Ratio of Magnetic Fields in Bohr Model Orbits

To find the ratio of the magnetic fields generated at the nucleus due to the motion of the electron in the $2^{\text{nd}}$ and $4^{\text{th}}$ orbits of a hydrogen atom using Bohr's model, we analyze the dependence of the magnetic field on the principal quantum number $n$.

The magnetic field $B$ produced at the center of a circular loop of radius $r$ carrying a current $I$ is given by: $B = \frac{\mu_0 I}{2r}$

The equivalent current $I$ due to an electron of charge $e$ revolving in an orbit with time period $T$ and velocity $v$ is: $I = \frac{e}{T} = \frac{e v}{2\pi r}$

Substituting $I$ into the formula for magnetic field, we get: $B = \frac{\mu_0}{2r} \left(\frac{e v}{2\pi r}\right) = \frac{\mu_0 e v}{4\pi r^2}$

According to Bohr's model of the hydrogen atom, for an orbit with principal quantum number $n$:

1. The orbital radius $r_n$ is directly proportional to $n^2$: $r_n \propto n^2$

2. The orbital speed $v_n$ is inversely proportional to $n$: $v_n \propto \frac{1}{n}$

Substituting the proportionalities of $v_n$ and $r_n$ into the equation for $B_n$: $B_n \propto \frac{v_n}{r_n^2} \propto \frac{1/n}{(n^2)^2} = \frac{1}{n^5}$

Thus, the ratio of the magnetic fields in the $2^{\text{nd}}$ orbit ($n_1 = 2$) and $4^{\text{th}}$ orbit ($n_2 = 4$) is: $\frac{B_2}{B_4} = \left(\frac{n_2}{n_1}\right)^5 = \left(\frac{4}{2}\right)^5 = 2^5 = 32$