Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be

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

Answer 1

Answer: (A)

$\sqrt { 3 }$W

Answer 2

τ = $M B \sin \theta$ and $W = M B ( 1 - \cos \theta )$⇒

$W = M B \left( 1 - \cos 60 ^ { \circ } \right) = \frac { M B } { 2 }$. Hence τ =