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A bar magnet of magnetic moment 1.5 JT^(-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 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 :`m = 1.5 "JT"^(-1), B = 0.22T` a. i mB = 0.330 J `""ii.2mB=0.66J` b. i. Maximum = 0.33 , TENDING to turn it towards the direction of the field. ii.Zero`(SIN 180^@=0)` |
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