Q. A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius $0.8\ metre$. The value of the magnetic field produced at the centre will be $(\mu_{0} =$ permeability for vacuum)
A) $\frac{10^{- 7}}{\mu_{0}}$
B) $10^{- 17}\mu_{0}$
C) $10^{- 6}\mu_{0}$
D) $10^{- 7}\mu_{0}$
Explanation
(b) $i = \frac{q}{t} = 100 \times e$
$B_{centre} = \frac{\mu_{o}}{4\pi}.\frac{2\pi i}{r} = \frac{\mu_{o}}{4\pi}.\frac{2\pi \times 100e}{r}$
$= \frac{\mu_{o} \times 200 \times 1.6 \times 10^{- 19}}{4 \times 0.8} = 10^{- 17}\mu_{o}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.