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. `2W`
B. `W`
C. `W/sqrt2`
D. `W/sqrt3`

Explanation

Here, `theta_1=0^@`, `theta_2=60^@`
Work done, `W=MB[cod theta_1-cos theta_2]`
`=MB[cos 0^@-cos 60^@]`
`=MB[1-1/2]=(MB)/(2)` …(i)
Torque, `tau=MB sin 60^@=MB sqrt3//2=sqrt3W`