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A charged particle of charge q and mass m enters perpendicularly in a magnetic field \mathrm{\underset{B}{\rightarrow}} . Kinetic energy of the particle is E; then frequency of rotation is

Option: 1

\frac{\mathrm{qB}}{\mathrm{m} \pi}


Option: 2

\frac{\mathrm{qB}}{2 \pi \mathrm{m}}


Option: 3

\mathrm{\frac{q B E}{2 \pi m} }


Option: 4

\frac{\mathrm{qB}}{2 \pi \mathrm{E}}


Answers (1)

best_answer

For circular path in magnetic field,
\mathrm{ \mathrm{mr} \omega^2=q v B \\ }

\mathrm{ \Rightarrow \omega^2=\frac{\mathrm{qvB}}{\mathrm{mr}} \text { As } v=r \omega \\ }

\mathrm{ \omega^2=\frac{\mathrm{q}(\mathrm{r} \omega) \mathrm{B}}{\mathrm{mr}} \Rightarrow \omega=\frac{\mathrm{qB}}{\mathrm{m}} }
$$
\mathrm{ \therefore } If v is frequency of roatation, then
\mathrm{ \mathrm{v}=\frac{\omega}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{qB}}{2 \pi \mathrm{m}} }

Posted by

shivangi.bhatnagar

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