InterviewSolution
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InterviewSolution
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A magnet of magnetic dipole moment M is released in a uniform magnetic field of induction B from the position shown in the figure.Find (i) Its kinetic energy at theta=90^(@) (ii)its maximum kinetic energy during the motion. (iii) will it perform SHM? oscillation? Periodic motion? What is its amplitude ? |
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Answer» Solution :(i) Apply energy conservation at `theta=120^(@)` and `theta=90^(@)` `-MB cos 120^(@)+0` `=-MB cos 90^(@)+(K.E)` `KE=(MB)/2` (ii) `K.E.` will be maximum where `P.E.` is minimum. `P.E` is minimum at `theta=0^(@)`.Now apply energy conservation between `theta=120^(@)` and `theta=0^(@)` `-mB cos 120^(@)+0` `=-mB cos 0^(@)+(K.E)_(max)` `(KE)_(max)=3/2MB` 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 increase till `theta=0^(@)`, Beyond that reverses its direction and then `K.E.` starts decreasing `:. theta=0^(@)` is the orientation of `M` to here the maximum `K.E.` (iii) Since `theta` is not small. `:.`the motion is not `S.H.M` but it is OSCILLATORY and periodic amplitude is `120^(@)` |
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