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

The magnetic field $B$ at the center of a circular current loop carrying current $I$ is given by Biot-Savart's law application: $$B = \frac{\mu_0 I}{2r}$$ The effective electric current $I$ generated by an electron revolving in a orbit with a time period $T$ is: $$I = \frac{e}{T}$$ Since the electron moves with a uniform tangential speed $V$ along a circular track of perimeter $2\pi r$, the time period is $T = \frac{2\pi r}{V}$. Substituting this back into the current expression: $$I = \frac{e}{\left(\frac{2\pi r}{V}\right)} = \frac{eV}{2\pi r}$$ Now, substitute this current value into our center magnetic field equation: $$B = \frac{\mu_0}{2r} \left(\frac{eV}{2\pi r}\right) = \frac{\mu_0 e V}{4\pi r^2} = \frac{\mu_0 e}{4\pi} \left(\frac{V}{r^2}\right)$$
Final Answer:
The magnetic field at the center is matching option (A).