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A magnt of magnetic dipole moment M is released in a uniform magnetic field of induction B from the position shown in the figure. Find : its maximum kinetic energy during the motion. |
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Answer» Solution : K.E. will be maximum where P.E. is MINIMUM. P.E. is minimum at `theta=0^(@)`. Now apply energy conservation between `theta=120^(@)andtheta=0^(@)`. `-mBcos120^(@)+0` `=-mBcos0^(@)+(KE)_(max)` `(KE)_(max)=(3)/(2)MB`. The K.E. is max at `theta = 0°` can also be proved by TORQUE METHOD. From `theta = 120°` to `theta = 0°` the torque always acts on the dipole in the same direction (here it is clockwise) so its K.E. keeps on increases till `theta=0°`. Beyond that reverses its direction and then K.E. starts decreasing `THEREFORE theta=0^(@)`is the orientation of M to here the maximum K.E. |
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