A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60^(@) is W. Now the torrue required to keep the magnet in this new position is

Options

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

Answer

B

Explanation

`W=MB(cos^(@)-cos60^(@))`
`W=MB(1-(1)/(2))=(MB)/(2)`
required torque for this position
`tau=Mbsintheta`
`=MBsin60^@`
`=(sqrt(3))/(2)MB=sqrt(3)W`