Question

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

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

Answer 1

The correct option is B 2MB

Work done in rotating the dipole
$W_{e_1}-E_{2_2}=MB[cos{\theta }_1-cos{\theta }_2]$
Initially dipole is in stable position $\theta =0^{\circ }$
$U_1=-MBcos{0}^{\circ }=-MB$
$U_2=-MBcos{80}^{\circ }=MB$
$W_{ext}$ (We rotate it) $=U_2-U_1=2MB$