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(C) The magnetic field $B$ at the center of the orbit due to an electron moving in a circular path is given by $B = \frac{\mu_0 I}{2r}$.
The current

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$32$

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(C) The magnetic field $B$ at the center of the orbit due to an electron moving in a circular path is given by $B = \frac{\mu_0 I}{2r}$.
The current $I$ is given by $I = \frac{ev}{2\pi r}$, where $e$ is the charge of the electron, $v$ is the velocity, and $r$ is the radius.
According to Bohr's model, $v \propto \frac{1}{n}$ and $r \propto n^2$.
Substituting these into the expression for current: $I \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.
Now, substituting $I$ and $r$ into the expression for $B$: $B \propto \frac{I}{r} \propto \frac{1/n^3}{n^2} = \frac{1}{n^5}$.
Therefore, the ratio of the magnetic fields for the $2^{nd}$ and $4^{th}$ orbits is $\frac{B_2}{B_4} = \left( \frac{4}{2} \right)^5 = 2^5 = 32$.
Thus, the ratio is $32:1$.

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom. (in $32$ : $1$)

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