Physics - Magnetism Question with Solution | TestHub

The magnetic field at the center of a circular loop of radius $r$ carrying current $i$ is given by:

$$B=\frac{{\mu }_{0}i}{2r}$$

The current $i$ due to the electron is given by $i=\frac{e}{T}$, where $T$ is the time period of revolution. Also, $T=\frac{2\pi r}{v}$, where $v$ is the velocity of the electron. Therefore, $i=\frac{ev}{2\pi r}$.

$$B=\frac{{\mu }_0}{2r}\cdot \frac{ev}{2\pi r}=\frac{{\mu }_{0}ev}{4\pi r^2}$$

We need to find $v$. The electrostatic force between the electron and proton provides the centripetal force for the circular motion:

$$\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r^2}=\frac{m{v}^2}{r}$$

$$v^2=\frac{e^2}{4\pi {ϵ}_{0}mr}$$

$$v=\sqrt{\frac{e^2}{4\pi {ϵ}_{0}mr}}$$

Substituting this value of $v$ into the expression for $B$:

$$B=\frac{{\mu }_{0}e}{4\pi r^2}\sqrt{\frac{e^2}{4\pi {ϵ}_{0}mr}}=\frac{{\mu }_0{e}^2}{4\pi r^2}\sqrt{\frac{1}{4\pi {ϵ}_{0}mr}}$$

$$B=\frac{{\mu }_0{e}^2}{4\pi r^2\sqrt{4\pi {ϵ}_{0}mr}}=\frac{{\mu }_0{e}^2}{8{\pi }^{3/2}{r}^2\sqrt{{ϵ}_{0}mr}}=\frac{{\mu }_0{e}^2}{8{\pi }^{2}r}\sqrt{\frac{\pi }{m{ϵ}_{0}r}}$$

Therefore, the correct option is B.

Final Answer: The final answer is $\boxed{B}$