Question

A deuteron of kinetic energy \(50 \mathrm{keV}\) is describing a circular orbit of radius \(0.5 \mathrm{~m}\), in a planeperpendicular to magnetic field \(B\). The kinetic energy of a proton that describes circular orbit of radius \(0.5 \mathrm{~m}\) in the same plane with the same magnetic field is

(a) \(200 \mathrm{keV}\)

(b) \(50 \mathrm{keV}\)

(c) \(100 \mathrm{keV}\)

(d) \(25 \mathrm{keV}\)

 

Answer 1

The correct option of this question will be (c).

Solution —

In this case, magnetic force provides necessary centripetal foree $i$.e.

$$q v B=\frac{m v^{2}}{r}$$

Radius of path, $r=\frac{m v}{B q}=\frac{\sqrt{2 m E}}{q B}$

$\therefore \quad r=\frac{\sqrt{2 m E}}{B q}=\frac{\sqrt{2 m_{1} E_{1}}}{B q}$

or $E_{1}=\frac{m E}{m_{1}}=\frac{\left(2 m_{1}\right)}{m_{1}} \times 50 \mathrm{keV} \quad\left[\because m=2 m_{1}\right]$ $=100 \mathrm{keV}$