Question 289: Electromagnetic Induction

Question

An electron (mass $$\mathrm{m}$$ ) is accelerated through a potential difference of '$$V$$' and then it enters in a magnetic field of induction '$$B$$' normal to the lines. The radius of the circular path is ($$\mathrm{e}=$$ electronic charge)

Accepted Answer

Radius of circular path in a cyclotron is given by
$$\mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}$$
Here $$\mathrm{q}=\mathrm{e}$$,
$$	herefore \quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}$$ ..... (i)
On entering the field,
$$\begin{aligned}
& \mathrm{KE}=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2 \
& 	herefore \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \quad 	ext{.... (ii)}\
& 	ext { Putting (ii) into (i), } \
& \mathrm{R}=\frac{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{eB}}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^2}}
\end{aligned}$$

Answer

$$\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$$

Answer

$$\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^2}}$$

Answer

$$\sqrt{\frac{2 V m}{e B}}$$

Answer

$$\sqrt{\frac{2 V m}{\mathrm{e}^2 \mathrm{~B}}}$$

Electromagnetic Induction

An electron (mass $\mathrm{m} ) is accelerated through a potential difference of 'V' and then it enters in a magnetic field of induction 'B' normal to the lines. The radius of the circular path is (\mathrm{e}=$ electronic charge)

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