Question

An electron is moving around a proton in an orbit of radius 1Å and produces $16Wb/{{m}^{2}}$ of magnetic field at the center, then finds the angular velocity of the electron.
A. $20\pi \times {{10}^{16}}rad/\sec $
B. ${{10}^{17}}rad/\sec $
C. $\dfrac{5}{2\pi }\times {{10}^{16}}rad/\sec $
D. $\dfrac{5}{4\pi }\times {{10}^{16}}rad/\sec $

Answer 1

Hint: This problem can be solved by considering the electron as a current carrying circular coil and using the formula for the magnetic field at the center of a circular current carrying coil. The electron constitutes a current since it is undergoing uniform circular motion and it is a charged body.

Formula used:
If an electron goes around in a circle of radius $R$, at angular speed $\omega $, the current $I$ constructed is
$I=\dfrac{e}{\dfrac{2\pi }{\omega }}=\dfrac{e\omega }{2\pi }$
Where $e=1.6\times {{10}^{-19}}C$ is the magnitude of the charge carried by the electron.
The magnetic field due to a current carrying coil of radius $R$ at its center is given by
$B=\dfrac{{{\mu }_{0}}I}{2R}$
where ${{\mu }_{0}}=4\pi \times {{10}^{7}}m.kg.{{s}^{-2}}{{A}^{-2}}$ and $I$ is the current in the coil.

Complete step by step answer:
We can solve this problem by considering that as the electron goes in a circular path, it constitutes a current and behaves as a circular current carrying coil. Then we can apply the formula for the magnetic field at the center of a circular current carrying coil to get the required answer.
Now, current is nothing but net charge passing per unit time. Therefore, in the circle, charge carried by the electron is passing through a specific location in one time period.
Let the time period of the rotation be $T$ and the angular speed be $\omega $. Therefore, obviously,
$T=\dfrac{2\pi }{\omega }$ (by definition) -(1)
Therefore, the current $I$ will be
$\dfrac{e}{T}$
Where $e=1.6\times {{10}^{-19}}C$ is the magnitude of the charge carried by the electron.
Using (1), we get,
$I=\dfrac{e}{\dfrac{2\pi }{\omega }}=\dfrac{e\omega }{2\pi }$ --(2)
Now,
the magnetic field due to a current carrying coil of radius $R$ at its center is given by
$B=\dfrac{{{\mu }_{0}}I}{2R}$ --(3)
where ${{\mu }_{0}}=4\pi \times {{10}^{7}}m.kg.{{s}^{-2}}{{A}^{-2}}$ and $I$ is the current in the coil.
According to the question,
$B=16Wb/{{m}^{2}}$
$R=1$Å $={{10}^{-10}}m$ (1Å$={{10}^{-10}}m$ )
Therefore, using these values and (2) in (3), we get,
$16=\dfrac{4\pi \times {{10}^{-7}}}{2\times {{10}^{-10}}}\times \dfrac{1.6\times {{10}^{-19}}\omega }{2\pi }$
$\therefore 16=\dfrac{{{10}^{-7}}}{{{10}^{-10}}}\times 1.6\times {{10}^{-19}}\omega $
$\therefore 16=1.6\times {{10}^{-16}}\omega $
$\therefore \omega =\dfrac{16}{1.6\times {{10}^{-16}}}=\dfrac{10}{{{10}^{-16}}}={{10}^{17}}rad/\sec $
Hence, the required value of the angular velocity is ${{10}^{17}}rad/\sec $.
Therefore, the correct option is B) ${{10}^{17}}rad/\sec $.

Note: It is not quite intuitive at first reading that an electron revolving around a nucleus can be treated as a circular current carrying coil, but when thought about carefully and understanding the explanation in the answer, it must be very clear to students why the electron produces an electric current and can be considered a current carrying circular coil. The magnetic field thus produced is responsible for giving certain magnetic moments to the atom. This effect is one of the characteristics of the electron in the quantum mechanical model of the atom and is known as the electron’s magnetic spin or magnetic quantum number.