InterviewSolution
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InterviewSolution
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A short bar magnet placed with its axis at 30^@ to a uniform magnetic field of 0.2 T experiences a torque of 0.060 Nm. i. Calculate magnetic moment of the magnet ii. Find out what orientation of the magnet corresponds to its stable equilibrium in the magnetic field. |
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Answer» Solution :Here `TAU =0.060 Nm, theta = 30^@ , B = 0.2 T` As `tau =m B sin theta "" 0.060 = mxx 0.2 XX sin 30^@` `therefore m = (0.060)/(0.2 xx 0.5) = 0.06JT^(-1)` ii PE of MAGNET in field `vec B` is GIVEN by U=-m cos `theta` When magnet is paced parallel to `vec B , theta = 0^@` and `U = mB cos 0^(@)`=- mB i.e , the PE of the dipole is minimum. This is the STATE of stable equilibrium. |
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