A proton of energy 8, eV is moving in a circular path in a uniform magnetic field. energy of an alph

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4.Moving Charges and Magnetism

medium

A proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

A

$4$

B

$2$

C

$8$

D

$6$

Solution

(c) $r = \frac{{\sqrt {2mK} }}{{qB}} \Rightarrow q \propto \sqrt {mK} \Rightarrow K \propto \frac{{{q^2}}}{m}$
$ \Rightarrow \frac{{{K_\alpha }}}{{{K_p}}} = {\left( {\frac{{{q_\alpha }}}{{{q_p}}}} \right)^2} \times \frac{{{m_p}}}{{{m_\alpha }}} \Rightarrow \frac{{{K_\alpha }}}{8} = {\left( {\frac{{2{q_p}}}{{{q_p}}}} \right)^2} \times \frac{{{m_p}}}{{4{m_p}}} = 1$
$ \Rightarrow {K_\alpha } = 8\;eV$

Standard 12

Physics

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