Question 11 Mark
An electron is revolving around a proton in a circular orbit of diameter $0.1 \ nm$ . It produces a magnetic field of $14 Wb / m ^2$ at the proton. What is angular speed of the electron?
Answer
View full question & answer→$(a): 4.4 \times 10^{16} rads ^{-1}$
Explanation : The revolving electron is similar to a loop carrying current.
Field at the center of the loop of radius $r$ is $B=\frac{\mu_0 I}{2 r}$.
The current due to the revolving electron $I=\frac{B(2 r)}{10}$
$=\frac{14 \times 0.1 \times 10^{-9}}{4 \times 10^{-7}}$
$=\frac{7 \times 10^{-3}}{2 \pi}$
The current can also be written as, $I=\frac{e}{T}$
where $,T$ is the time taken to complete one revolution.
Since $T=\frac{2 \pi}{\omega}$ where $\omega$ is the angular speed of the electron,
$I=\frac{\epsilon}{T}=\frac{\epsilon \omega}{2 \pi}$
$\frac{\epsilon \omega}{2 \pi}=\frac{7 \times 10^{-3}}{2 \pi}$
$\omega=\frac{7 \times 10^{-3}}{\epsilon}$
$=\frac{7 \times 10^{-3}}{1.5 \times 10^{-19}}$
$=4.38 \times 10^{16} \approx 4.4 \times 10^{16} rad / s $
Explanation : The revolving electron is similar to a loop carrying current.
Field at the center of the loop of radius $r$ is $B=\frac{\mu_0 I}{2 r}$.
The current due to the revolving electron $I=\frac{B(2 r)}{10}$
$=\frac{14 \times 0.1 \times 10^{-9}}{4 \times 10^{-7}}$
$=\frac{7 \times 10^{-3}}{2 \pi}$
The current can also be written as, $I=\frac{e}{T}$
where $,T$ is the time taken to complete one revolution.
Since $T=\frac{2 \pi}{\omega}$ where $\omega$ is the angular speed of the electron,
$I=\frac{\epsilon}{T}=\frac{\epsilon \omega}{2 \pi}$
$\frac{\epsilon \omega}{2 \pi}=\frac{7 \times 10^{-3}}{2 \pi}$
$\omega=\frac{7 \times 10^{-3}}{\epsilon}$
$=\frac{7 \times 10^{-3}}{1.5 \times 10^{-19}}$
$=4.38 \times 10^{16} \approx 4.4 \times 10^{16} rad / s $
