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. `2sqrt(3) W`
B. `sqrt(3) W`
C. `sqrt(3)/2 W`
D. `sqrt(3)/4 W`

Answer

B

Explanation

Given, work done = W and `theta = 60^(@)`
`W = MB(1- cos theta) rArr MB(1- cos 60^(@))`
`= (MB)/2`
Hence, torque
`|tau| = MB sin 60^(@) = sqrt(3) W`