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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)/(sqrt3)`

B

`sqrt3W`

C

`(sqrt3W)/(2)`

D

`(2W)/(sqrt3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Work done in rotating the magnet
`" " W=MB(cos theta_(0)-cos theta)`
where, M =magnetic moment of the magnet and
`" " "B=magnetic field"`
`W=MB(cos 0^(@)-cos60^(@))=MB(1-(1)/(2)) Rightarrow (MB)/(2)`
`therefore MB=2W`
Torque on a magnet in this position is given by, `tau =MxxB=MB.sin theta=2W.sin60 " " "["From Eq. (i)"]" `
`=2W (sqrt3)/(2)=Wsqrt3`
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