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 of 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 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 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 \[E=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m{{(R\omega )}^{2}}\] \[=\frac{1}{2}m{{R}^{2}}{{B}^{2}}\frac{{{q}^{2}}}{{{m}^{2}}}\] 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 \[qV=\frac{1}{2}\frac{{{R}^{2}}{{B}^{2}}{{q}^{2}}}{m}\] If \[V\] and \[B\] are kept constant, then \[\frac{q}{m}\propto \frac{1}{{{R}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
