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A short bar magnet is kept in uniform magnetic field of `0.16` T such that dipole moment vector is at angle `30^(@)` with the field . The torque acting on the bar magnet is 0.032 N m . What is the magnitude of magnetic dipole moment of the magnet ? |
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Answer» `tau = MB sin theta` `0.032 = M xx 0.16 xx (1)/(2)` `0.064 = M xx 0.16` `M = (6.4)/(16) = 0.4 Am^(2)` |
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