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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) $25 \mathrm{keV}$

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

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