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^o
• B. 45^o
• C. 60^o
• D. 90^o

Answer 1

Answer: (C)

60^o

Answer 2

$\tau = M B \sin \theta \Rightarrow \tau \propto \sin \theta$

$\Rightarrow \frac { \tau _ { 1 } } { \tau _ { 2 } } = \frac { \sin \theta _ { 1 } } { \sin \theta _ { 2 } } \Rightarrow \frac { \tau } { \tau / 2 } = \frac { \sin 90 } { \sin \theta _ { 2 } }$

$\Rightarrow \sin \theta _ { 2 } = \frac { 1 } { 2 } \Rightarrow \theta _ { 2 } = 30 ^ { \circ }$

⇒ Angle of rotation