Physics - Magnetism Question with Solution | TestHub

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 or $$\begin{array}{rl}E& =\frac{1}{2}m{v}^2=\frac{1}{2}m(R\omega {)}^2\\ & =\frac{1}{2}m{R}^2{B}^2\frac{q^2}{m^2}\\ E& =\frac{1}{2}\frac{R^2{B}^2{q}^2}{m}\end{array}$$ ... (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 $$\frac{q}{m}\propto \frac{1}{R^2}$$