A) \[\frac{1}{R}\]
B) \[\frac{1}{{{R}^{2}}}\]
C) \[{{R}^{2}}\]
D) \[R\]
Correct Answer: B
Solution :
The radius of the orbit in which ions moving is determined by the relation as given below \[\frac{m{{v}^{2}}}{R}=qvB\] where \[m\]is the mass, \[v\]is velocity, \[q\]is charge of ion and B is the flux density of the magnetic field, so that \[qvB\] is the magnetic force acting on the ion, and \[\frac{m{{v}^{2}}}{R}\] is the centripetal force on the ions moving in a curved path of radius \[R.\] The angular frequency of rotation of the ions about the vertical field B is given by \[\omega =\frac{v}{R}=\frac{qB}{m}=2\pi v\] where \[v\] is frequency. Energy of ion is given by \[E=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m{{(R\omega )}^{2}}\] \[=\frac{1}{2}m{{R}^{2}}{{B}^{2}}\frac{{{q}^{2}}}{{{m}^{2}}}\] or \[E=\frac{1}{2}\frac{{{R}^{2}}{{B}^{2}}{{q}^{2}}}{m}\] ?(i) If ions are accelerated by electric potential \[V,\] then energy attained by ions \[E=qV\] ?(ii) From Eqs. (i) and (ii), we get \[qV=\frac{1}{2}\frac{{{R}^{2}}{{B}^{2}}{{q}^{2}}}{m}\] or \[\frac{q}{m}=\frac{2V}{{{R}^{2}}{{B}^{2}}}\] If V and B are kept constant, then \[\frac{q}{m}\propto \frac{1}{{{R}^{2}}}\]
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