Using Bohr's model, calculate the ratio of the magnetic fields...

The magnetic field at the center of a circular orbit is given by $B = \dfrac{\mu_0 I}{2r}$.

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

Substituting $I$ into the expression for $B$, we get $B = \dfrac{\mu_0 ev}{4\pi r^2}$.

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

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

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

Answer: $32$