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A short bar magnet placed with its axis at 30^@ with an external field of0.25 T experiences a torque of magnitude to 4.5xx10^(-2)J . What is the magnitude of magnetic moment of the magnet ? |
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Answer» SOLUTION :`theta=30^@, B = 0.25T, tau=4.5xx10^(-2)J, ""m=? ""tau=mBsintheta` `:.m=tau/(BSINTHETA)=(4.5xx10^(-2))/(0.25xxsin30^@)=(4.5xx10^(-2)xx2)/(0.25)=36.0xx10^(-2)=0.36Am^2(" or" JT^(-1))` |
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