Magnetic field at the centre of the hydrogen | vedclass

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(D) The radius of the $n^{	ext{th}}$ Bohr orbit is given by $r_n \propto n^2$.The velocity of the electron in the $n^{	ext{th}}$ orbit is $v_n \prop

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$n^{-5}$

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(D) The radius of the $n^{	ext{th}}$ Bohr orbit is given by $r_n \propto n^2$.The velocity of the electron in the $n^{	ext{th}}$ orbit is $v_n \propto \frac{1}{n}$.The angular velocity is $\omega_n = \frac{v_n}{r_n} \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.The current $I_n$ produced by the revolving electron is $I_n = \frac{e}{T_n} = \frac{e \omega_n}{2 \pi} \propto \omega_n \propto \frac{1}{n^3}$.The magnetic field at the centre of the orbit is $B_n = \frac{\mu_0 I_n}{2 r_n}$.Substituting the proportionalities: $B_n \propto \frac{I_n}{r_n} \propto \frac{1/n^3}{n^2} = \frac{1}{n^5}$.Therefore,$B_n \propto n^{-5}$.

Magnetic field at the centre of the hydrogen atom due to the motion of an electron in the $n^{\text{th}}$ orbit is proportional to:

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