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A charged particle of charge e and mass $m$ is moving in an electric field $\mathrm{E}$ and magnetic field $\mathrm{B}$. Construct dimensionless quantities and quantities of dimension $[\mathrm{T}]^{-1}$.
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Verified Answer
When a charge particle in an electric and magnetic field its motion will be circular.
The charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
Centripetal force $=\frac{\mathrm{mv}^2}{\mathrm{R}}$ is balanced by magnetic force
$$
\left(F_m\right)=q v B \sin 90=q v B
$$
So, $\quad \frac{\mathrm{mv}^2}{\mathrm{R}}=\mathrm{qvB}$
By solving, we get
$$
\therefore \quad \frac{\mathrm{qB}}{\mathrm{m}}=\frac{\mathrm{v}}{\mathrm{R}}=\omega,
$$
Then dimensional formula of angular velocity $(\omega)$
$$
\therefore \quad[\omega]=\left[\frac{\mathrm{eB}}{\mathrm{m}}\right]=\left[\frac{\mathrm{v}}{\mathrm{R}}\right]=[\mathrm{T}]^{-1}
$$
This is the required expression.
The charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
Centripetal force $=\frac{\mathrm{mv}^2}{\mathrm{R}}$ is balanced by magnetic force
$$
\left(F_m\right)=q v B \sin 90=q v B
$$
So, $\quad \frac{\mathrm{mv}^2}{\mathrm{R}}=\mathrm{qvB}$
By solving, we get
$$
\therefore \quad \frac{\mathrm{qB}}{\mathrm{m}}=\frac{\mathrm{v}}{\mathrm{R}}=\omega,
$$
Then dimensional formula of angular velocity $(\omega)$
$$
\therefore \quad[\omega]=\left[\frac{\mathrm{eB}}{\mathrm{m}}\right]=\left[\frac{\mathrm{v}}{\mathrm{R}}\right]=[\mathrm{T}]^{-1}
$$
This is the required expression.
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