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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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Knowledge Check

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