Question

An electron (e) moves in a circular orbit of radius 'r' with uniform speed 'V'. It produces a magnetic field 'B' at the centre of the circle. The magnetic field B is ($\mu_0$ = permeability of free space)

• A. $\frac{\mu_0 e}{4\pi} \left(\frac{V}{r^2}\right)$
• B. $\frac{\mu_0 e}{4\pi} \frac{V}{r}$
• C. $\frac{\mu_0 e}{2\pi} \left(\frac{V}{r^2}\right)$
• D. $\frac{\mu_0 e}{2\pi} \frac{V}{r}$

Hint:

You can also use the moving point charge vector form of Biot-Savart Law directly: $B = \frac{\mu_0}{4\pi}\frac{e(\vec{V} \times \hat{r})}{r^2}$. Since velocity is perpendicular to the radial unit vector anywhere along the circle, the scalar magnitude yields $\frac{\mu_0 e V}{4\pi r^2}$ instantly!

Answer 1

The Correct Option is A

Step 1: Field at the centre of a loop.
A current $I$ in a circular loop of radius $r$ makes a central field $B=\frac{\mu_0 I}{2r}$.

Step 2: Find the electron's current.
The electron goes round once in time $T=\frac{2\pi r}{V}$, so the current is $I=\frac{e}{T}=\frac{eV}{2\pi r}$.

Step 3: Combine.
\[ B=\frac{\mu_0}{2r}\cdot\frac{eV}{2\pi r}=\frac{\mu_0 e}{4\pi}\cdot\frac{V}{r^2} \]
\[ \boxed{\frac{\mu_0 e}{4\pi}\left(\frac{V}{r^2}\right)} \]