An electron in an atom revolves around the nucleus in an orbit of radius $0.53A^\circ $. Calculate the equivalent magnetic moment if the frequency of revolution of electron is $6.8 \times {10^9}MHz$.

Hint: The magnetic moment is a determination of its tendency to get arranged through a magnetic field. Electric charges when in motion produce a magnetic moment, and this is called induced magnetism.
Formulas used: We will be using the formula of Magnetic moment of a loop given by $M = NIA$ where $M$ is the magnetic moment induced by the loop, $N$ is the number of turns in the loop, $I$ is the electric current in the loop, and $A$ is the area of the loop.
Here since the loop is a circular orbit of the electron, $A = \pi {R^2}$ where $R$ is the radius of the loop/ the circular orbit of the electron around the nucleus.
We will also be using the formula, $I = \dfrac{q}{T}$ where $I$ is the electric current in the loop, $q$ is the charge flowing through the conductor, $T$ is the time period required for the charge to flow through the conductor.
We also know that the charge $q$ here is the charge of the electron hence, $q = e = 1.602 \times {10^{ - 19}}C$ .

Complete Step by Step answer:
We know that a charge carrying conductor, or a charge itself moving along a path has the tendency to induce magnetism and vice versa. If you look at the case in this situation, we can see that an electron, which carries a negative charge of magnitude, $q = e = 1.602 \times {10^{ - 19}}C$ is revolving around a nucleus. It follows a circular orbit of radius $R = 0.5A^\circ $ . Thus we can say that there is magnetism induced due to the revolving electron.
The magnetic moment thus produced can be found by the formula, $M = NIA$. Here we know that the loop is a single circular orbit, thus the number of turns of the loop will be, $N = 1$. Thus altering the formula to be, $M = IA$
We also know that electric current flowing through a conductor is given by, $I = \dfrac{q}{T}$. We also know that the frequency is the inverse of time, $v = \dfrac{1}{T}$. Substituting it in the above equation we get, $I = qv$ .
Now solving to find a magnetic moment, $M$ and substituting the values of, $M = qv \times A$. We also know that the area of a circular loop is given by, $A = \pi {R^2}$ and $R = 0.5A^\circ = 0.5 \times {10^{ - 10}}m$.
$M = 1.6 \times {10^{ - 19}} \times 6.8 \times {10^9} \times {10^6} \times \pi \times {\left( {0.5} \right)^2} \times {\left( {{{10}^{ - 10}}} \right)^2}$
Solving the above equation, we get,
$M = 9.6 \times {10^{ - 24}}A{m^{ - 2}}$


Note: Magnetic moment is a vector quantity and denotes both the strength and orientation of the object that produces the magnetic field, which in this case can be found out by the