Question

An electron is revolving around a proton, producing a magnetic field of $16{\text{Wb m}}^{-2}$ in a circular orbit of radius $1\mathring{\text{A}}.$ It's angular velocity will be

• A. ${10}^{17}{\text{rad s}}^{-1}$
• B. $\frac{1}{2\pi }{10}^{12}{\text{rad s}}^{-1}$
• C. $2\pi \times {10}^{12}{\text{rad s}}^{-1}$
• D. $4\pi \times {10}^{12}{\text{rad s}}^{-1}$

Answer 1

The correct option is A ${10}^{17}{\text{rad s}}^{-1}$

Given, $B=16{\text{Wb m}}^{-2}r=1\mathring{A}$

Magnetic field due to revolution of electron,

$B=\frac{{\mu }_{0}i}{2r}=\frac{{\mu }_0\left(\frac{e\omega }{2\pi }\right)}{2r}\left\{\begin{array}{c}\because I=\frac{q}{t}\text{and}t=\frac{2\pi }{\omega }\&\\ {\mu }_0=4\pi \times {10}^{-7}{\text{Hm}}^{-1}\end{array}\right\}$

$\Rightarrow B=\frac{{10}^{-7}\times e\times \omega }{r}$

$16=\frac{{10}^{-7}\times 1.6\times {10}^{-19}\times \omega }{1\times {10}^{-10}}$

$\therefore \omega ={10}^{17}{\text{rad s}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.