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(B) The magnetic field $B$ at the centre of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.Here,the current $I$ is the charge $e$ divide

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$n^{-5}$

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(B) The magnetic field $B$ at the centre of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.Here,the current $I$ is the charge $e$ divided by the time period $T$,so $I = \frac{e}{T}$.The time period $T$ is given by $T = \frac{2\pi r}{v}$,where $v$ is the orbital velocity.Thus,$I = \frac{ev}{2\pi r}$.Substituting this into the magnetic field formula: $B = \frac{\mu_0 ev}{4\pi r^2}$.According to Bohr's theory,the radius $r \propto n^2$ and the velocity $v \propto n^{-1}$.Substituting these proportionalities: $B \propto \frac{n^{-1}}{(n^2)^2} = \frac{n^{-1}}{n^4} = n^{-5}$.Therefore,the magnetic field is proportional to $n^{-5}$.

According to Bohr's theory,the time-averaged magnetic field at the centre (i.e.,nucleus) of a hydrogen atom due to the motion of electrons in the $n^{th}$ orbit is proportional to ($n =$ principal quantum number).

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