Q. An electron is revolving round a proton, producing a magnetic field of $16\ weber/m^{2}$ in a circular orbit of radius $1\mathring{\mathrm{A}}$. It’s angular velocity will be
A) $10^{17}rad/sec$
B) $1/2\pi \times 10^{12}rad/sec$
C) $2\pi \times 10^{12}rad/sec$
D) $4\pi \times 10^{12}rad/sec$
Explanation
(a) Magnetic field due to revolution of electron
$B = \frac{\mu_{0}}{4\pi}.\frac{2\pi i}{r} = \frac{\mu_{0}}{4\pi}.\frac{2\pi.\left( \frac{e\omega}{2\pi} \right)}{r} = 10^{- 7} \times \frac{e\omega}{r}$
$\Rightarrow 16 = 10^{- 7} \times \frac{1.6 \times 10^{- 19}\omega}{1 \times 10^{- 10}} \Rightarrow \omega = 10^{17}rad/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.