JEE Main 2026 — Atomic Physics Question with Solution

The magnetic field at the center of a circular orbit is given by $B=\frac{{\mu }_{0}I}{2r}$.

The equivalent current due to the revolving electron is $I=\frac{e}{T}=\frac{ev}{2\pi r}$.

Substituting $I$ into the expression for $B$, we get $B=\frac{{\mu }_{0}ev}{4\pi r^2}$.

According to Bohr's model for a hydrogen atom, the velocity of the electron in the $n^{\text{th}}$ orbit is $v\propto \frac{1}{n}$ and the radius of the orbit is $r\propto n^2$.

Therefore, the magnetic field $B$ depends on the principal quantum number $n$ as $B\propto \frac{1/n}{(n^2{)}^2}\Rightarrow B\propto \frac{1}{n^5}$.

The ratio of the magnetic fields in the $2^{\text{nd}}$ and $4^{\text{th}}$ orbits is $\frac{B_2}{B_4}={\left(\frac{n_4}{n_2}\right)}^5={\left(\frac{4}{2}\right)}^5=2^5=32$.

Answer: $32$