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A short bar magnet placed with its axis at 30^@ to a uniform magnetic field of 0.02 T experiences a torque of 0.060 Nm. (i) Calculate magnetic moment of the magnet, and (ii) find out what orientation of the magnet corresponds to its stable equilibrium in the magnetic field. |
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Answer» Solution :Here B= 0.02 T, `THETA= 30^@`and `tau = 0.060 Nm` (i) `THEREFORE ` Magnetic moment of the magnet `m =(tau)/(B sin theta) = (0.060)/(0.02 xx 0.5000) = 6 A m^2` (II) For stable equilibrium of the magnet in the magnetic field `theta = 0^@` |
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