Q. An electron is revolving around a proton in a circular path of diameter 0.1 nm. It produces a magnetic field 14 T at a proton. Then the angular speed of the electron is
A) $8.8\ \times 10^{6}\ rad\ s^{- 1}$
B) $4.4\ \times 10^{16}\ rad\ s^{- 1}$
C) $2.2\ \times 10^{16}\ rad\ s^{- 1}$
D) $1.1\ \times 10^{16}\ rad\ s^{- 1}$
Explanation
(b) Here, $2r = 0.1\ nm = 0.1\ \times 10^{- 9}\ m = 10^{- 10}\ m;$
$i = \frac{e}{T} = \frac{e\omega}{2\pi}\ $
Now, $B = \frac{\mu_{0}}{4\pi}\frac{2\pi ni}{r} = \frac{\mu_{0}}{4\pi}\frac{2\pi n}{r}\ \left( \frac{e\omega}{2\pi} \right)$
$= \frac{\mu_{0}}{4\pi}\frac{ne\omega}{r}$
Or $\ \omega = B.\left( \frac{4\pi}{\mu_{0}} \right) \times \frac{r}{ne}$
$= 14\ \times \frac{1}{10^{- 7}} \times \ \frac{\left( 10^{- 10} \right)/2}{1 \times 1.6 \times 10^{- 19}}$
$= 4.4 \times 10^{16}\ rads^{- 1}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.