Physics - Magnetism Question with Solution | TestHub

As charge on both proton and deuteron is same i.e. ' $e$ ' Energy acquired by both, $E=\mathrm{eV}$ For Deuteron. Kinetic energy, $\frac{1}{2}m{V}^2=eV$ [ $V$ is the potential difference] $$v=\sqrt{\frac{2eV}{m_d}}$$ But $m_d=2m$ Therefore, $v=\sqrt{\frac{2eV}{2m}}=\sqrt{\frac{eV}{m}}$ Radius of path, $R=\frac{mv}{eB}$ Substituting value of ' $v$ ' we get $$\begin{array}{r}R=\frac{2m\sqrt{\frac{ev}{m}}}{eB}\\ \frac{R}{2}=\frac{m\sqrt{\frac{ev}{m}}}{eB}\end{array}$$ For proton : $$\begin{array}{rl}& \frac{1}{2}m{V}^2=eV\\ & V=\sqrt{\frac{2eV}{m}}\\ & \text{ Radius of path, }{R}^{\prime }=\frac{mV}{eB}=\frac{m\sqrt{\frac{2eV}{m}}}{eB}\\ & R^{\prime }=\sqrt{2}\times \frac{R}{2}\text{ [From eq. (i)] }\\ & R^{\prime }=\frac{R}{\sqrt{2}}\end{array}$$