Q. An electron revolves in a circle of radius $0.4\ \mathring{\mathrm{A}}$ with a speed of $10^{5}ms^{- 1}$. The magnitude of the magnetic field, produced at the center of the circular path due to the motion of the electron, in $weber\ metre^{- 2}$ is
A) 0.01
B) 10.0
C) 1.0
D) 0.005
Explanation
(c) $B = \frac{\mu_{0}I}{2r} = \frac{\mu_{0}}{2r} \times \frac{eV}{2\pi r}$
$= \frac{4\pi \times 10^{- 7} \times 1.6 \times 10^{- 19} \times 10^{5}}{2\pi(2 \times 0.4 \times 10^{- 10})(0.4 \times 10^{- 10})}$
$= \frac{4 \times 10^{- 26 + 10 + 5}}{4 \times 10^{- 11}} = \frac{4 \times 10^{- 26 + 15}}{4 \times 10^{- 11}} = \frac{4 \times 10^{- 11}}{4 \times 10^{- 11}} = 1\omega b/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.