Question

A magnet of magnetic moment M is lying along a uniform magnetic field B the work done in rotating the magnet by 90 is

Answer 1

Answer : W = M B

Explanation :

It is given that a magnet if magnetic moment M is lying along a uniform magnetic field B. the work done in rotating the magnet by $90^0$ is :

$W=\int\limits^a_b {MB\ sin\theta} \, d\theta$

$W=\int\limits^{90^0}_0 {M.Bsin\theta} \, d\theta$

$W=-MB[cos\ 90^0-cos\ 0]$

$W=MB$

So, the work done in rotating the magnet by $90^0$ is MB.

Hence, this is the required solution.

Answer 2

Answer: The work done in rotating the magnet by ninety degree is MB.

Explanation:

In the given problem, a magnet of magnetic moment M is lying along a uniform magnetic field B.

The expression for torque in terms of magnetic moment and magnetic field is as follows;

$\tau =MBsin\theta$

Here, tau is torque, M is the magnetic moment, B is the magnetic field and theta is the angle.

The expression for the work done in rotating the magnet is as follows;

$W = \int_{0}^{\theta} \tau\ d\theta$

Put $\theta =90\degree$ and $\tau =MBsin\theta$.

$W = \int_{0}^{90\degree} \MBsin\theta\ d\theta$

$W = MB\ [-cos\theta]_{0}^{90^{o}}W = MB [-cos90^{o} + cos0^{0}]$

$W =MB$

Therefore, the work done in rotating the magnet by ninety degree is MB.