Question

The work done in moving a dipole from its most stable to most unstable position in a $0.09\, T$ uniform magnetic field is
(dipole moment of this dipole $= 0.5\, A\, m^2$)

• A. $0.07 \,J$
• B. $0.08 \,J$
• C. $0.09 \,J$
• D. $0.1 \,J$

Answer 1

Answer: (C)

Since the most stable position is at $\theta=0$ and the most unstable position is at $\theta=180^{\circ}$, then the work done is given by
$ W=\int\limits_{\theta=0^{\circ}}^{\theta=180^{\circ}} \tau(\theta) d \theta=\int\limits_{0^{\circ}}^{180^{\circ}} m B \sin \theta d \theta=$
$-m B[\cos \theta]_0^{180^{\circ}}$
$ =-m B\left[\cos 180^{\circ}-\cos 0^{\circ}\right]=-m B[-1-1]$
$=-m B[-2]=2 m B$
Here, $ m=0.5 A m^2 \text { and } B=0.09 T $
$ \therefore W=2 \times 0.50 \times 0.09=0.09 J$