Q. In a hydrogen atom, the electron is making $6.6 \times 10^{15}$ rps in a circular path of radius $0.53\ \mathring{\mathrm{A}}.$ What is the magnetic induction produced at the centre of orbit?
A) $6.3\ T$
B) $12.6\ T$
C) $18.9\ T$
D) $25.2\ T$
Explanation
(b) Magnetic field $B_{centre} = \frac{\mu_{0}I}{2r}$
But $\ \ \ \ \ \ \ \ I = ne$
$\therefore\ B_{centre} = \frac{\mu_{0}ne}{2r}$
$\Rightarrow B_{centre} = \frac{4\pi \times 10^{- 7} \times 6.6 \times 10^{15} \times 1.6 \times 10^{- 19}}{2 \times 0.53 \times 10^{- 10}} = 12.6\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.