Question

The ratio of magnetic force $\left( {{F}_{m}} \right)$ and electric force $\left( {{F}_{e}} \right)$ acting on a moving charge is
$\begin{aligned} & \text{A}\text{. }{{\left( \dfrac{V}{C} \right)}^{2}} \\\ & \text{B}\text{. }{{\left( \dfrac{C}{V} \right)}^{2}} \\\ & \text{C}\text{. }\dfrac{V}{C} \\\ & \text{D}\text{. }\dfrac{C}{V} \\\ \end{aligned}$

Answer 1

If a charge is moving in a region having both electric field and magnetic field then the charge will experience two forces one is due to electric field and one is due to magnetic field. If the region has an electric field and magnetic field then the electric force and magnetic force can be calculated. And also their ratio can be calculated.

Formulas used:
Electric field due to a charge $q$ at a distance $r$ is $E=\dfrac{1}{4\pi {{\epsilon }_{0}}}\dfrac{q}{{{r}^{2}}}$
Magnetic field is given by $B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{qv\sin \theta }{{{r}^{2}}}$
Electric force, ${{F}_{e}}=qE$
And magnetic force ${{F}_{m}}=qvB\sin \theta$

Complete step by step answer:
If a charge of charge $q$ is moving inside the field of a charge ${{q}_{2}}$
The electric field due to charge ${{q}_{2}}$ at a distance $r$ is given by
$E=\dfrac{1}{4\pi {{\epsilon }_{0}}}\dfrac{{{q}_{2}}}{{{r}^{2}}}$
Then the force on charge $q$due to charge ${{q}_{2}}$ which is at a distance $r$ from $q$ is given by
${{F}_{e}}=qE=\dfrac{1}{4\pi {{\epsilon }_{0}}}\dfrac{q{{q}_{2}}}{{{r}^{2}}}$ .
And the magnetic force is given by
$\begin{aligned} & {{F}_{m}}=qvB\sin \theta =qv\times \dfrac{{{\mu }_{0}}}{4\pi }\dfrac{{{q}_{2}}v\sin \theta }{{{r}^{2}}}=q{{q}_{2}}{{v}^{2}}\times \dfrac{{{\mu }_{0}}}{4\pi }\dfrac{\sin 90{}^\circ }{{{r}^{2}}} \\\ & \Rightarrow {{F}_{m}}=q{{q}_{2}}{{v}^{2}}\times \dfrac{{{\mu }_{0}}}{4\pi }\dfrac{1}{{{r}^{2}}} \\\ \end{aligned}$
Now,
$\dfrac{{{F}_{e}}}{{{F}_{m}}}=\dfrac{\left( \dfrac{1}{4\pi {{\epsilon }_{0}}}\dfrac{q{{q}_{2}}}{{{r}^{2}}} \right)}{\left( q{{q}_{2}}{{v}^{2}}\times \dfrac{{{\mu }_{0}}}{4\pi }\dfrac{1}{{{r}^{2}}} \right)}=\dfrac{1}{{{\mu }_{o}}{{\epsilon }_{0}}}\times \dfrac{1}{{{v}^{2}}}$
But $\dfrac{1}{{{\mu }_{o}}{{\epsilon }_{0}}}={{c}^{2}}$ where ${{c}^{2}}$ is the velocity of light in vacuum.
So
$\dfrac{{{F}_{e}}}{{{F}_{m}}}=\dfrac{1}{{{\mu }_{o}}{{\epsilon }_{0}}}\times \dfrac{1}{{{v}^{2}}}=\dfrac{{{c}^{2}}}{{{v}^{2}}}={{\left( \dfrac{c}{v} \right)}^{2}}$
So the correct option is A.
Additional Information: A particle moving in a region having both electric field and magnetic field will experience a force called Lorentz force which is the sum of electric force and magnetic force. If the particle have charge $q$ and moving with velocity $v$ in a region having electric field $E$ and magnetic field $B$ then the Lorentz force is given by
$\overrightarrow{F}=q\left( \overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B} \right)$

Note:
The direction of Lorentz force depends upon both the electric field and magnetic field. The direction of magnetic force is perpendicular to the plane containing the velocity of the particle and the magnetic field. If the angle between the velocity and the magnetic field is $90{}^\circ$ then the particle will follow a circular path and if the angle is less than $90{}^\circ$ then the particle will follow a helical path. If the angle is zero then the particle will move along the field.