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. `sqrt3W`
B. W
C. `(sqrt3)/(2)W`
D. 2W

Answer

A

Explanation

`W= Delta U= - MB (cos theta_(f)- cos theta_(i))`
`= - MB (cos 60^(@)- cos 0^(@))`
`W= (MB)/(2)`
`tau = MB sin theta rArr tau = MB (sqrt3)/(2)`
`therefore tau= sqrt3 W`