A magnetic needle lying parallel to a magnetic field is turned through `60^(@)`. The work done on it is W. The torque required to maintain the magnetic needle in the position mentioned above is

Options

A. `sqrt3w`
B. `(sqrt3)/(2)W`
C. `(W)/(2)`
D. `2w`

Answer

(a)

Explanation

Work done,
  `W = MB (1-cos theta) MB (1 - cos 60^(@)) = MB//2` Torque,
  `tau=MB sin theta=MB sin 60^(@)`
`=MBsqrt3//2`
`tau=sqrt 3W`