Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the \(2^{\text{nd}}\) and \(4^{\text{th}}\) orbits of hydrogen atom _______.

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: \(32\)