Question

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 _________.

Answer 1

Step 1: Given data

An electron moves in a circular orbit with a uniform speed = $v$.

It produces a magnetic field $B$ at the centre of the circle.

Step 2: Determine the time period

We can find the time period of an electron moving in a circular path by rearranging the formula for speed, i.e.,
$T=\frac{2\mathrm{\pi }r}{v}$
where $T$ is the time period, $2\mathrm{\pi }r$ is the distance covered (i.e., the circumference of the circle along which the electron is moving) and $v$ is the speed (given)

Step 3: Determine the current produced

The equivalent current that is produced in the conducting wire (which is circular, and in which the electron is moving) is,
$I=\frac{e}{T}=\frac{e}{{\displaystyle \frac{2\mathrm{\pi }r}{v}}}=\frac{ev}{2\mathrm{\pi }r}\left[\because \text{current}=\frac{\text{charge}}{\text{time}}\right]$
Where, $e$ is the charge of the electron.

Step 4: Substitute in Biot-Savart Law

The magnetic field ($B$) at a point induced by a flowing current ($I$) is given as, $B=\frac{{\mu }_{0}I}{2r}$
Where ${\mu }_0$ is the permeability of free space, and $r$ is the radius of the circular path.

Substituting for $I$ and rearranging,
$\begin{array}{rcl}& & \begin{array}{cc}& B=\frac{{\mu }_0\left({\displaystyle \frac{ev}{2\mathrm{\pi }r}}\right)}{2r}\\ \Rightarrow & B=\frac{{\mu }_{0}ev}{4{\mathrm{\pi r}}^2}\\ \Rightarrow & r^2=\frac{{\mu }_{0}ev}{4\mathrm{\pi }B}\\ \Rightarrow & r^2\propto \frac{v}{B}\\ \therefore & r\propto \sqrt{\frac{v}{B}}\end{array}\end{array}$

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 $\sqrt{\frac{v}{B}}$.