JEE Main & Advanced Physics Magnetism JEE PYQ-Magnetism

A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through\[{{60}^{o}}\]. The torque needed to maintain the needle in this position will be [AIEEE 2003]

A) \[\sqrt{3}\] W

B) W

C) \[(\sqrt{3}/2)\]W

D) 2 W

Correct Answer: A

Solution :

[a] As the work \[W=MB\,(1-\cos \theta )\]

\[\Rightarrow \] \[W=MB\,(1-\cos \,{{60}^{o}})\] \[(\because \theta ={{60}^{o}})\]

\[\Rightarrow \] \[W=\frac{MB}{2}\]

\[\therefore \] MB = 2 W

Torque, \[\tau =MB\sin {{60}^{o}}\]

\[=\frac{MB\sqrt{3}}{2}=\frac{2W\sqrt{3}}{2}=W\sqrt{3}\]