A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through `60^(@)` . How much torque should be needed to maintain the needle in this position ?

Explanation

Work done in rotating the magnetic needle from `0^(@)` position through an angle `theta` in a magnetic field B is ,
`W=p_(m)B(1-costheta)`
` " [ where" p_(m)` = magnetic moment of the needle ]
i.e., `p_(m)B=(W)/(1-costheta)`
`therefore` Required torque ,
`tau =p_(m)Bsintheta=(Wsintheta)/(1-costheta)=(Wsin60^(@))/(1-cos60^(@))=sqrt(3)W`