Q 12-12-123JEE MainJEE Main 2017 (8 Apr)Medium
According to Bohr's theory, the time averaged magnetic field at the centre (i.e., nucleus) of a hydrogen atom due to the motion of electrons in the $n^{\text{th}}$ orbit is proportional to: ($n$ = principal quantum number)
Answer: (D) $n^{-5}$
The orbiting electron is a current loop: $I = \dfrac{ev}{2\pi r}$, giving at the centre
$$B = \frac{\mu_0I}{2r} = \frac{\mu_0ev}{4\pi r^2}$$
In Bohr's model $v\propto\dfrac1n$ and $r\propto n^2$, so
$$B \propto \frac{n^{-1}}{n^4} = n^{-5}$$
Solution by Sreeraj P, M.Sc Physics