If an electron revolves in the path of a circle of radius \[0.5 \times {10^{ - 10}}m\] at a frequency of \[5 \times {10^{15}}\] cycles/s, the electric current in the circle is (charge of an electron = \[1.6 \times {10^{ - 19}}\] )
A. \[0.4\,mA\]
B. \[0.8\,mA\]
C. \[1.2\,mA\]
D. \[1.6\,mA\]

Hint:The current is calculated by dividing charge by time where time is defined as the distance /speed. The magnetic induction at the centre of the circular path is given by the formula μ0times the current divided by twice the radius.

Formula Used:
Current, $I = \dfrac{q}{t}$
where $q$ is the charge flowing in $t$ period of time.
Total charge, $q = ne$
where $n$ is the number of electrons and $e$ is the charge on a single electron.
And $\text{time} = \dfrac{1}{\text{frequency}}$

Complete step by step solution:
Given: Radius of the circular path, $r = 0.5 \times {10^{ - 10}}m$
Frequency of the electron, $\nu = 5 \times {10^{15}}\,cycles/s$
Since, $\text{time} = \dfrac{1}{\text{frequency}}$ therefore,
$t = \dfrac{1}{{5 \times {{10}^{15}}}}$
Also, the number of electrons given is, $n = 1$.

Electrons can move freely inside the confines of the body when the nucleus is only lightly holding them. Because electrons are negatively charged particles, they cause a number of charges to flow when they move.
We know that, $I = \dfrac{q}{t}$ and $q = ne$
Therefore, we get, $I = \dfrac{{ne}}{t}$

Putting the known values in the above expression, we get,
$I = \dfrac{{1 \times \left( {1.6 \times {{10}^{ - 19}}} \right)}}{{\left( {\dfrac{1}{{5 \times {{10}^{15}}}}} \right)}} \\ $
Simplifying this, we get,
$I = \left( {5 \times {{10}^{15}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}} \right) \\ $
Thus, $I = 8 \times {10^{ - 4}}A$
That is, $I = 0.8\,mA$

Hence, option B is the correct answer.

Note: The magnetic induction produced at the centre of the circular path is due to charge carrying particles i.e., electrons. The electron revolves in a circular path. So, the distance covered is equal to the circumference of the circle.