An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to:

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Aicharya · Accepted Answer

The magnetic field $(B)$ at a point induced by a flowing current $(I)$ is given as, $B=\frac{\mu_0 I}{2 r}$ Where $\mu_0$ is the permeability of free space, and $r$ is the radius of the circular path. Substituting for $I$ and rearranging, $\begin{aligned} & B=\frac{\mu_0\left(\frac{e v}{2 \pi r}\right)}{2 r} \ \Rightarrow & B=\frac{\mu_0 e v}{4 \pi r^2} \ \Rightarrow & r^2=\frac{\mu_0 e v}{4 \pi B} \ \Rightarrow & r^2 \propto \frac{v}{B} \ \therefore & r \propto \sqrt{\frac{v}{B}} \end{aligned}$ Therefore, when an electron moves in a circular orbit with a uniform speed $v$ and produces a magnetic field $B$ at the centre of the circle, the radius of the circle is proportional to $\sqrt{\frac{v}{B}}$.

An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the…

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