Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by ${60}^{\circ }$ is $W$. Now the torque required to keep the magnet in this new position is

• A. $\frac{2W}{\sqrt{3}}$
• B. $\frac{\sqrt{3}W}{2}$
• C. $\sqrt{3}W$
• D. $\frac{W}{\sqrt{3}}$

Answer 1

The correct option is C $\sqrt{3}W$

Work done by the earth's magnetic field to rotate a bar magnet is,

$W=M{B}_H(\cos {\theta }_1-\cos {\theta }_2)$

$W=M{B}_H(1-\cos {60}^{\circ })[\because {\theta }_1=0^{\circ };{\theta }_2={60}^{\circ }]$

$W=M{B}_H(1-\frac{1}{2})$

$W=\frac{M{B}_H}{2}$

$\Rightarrow M{B}_H=2W$

The required torque to keep the magnet in equilibrium is,

$\tau =M{B}_H\sin {60}^0$

=$2W\times \frac{\sqrt{3}}{2}$

=$\sqrt{3}W$

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