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 ${60}^{\circ }$

$\text{Given:-}$ Couple acting on the magnet is halved by rotating it

$\text{To find:-}$Angle of rotation

$\text{Solution:-}$

$\tau =MBsin\theta$

$⟹\tau \propto sin\theta$

$⟹\frac{{\tau }_1}{{\tau }_2}=\frac{sin{\theta }_1}{sin{\theta }_2}$

$⟹\frac{\tau }{\frac{\tau }{2}}=\frac{sin{90}^0}{sin{\theta }_2}$

$⟹sin{\theta }_2=1/2$

$⟹{\theta }_2={30}^0$ `

$⟹$ Angle of rotation =${90}^0-30={60}^0$

$\text{Hence the correct option is C}$