Question

A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle ${180}^o$, then the work is (M$=$ magnetic dipole moment of bar magnet).

• A. MB
• B. $2$MB
• C. $\frac{MB}{2}$
• D. Zero

Answer 1

Answer: (B)

$2$MB

The work done in rotating the bar magnet is given by:

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

Since the magnet is rotated from stable position $0^o$ to ${180}^o$.
$=MB(1-\cos {180}^o)$
$=2MB$.