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A bar magnet of magnetic moment 6 J/T is aligned at 60^@ with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii). |
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Answer» Solution :Here MAGNETIC moment of bar magnet m = 6 J/T , uniformexternal magnetic field B = 0.44 T and `theta_1 = 60^@` . (a) (i) The work done in turning the magnet to ALIGN its magnetic normal to the magnetic field [i.e. , `theta_2 =90^@`) will be W = mB `[cos theta_1 - cos theta_2] =6xx 0.44 XX [ cos 60^@ - cos 90^@] =6xx0.44 xx[1/2-0]` (ii) The work done in turning the magnet to align its magnetic moment opposite to the magentic field i.e., `THETA=180^@`] will be `W= mB [ cos theta_1 - cos theta_2] = 6 xx 0.44 [cos 60^@ - cos 180^@] = 6 xx 0.44 xx[1/2 -(-1)]=3.96 J` (b) The torque on the magnet in the final orientation in case (ii) [i.e. , when `theta=180^@`] `tau= mB sib theta =6 xx0.44 xx sin 180^@=0` |
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