Question

The couple acting on a bar magnet of pole strength $4\text{Am}$, when placed in a uniform magnetic field of intensity $10\sqrt{3}{\text{Am}}^{-1}$, such that axis of the magnet makes an angle ${60}^{\circ }$ with the direction of the field is $80\times {10}^{-6}\text{Nm}$. Find the distance between the poles of the magnet is,

• A. $0.6\mu \text{m}$
• B. $0.3\mu \text{m}$
• C. $1.3\mu \text{m}$
• D. $0.9\mu \text{m}$

Answer 1

The correct option is C $1.3\mu \text{m}$

Given:
Pole strength, $m=4\text{Am};\theta ={60}^{\circ }$
$B=10\sqrt{3}\text{A/m};\tau =80\times {10}^{-6}\text{Nm and}$

Magnetic dipole moment of a magnet is given by

$M=m\left(2l\right)$

Here, $\left(2l\right)$ is the length of the bar magnet.

Now using relation of torque,

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

$\tau =MB\sin \theta =m\left(2l\right)B\sin \theta$

Substituting the given values, we get

$80\times {10}^{-6}=4\times \left(2l\right)\times 10\sqrt{3}\times \frac{\sqrt{3}}{2}$

$80\times {10}^{-6}=\left(2l\right)\times 60$

$\therefore 2l=\frac{4}{3}\times {10}^{-6}\text{m}=1.3\mu \text{m}$

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