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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 isA. 1.36 J/TB. 0.06 J/TC. 0.36 J/TD. 0.60 J/T |
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Answer» Correct Answer - C `tau=MB sintheta` `M=(tau)/(Bsintheta)=(4.5xx10^(-2))/(0.25xxsin30)=0.36 Am^(2)` |
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