Question

If a bar magnet of magnetic moment $M$ is freely suspended in a uniform magnetic field of strength $B$, the work done in rotating the magnet through an angle $\theta$ is

• A. $MB(1-\sin \theta )$
• B. $MB\sin \theta$
• C. $MB\cos \theta$
• D. $MB(1-\cos \theta )$

Answer 1

The correct option is D $MB(1-\cos \theta )$

Given that,

Magnetic moment = $M$

Magnetic field of strength = $B$

We know that,

$dW=MB\sin \theta d\theta$

Now, the work done in rotating the magnet through an angle θ from initial position$({\theta }_1=0^0)$is $W=MB(\cos {\theta }_1-\cos \theta )$

$W=MB(\cos 0^0-\cos \theta )$

$W=MB(1-\cos \theta )$

Hence, the work done in rotating the magnet through an angle $\theta$ is $MB(1-cos\theta )$