A) \[\frac{1}{R}\]
B) \[\frac{1}{{{R}^{2}}}\]
C) \[{{R}^{2}}\]
D) R
Correct Answer: B
Solution :
Key Idea: Centripetal force is provided by the magnetic force qvB. 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 die 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 ion 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 \[E=\frac{1}{2}\,m{{v}^{2}}=\frac{1}{2}m{{(R\omega )}^{2}}\] \[=\frac{1}{2}\,m{{R}^{2}}{{B}^{2}}\frac{{{q}^{2}}}{{{m}^{2}}}\] \[orE=\frac{1}{2}\,\frac{{{R}^{2}}\,{{B}^{2}}\,{{q}^{2}}}{m}....(i)\] 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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