A magnetic needle lying parallel to a magnetic field requires `W units` of work to turn it through `60^(@)`. The torque needed to maintain the needle in this position will be

Options

A. `sqrt3 W`
B. W
C. `W/sqrt2`
D. `W/sqrt3`

Answer

A

Explanation

Potential energy of magnetic dipole is given by
`U = - MB cos theta` .
`W = U_(2) - U_(1)`
`W = (- MB cos 60) - (- MB cos 0) = MB//2`
` rArr " " MB = 2W`
Torque is given by formula : `tau = MB sin theta`
`rArr " " tau = (2W) sin 60^(@) = W sqrt3`
Hence, option (a) is correct.