Question

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

• A. $\dfrac{1}{vr^2}$
• B. $\dfrac{r^2}{v}$
• C. $\dfrac{v}{r^2}$
• D. $vr^2$

Hint:

Smaller orbit radius and higher speed produce stronger magnetic field at the centre.

Answer 1

The Correct Option is C

Step 1: Write expression for current due to orbiting electron.
An electron revolving in a circular orbit constitutes a current given by: \[ I = \dfrac{e}{T} \] where time period \[ T = \dfrac{2\pi r}{v} \] \[ I = \dfrac{ev}{2\pi r} \] Step 2: Magnetic field at centre of circular loop.
Magnetic field at the centre of a circular current loop is: \[ B = \dfrac{\mu_0 I}{2r} \] Step 3: Substitute current expression.
\[ B = \dfrac{\mu_0}{2r} \cdot \dfrac{ev}{2\pi r} \] Step 4: Identify proportionality.
\[ B \propto \dfrac{v}{r^2} \]