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A magnet suspended in a uniform magnetic...

A magnet suspended in a uniform magnetic field of induction `0.2" Wb/m"^(2)`, makes an angle of `60^(@)` with the normal to the direction of the field. What is the magnetic moment of the magnet, if a torque of moment `15xx10^(-2)` Nm acts on it?

A

`1 Am^(2)`

B

`1.25 Am^(2)`

C

`1.5 Am^(2)`

D

`2 Am^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`theta=90^(@)-60^(@)=30^(@)=` angle between magnetic field and magnetic moment
`tau =MB sin theta`
`M=tau/(B sin theta)=(15xx10^(-2))/(2xx10^(-1) xx sin 30^(@))`
`=15xx10^(1)=1.5 Am^(2)`
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