moment 5.0 A m2 is lying in a uniform magnetic field of 7 x 10⁻⁴ T such that its dipole moment vector makes an angle of 30° with the field. The work done in increasing this angle from 30° to 45° is about _______ J.(A) 5.56 x 10⁻⁴
(B) 24.74 x 10⁻⁴
(C) 30.3 x 10⁻⁴
(D) 5.50 x 10⁻³

Answer 1

Answer:

The total work done in rotating the dipole from $30^\circ$ to $45^\circ$ is $5.56\times 10^{-4}\ J$.

Option (A) is correct.

Explanation:

The work done in rotating from initial angle $\theta_o$ to a final angle $\theta_1$ is given by

$W=\int^{\theta_1}_{\theta_o} \tau d\theta$.

where,

$\tau$ = torque exerted on the magnet due to the magnetic field.

We know,

$\vec \tau = \vec \mu \times \vec B\\\tau = \mu B\sin\theta$

where,

• $\vec \mu$ = dipole moment of the magnet.

• $\vec B$ = magnetic field in which the magnet is placed.

• $\theta$ = angle between $\vec \mu$ and $\vec B$.

Therefore, the work done in rotating from initial angle $\theta_o$ to a final angle $\theta_1$ is given by

$W=\int^{\theta_1}_{\theta_o} \mu B\sin\theta d\theta = \mu B (-\cos\theta)^{\theta_1}_{\theta_o}=\mu B(-\cos\theta_1+\cos\theta_o).$

Given:

• $\mu = 5.0\ Am^2$.

• $B=7\times 10^{-4}\ T$.

• $\theta_o = 30^\circ$.

• $\theta_1 = 45^\circ$.

Putting these values in above expression of required work done, we get,

$W=(5.0)\times (7\times 10^{-4})\times (-\cos(45^\circ)+\cos(30^\circ))=5.56\times 10^{-4}\ J.$

Thus, the correct option is (A).