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 |
You need to login to perform this action.
You will be redirected in
3 sec