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| 1. |
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 ? |
| Answer» <html><body><p></p>Solution :(i) Apply energy conservation at `theta=120^(@)` and `theta=90^(@)` <br/> `-<a href="https://interviewquestions.tuteehub.com/tag/mb-546939" style="font-weight:bold;" target="_blank" title="Click to know more about MB">MB</a> cos 120^(@)+<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>` <br/> `=-MB cos 90^(@)+(K.E)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/ke-527890" style="font-weight:bold;" target="_blank" title="Click to know more about KE">KE</a>=(MB)/2` <br/> (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^(@)` <br/> `-mB cos 120^(@)+0` <br/> `=-mB cos 0^(@)+(K.E)_(max)` <br/> `(KE)_(max)=3/2MB` <br/> The `K.E` is max at `theta=0^(@)` can also be proved by torque method.From `theta=120^(@)` to `theta=0^(@)` the torque always <a href="https://interviewquestions.tuteehub.com/tag/acts-848461" style="font-weight:bold;" target="_blank" title="Click to know more about ACTS">ACTS</a> 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 <br/> `:. theta=0^(@)` is the orientation of `M` to here the maximum `K.E.` <br/> (iii) Since `theta` is not small. <br/> `:.`the motion is not `S.H.M` but it is <a href="https://interviewquestions.tuteehub.com/tag/oscillatory-2208020" style="font-weight:bold;" target="_blank" title="Click to know more about OSCILLATORY">OSCILLATORY</a> and periodic amplitude is `120^(@)`</body></html> | |