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. `sqrt(3)W`
B. W
C. `(sqrt(3))/2W`
D. `2W`

Answer

A

Explanation

`W=MB(cos theta_(2)-cos theta_(1))`
`=-MB(cos 60^(@)-cos 0^(@))=(MB)/2" ":. MB=2W`……..i
Torque `=MB sin 60^(@)=(2W)sin 60^(@)=(2Wxxsqrt(3))/2=sqrt(3)W`