Question

A magnetic dipole of dipole moment $M=10{\text{Am}}^2$ is suspended at angle ${45}^{\circ }$ in an external magnetic field of $\sqrt{2}\times {10}^{-5}\text{T}$. Find the work done by the magnetic field to bring the dipole moment to be in stable equilibrium ?

• A. $\sqrt{2}\times {10}^{-4}\text{J}$
• B. $(1-\sqrt{2})\times {10}^{-4}\text{J}$
• C. $5\times {10}^{-4}\text{J}$
• D. $\frac{1}{\sqrt{2}}\times {10}^{-4}\text{J}$

Answer 1

The correct option is B $(1-\sqrt{2})\times {10}^{-4}\text{J}$

Given:
$M=10{\text{Am}}^2;{\theta }_1={45}^{\circ };T=\sqrt{2}\times {10}^{-5}\text{T}$

As we know that, in the case of stable equilibrium, $\overrightarrow{M}$ and $\overrightarrow{B}$ are parallel and the angle between them is zero.

$\therefore {\theta }_2=0$

Work done by the magnetic field is given by

$W=MB(\cos {\theta }_1-\cos {\theta }_2)=MB(\cos {45}^{\circ }-\cos 0^{\circ })$

$W=10\times \sqrt{2}\times {10}^{-5}\times (\frac{1}{\sqrt{2}}-1)$

$\therefore W=(1-\sqrt{2})\times {10}^{-4}\text{J}$

Therefore, option $\left(B\right)$ is correct.