Q. Imaging that an electron revolves round a circle of radius $5.3 \times 10^{- 11}$ m with a linear velocity of $7.5\ \times 10^{4}\ ms^{- 1}$ in a hydrogen atom. The magnetic field produced at the centre of the circle due to the electron is
A) $43\ Wbm^{- 2}$
B) $43 \times 10^{2}\ Wbm^{- 2}$
C) $0.43\ Wbm^{- 2}$
D) $43 \times 10^{- 4}\ Wbm^{- 2}$
Explanation
(c) $B = \frac{\mu_{0}}{4\pi}\frac{2\pi i}{r}\frac{\mu_{0}}{4\pi}\frac{2\pi}{r}\ \frac{e}{(2\pi r/v)}$=$\frac{\mu_{0}}{4\pi}\frac{ev}{r^{2}}$
$= \ \frac{10^{- 7} \times 1.6\ \times 10^{- 19} \times 7.5 \times 10^{+ 4}}{\left( 5.3 \times 10^{- 11} \right)^{2}}$
On solving $B = 0.43\ \ Wb\ m^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.