A charged particle is moving in a uniform magnetic field in a circular path with radius ' $R$ '. When the…
A charged particle is moving in a uniform magnetic field in a circular path with radius ' $R$ '. When the energy of the particle is doubled, then the new radius will be
$\frac{\mathrm{R}}{\sqrt{2}}$
2 R
$\frac{\mathrm{R}}{2}$
$\sqrt{2} R$
Solution
Force on a charged particle moving in a circular
$\begin{array}{ll}
& \text { path, } F=q v B=\frac{m v^2}{r} \\
\therefore \quad & r=\frac{m v}{q B}=\frac{\sqrt{2 m(K \cdot E)}}{q B} \\
\therefore \quad & r \propto \sqrt{K \cdot E} \\
\therefore & \frac{R}{R^{\prime}}=\sqrt{\frac{K \cdot E_1}{K \cdot E_2}} \\
\therefore \quad & \frac{R}{R^{\prime}}=\sqrt{\frac{K \cdot E_1}{2 K \cdot E_1}} \\
\therefore \quad & R^{\prime}=\sqrt{2} R
\end{array}$