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(D) The magnetic field $B$ at the center of a circular loop is given by $B = \frac{\mu_0 i}{2r}$.The current $i$ due to an electron moving with veloci

Vedclass · Accepted Answer

$\frac{Z^3}{n^5}$

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(D) The magnetic field $B$ at the center of a circular loop is given by $B = \frac{\mu_0 i}{2r}$.The current $i$ due to an electron moving with velocity $v$ in an orbit of radius $r$ is $i = \frac{e}{T} = \frac{ev}{2\pi r}$.Substituting $i$ into the magnetic field formula: $B = \frac{\mu_0 ev}{4\pi r^2}$.From Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$,so $v = \frac{nh}{2\pi mr}$.Substituting $v$ into the expression for $B$: $B = \frac{\mu_0 e}{4\pi r^2} \left( \frac{nh}{2\pi mr} \right) = \frac{\mu_0 enh}{8\pi^2 mr^3}$.For a hydrogen-like atom,the radius $r \propto \frac{n^2}{Z}$.Substituting $r^3 \propto \frac{n^6}{Z^3}$ into the expression for $B$: $B \propto \frac{n}{r^3} \propto \frac{n}{(n^2/Z)^3} = \frac{n}{n^6/Z^3} = \frac{Z^3}{n^5}$.Thus,$B \propto \frac{Z^3}{n^5}$.

Magnetic field at the centre of a hydrogen-like atom (atomic number $= Z$) due to the motion of an electron in the $n^{th}$ orbit is proportional to

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