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(B) The magnetic field $B$ at the center of a circular orbit is given by $B = \frac{\mu_0 I}{2r}$,where $I$ is the equivalent current and $r$ is the r

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$n^{-5}$

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(B) The magnetic field $B$ at the center of a circular orbit is given by $B = \frac{\mu_0 I}{2r}$,where $I$ is the equivalent current and $r$ is the radius of the orbit.From Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$. Since $v = \omega r$,we have $mr^2\omega = \frac{nh}{2\pi}$.For a hydrogen atom,the radius $r \propto n^2$. Substituting this into the angular momentum equation: $m(n^2)^2\omega \propto n$,which implies $\omega \propto \frac{n}{n^4} = n^{-3}$.The equivalent current $I$ is given by $I = \frac{e}{T} = \frac{e\omega}{2\pi}$,so $I \propto \omega \propto n^{-3}$.Now,substituting $I \propto n^{-3}$ and $r \propto n^2$ into the magnetic field formula: $B \propto \frac{I}{r} \propto \frac{n^{-3}}{n^2} = n^{-5}$.Therefore,the magnetic field induction is proportional to $n^{-5}$.

The magnetic field induction produced at the centre of orbit due to an electron revolving in $n^{\text{th}}$ orbit of a hydrogen atom is proportional to

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