Rankers Physics

Magnetic Field Due to Circular Current Carrying Wire: Practice Problem & Solution

An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has magnitude: (2015)
Zero
$\frac{\mu_0 n^2 e}{r}$
$\frac{\mu_0 n e}{2 r}$
$\frac{\mu_0 n e}{2 \pi r}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Magnetic Field Due to Circular Current Carrying Wire. Understanding the underlying formula is key to arriving at the correct answer below:

The current produced by the revolving electron is $i = qf = ne$. The magnetic field at the center of a circular loop of radius $r$ carrying current $i$ is $B = \frac{\mu_0 i}{2r} = \frac{\mu_0 n e}{2r}$.

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