Q 12-04-123JEE MainJEE Main 2023 (15 Apr, Shift 1)Medium
An electron in a hydrogen atom revolves around its nucleus with a speed of $6.76\times10^6\ \text{m s}^{-1}$ in an orbit of radius $0.52\ \text{Å}$. The magnetic field produced at the nucleus of the hydrogen atom is ______ T.
Numerical value type. Enter your answer.
Answer: 40
The orbiting electron is a current $I=\dfrac{ev}{2\pi r}$, so at the centre
$$B=\frac{\mu_0I}{2r}=\frac{\mu_0ev}{4\pi r^2}=10^{-7}\times\frac{1.6\times10^{-19}\times6.76\times10^6}{(0.52\times10^{-10})^2}=40\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics