Question

A magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength $B$. The work done is rotating it by an angle of ${180}^{\circ }$ will be

• A. $-MB$
• B. $+MB$
• C. Zero
• D. $+2MB$

Answer 1

Answer: (D)

$+2MB$

Let m be the pole strength of the magnet.

$\tau$ be the torque acting on magnet when it makes angle $0$ to $\theta$ in the uniform magnetic field of strength $B$.

$2l$ be the length of the magnet.

$M$ be the magnetic dipole moment of the magnet.

So, force acting on individual poles will be $mB$.

$\tau =mB\cdot 2lsin\theta =m2.lBsin\theta =\overrightarrow{M}\times \overrightarrow{B}$

where $M=m2l$

$W={\int }_0^{\theta }\tau d\theta$ = $MB(1-cos\theta )$

Here $\theta ={180}^o$

$W=2MB$