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(A) The magnetic field at the center of a circular current loop is given by $B = \frac{\mu_{0} I}{2r}$.For an electron revolving in an orbit,the equiv

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$\frac{\mu_{0} e^{7} \pi m^{2}}{8 \epsilon_{0}^{3} h^{5}}$

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(A) The magnetic field at the center of a circular current loop is given by $B = \frac{\mu_{0} I}{2r}$.For an electron revolving in an orbit,the equivalent current is $I = ef = e \left( \frac{v}{2 \pi r} \right) = \frac{ev}{2 \pi r}$.Substituting this into the magnetic field formula: $B = \frac{\mu_{0} ev}{4 \pi r^{2}}$.For the ground state of a hydrogen atom,the velocity $v = \frac{e^{2}}{2 \epsilon_{0} h}$ and the radius $r = \frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$.Substituting $v$ and $r$ into the expression for $B$:$B = \frac{\mu_{0} e}{4 \pi r^{2}} \left( \frac{e^{2}}{2 \epsilon_{0} h} \right) = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h r^{2}}$.Now,substituting $r = \frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$:$B = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h} \left( \frac{\pi m e^{2}}{h^{2} \epsilon_{0}} \right)^{2} = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h} \cdot \frac{\pi^{2} m^{2} e^{4}}{h^{4} \epsilon_{0}^{2}} = \frac{\mu_{0} \pi m^{2} e^{7}}{8 \epsilon_{0}^{3} h^{5}}$.

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