An electron moves in a circular orbit with uniform speed ' $v^{\prime}$. It produces a magnetic field 'B' at…
An electron moves in a circular orbit with uniform speed ' $v^{\prime}$. It produces a
magnetic field 'B' at the centre of the circle. The radius of the circle is
$\left[\mu_{0}=\right.$ permeability of free space, $\mathrm{e}=$ electronic charge $]$
The magnetic field at the center of a circular orbit of a moving electron is given by
$\begin{array}{l}
B=\frac{\mu_{0}}{4 \pi} \times \frac{e}{r^{3}}(v \times r) \\
B=\frac{\mu_{0}}{4 \pi} \times \frac{e v \sin \theta}{r^{2}}
\end{array}$
As, in a circular path, the angle between radius vector and velocity vector is $90^{\circ}$.
$\begin{array}{l}
\mathrm{B}=\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{ev}}{\mathrm{r}^{2}} \\
\therefore \mathrm{r}=\left(\frac{\mu_{0} \mathrm{ev}}{4 \pi \mathrm{B}}\right)^{1 / 2}
\end{array}$