A) \[\frac{qB}{m\pi }\]
B) \[\frac{qB}{2\pi m}\]
C) \[\frac{qBM}{2\pi m}\]
D) \[\frac{qB}{2\pi E}\]
Correct Answer: B
Solution :
| [b] Key idea: For a charged particle to move in a circular path in a magnetic field, the magnetic force on charge particle provides the necessary centripetal force. |
| Hence, magnetic force = centripetal force |
| i.e., \[qvB=\frac{m{{v}^{2}}}{r}\] |
| or \[qvB=mr{{\omega }^{2}}\] \[(v=n\omega )\] |
| or \[{{\omega }^{2}}=\frac{qvB}{mr}=\frac{q(r\omega )B}{mr}\] |
| or \[\omega =\frac{qB}{m}\] |
| If v is the frequency of rotation, then |
| \[\omega =2\pi v\,\Rightarrow \,\,v=\frac{\omega }{2\pi }\] |
| \[\therefore \] \[v=\frac{qB}{2\pi m}\] |
| Note: In the resultant expression \[\frac{q}{m}\] is known as specific change. It is sometimes denoted by \[\alpha \]. So, in terms of \[\alpha ,\]the above formula can be written as |
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