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A bar magnet is held perpendicular to a ...

A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is

A

`30^(@)`

B

`60^(@)`

C

`45^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
Couple acting on the magnet
`tau_(1) =MB sin theta_(1)=MB sin 90^(@) =MB`
`tau_(2) =MB sin theta_(2)=tau_(1)/2=(MB)/2`
`:. sin theta_(2) =1/2 :. theta_(2)=30^(@)`
Thus it should be rotated from `theta_(1)=90^(@)` to `theta_(2)=30^(@)` i.e. through `60^(@)`.
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