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A bar magnet is hung by a thin cotton th...

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 torque required to keep the magnet in this new position is

A

`(W)/(sqrt(3))`

B

`sqrt(3)W`

C

`(sqrt(3)W)/(2)`

D

`(2W)/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
B
`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`
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