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A short bar magnet placed with its axis ...

A short bar magnet placed with its axis at `30^(@)` with a uniform external magnetic field of 0.35 T experiences a torque of magnitude equal to `4.5xx10^(-2)Nm`. The magnitude of magnetic moment of the given magnet is

A

1.36 J/T

B

0.06 J/T

C

0.36 J/T

D

0.60 J/T

Text Solution

Verified by Experts

The correct Answer is:
C
`tau=MB sintheta`
`M=(tau)/(Bsintheta)=(4.5xx10^(-2))/(0.25xxsin30)=0.36 Am^(2)`
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