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Question

The kinetic energy acquired by a charged particle in a uniform magnetic field $B$ is 

$K=\frac{q^{2} B^{2} R^{2}}{2 m}\left(	ext { as } R=\fr

Vedclass · Accepted Answer

$1 $

Vedclass · Answer

The kinetic energy acquired by a charged particle in a uniform magnetic field $B$ is 

$K=\frac{q^{2} B^{2} R^{2}}{2 m}\left(	ext { as } R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}ight)$

where $q$ and $m$ are the charge and mass of the partic and $R$ is the radius of circular orbit

$	herefore $ The kinetic energy acquired by proton is

$K_{p}=\frac{q_{p}^{2} B^{2} R_{p}^{2}}{2 m_{p}}$

and that by the alpha particle is

$K_{\alpha}=\frac{q_{\alpha}^{2} B^{2} R_{\alpha}^{2}}{2 m_{\alpha}}$

Thus, $\frac{K_{\alpha}}{K_{p}}=\left(\frac{q_{\alpha}}{q_{p}}ight)^{2}\left(\frac{m_{p}}{m_{\alpha}}ight)\left(\frac{R_{\alpha}}{R_{p}}ight)^{2}$

or $\quad K_{\alpha}=K_{p}\left(\frac{q_{\alpha}}{q_{p}}ight)^{2}\left(\frac{m_{p}}{m_{\alpha}}ight)\left(\frac{R_{\alpha}}{R_{p}}ight)^{2}$

Here, $K_{p}=1$ $MeV$ $, \frac{q_{\alpha}}{q_{p}}=2, \frac{m_{p}}{m_{\alpha}}=\frac{1}{4}$

and $\frac{R_{\alpha}}{R_{p}}=1$

$	herefore \quad K_{\alpha}=(1\, \mathrm{MeV})(2)^{2}\left(\frac{1}{4}ight)(1)^{2}=1 \,\mathrm{MeV}$

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