Question

A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

$\tau =MBsin\theta \Rightarrow \tau \propto sin\theta$
$\Rightarrow \frac{{\tau }_1}{{\tau }_2}=\frac{sin{\theta }_1}{sin{\theta }_2}$ $\Rightarrow \frac{\tau }{\frac{\tau }{2}}=\frac{sin90}{sin{\theta }_2}$
$\Rightarrow sin{\theta }_2=\frac{1}{2}\Rightarrow {\theta }_2={30}^{\circ }$
$\Rightarrow$ angle of rotation $=0\stackrel{\circ }{A}-30={60}^{\circ }$