Question

A magnetic dipole of dipole moment $M$ is placed in a mutually perpendicular uniform magnetic field $B_1$ and $B_2$ respectively. If the magnetic dipole comes to stable equilibrium at an angle ${30}^{\circ }$ with $B_1$. Find the magnitude of $B_2$ ?
$[B_1=3\times {10}^{-5}\text{T}]$

• A. $\sqrt{3}\times {10}^{-5}\text{T}$
• B. $2\sqrt{3}\times {10}^{-5}\text{T}$
• C. $\sqrt{2}\times {10}^{-5}\text{T}$
• D. $3\sqrt{2}\times {10}^{-5}\text{T}$

Answer 1

The correct option is A $\sqrt{3}\times {10}^{-5}\text{T}$

Given:
$B_1=3\times {10}^{-5}\text{T}$


From diagram,
${\theta }_1={30}^{\circ };{\theta }_2={60}^{\circ }$

We know that at stable equilibrium, net torque acting on the bar magnet will be zero.

${\overrightarrow{\tau }}_1+{\overrightarrow{\tau }}_2=0$

$|{\overrightarrow{\tau }}_1|=|{\overrightarrow{\tau }}_2|$

$M{B}_1\sin {\theta }_1=M{B}_2\sin {\theta }_2$

$B_2=\frac{M{B}_1\sin {\theta }_1}{\sin {\theta }_2}=\frac{3\times {10}^{-5}\times \sin {30}^{\circ }}{\sin {60}^{\circ }}$

$\therefore B_2=\frac{3\times {10}^{-5}\times 2}{\sqrt{3}\times 2}=\sqrt{3}\times {10}^{-5}\text{T}$

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