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An electron (mass $\mathrm{m}$ ) is accelerated through a potential difference of ' $\mathrm{V}$ ' and then it enters in a magnetic field of induction ' $\mathrm{B}$ ' normal to the lines. The radius of the circular path is (e = electronic charge $)$
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Verified Answer
The correct answer is:
$\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^2}}$
Radius of circular path in a cyclotron is given by
$\mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}$
Here $\mathrm{q}=\mathrm{e}$,
$\therefore \quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}... (i)$
On entering the field,
$\begin{aligned}
& \mathrm{KE}=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2 \\
\therefore \quad \mathrm{V} & =\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}... (ii)
\end{aligned}$
Putting (ii) into (i),
$\mathrm{R}=\frac{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{eB}}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^2}}$
$\mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}$
Here $\mathrm{q}=\mathrm{e}$,
$\therefore \quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}... (i)$
On entering the field,
$\begin{aligned}
& \mathrm{KE}=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2 \\
\therefore \quad \mathrm{V} & =\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}... (ii)
\end{aligned}$
Putting (ii) into (i),
$\mathrm{R}=\frac{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{eB}}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^2}}$
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