InterviewSolution
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InterviewSolution
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A short bar magnet placed with its axis at 30^@ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 xx 10^(-2) J. What is the magnitude of magnetic moment of the magnet? |
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Answer» Solution :Here `THETA= 30^@ , B = 0.25 T and TAU = 4.5 xx 10^(-2) J` From the relation `tau= m B sin theta , ` we have Magnetic MOMENT of the MAGNET `m= (tau)/(B sin theta) = (4.5xx10^(-2))/(0.25xxsin 30^@) = 0.36 J T^(-1)` . |
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