Question

A bar magnet of length $10\text{cm}$ and pole strength $2\text{Am}$ makes an angle ${60}^{\circ }$ with a uniform magnetic field of induction $B=50\text{T}.$ The couple acting on it is-

• A. $20\sqrt{3}\text{Nm}$
• B. $10\sqrt{3}\text{Nm}$
• C. $\sqrt{3}\text{Nm}$
• D. $5\sqrt{3}\text{Nm}$

Answer 1

The correct option is D $5\sqrt{3}\text{Nm}$

Given, $2l=10\text{cm};m=2\text{Am}B=50\text{T};\theta ={60}^{\circ }$

Torque acting on the dipole is,

$\tau =MB\sin \theta [\because M=m(2l\left)\right]$

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

$=2\times 10\times {10}^{-2}\times 50\times \left(\frac{\sqrt{3}}{2}\right)$

$=5\sqrt{3}\text{Nm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.