A proton and an $\alpha - $particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25$ $\mu \, sec$ to make $5$ revolutions, then the periodic time for the $\alpha - $ particle would be........$\mu \, sec$
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(c) Time period of proton ${T_p} = \frac{{25}}{5} = 5\,\mu \,\sec $
By using $T = \frac{{2\pi \,m}}{{qB}}$$ \Rightarrow $$\frac{{{T_\alpha }}}{{{T_p}}} = \frac{{{m_\alpha }}}{{{m_p}}} \times \frac{{{q_p}}}{{{q_\alpha }}}$$ = \frac{{4{m_p}}}{{{m_p}}} \times \frac{{{q_p}}}{{2{q_p}}}$
$ \Rightarrow \,$ ${T_\alpha } = 2{T_p} = 10\,\mu \,\,\sec .$
By using $T = \frac{{2\pi \,m}}{{qB}}$$ \Rightarrow $$\frac{{{T_\alpha }}}{{{T_p}}} = \frac{{{m_\alpha }}}{{{m_p}}} \times \frac{{{q_p}}}{{{q_\alpha }}}$$ = \frac{{4{m_p}}}{{{m_p}}} \times \frac{{{q_p}}}{{2{q_p}}}$
$ \Rightarrow \,$ ${T_\alpha } = 2{T_p} = 10\,\mu \,\,\sec .$
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