InterviewSolution
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InterviewSolution
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A bar magnet of magnetic moment 1.5JT^(-1) lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii) ? |
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Answer» Solution :Here `m= 1.5 J T^(-1) , B = 0.22 T and theta_1 = 0^2` (a) (i) Work done to turn the MAGNET , so as to ALIGN its magnetic moment normal to the magnetic field i.e. `theta_2 =90^@ ` , is given as : `W= mB[cos theta_1 - cos theta_2] = 1.5xx0.22 [cos 0^@ - cos 90^@]` `=1.5xx0.22xx[1-0] =0.33J`. (ii) Work done to turn the magnet so as to align its magnetic moment opposite to the field direction i.e. `theta_2 = 180^@` `W = 1.5xx0.22xx[cos 0^@ - cos 180^@] =1.5xx0.22xx[1-(-2)] =0.66J` (b) TORQUE in case (i) `tau= m B sin theta_2 = 1.5xx0.22xx SIN90^@ = 1.5xx0.22xx1 = 0.33 ` N m Torque in case (ii) = `tau = m B sin theta_2 = 1.5xx0.22xxsin 180^@ = 1.5xx0.22xx0=0` |
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