Rankers Physics

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

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is: (2011 Mains)
$\frac{\mu_0 q f}{2 R}$
$\frac{\mu_0 q}{2 f R}$
$\frac{\mu_0 q}{2 \pi R}$
$\frac{\mu_0 q f}{2 \pi R}$

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 equivalent current is $i = qf$. The magnetic induction at the center of a circular ring of radius $R$ carrying current $i$ is $B = \frac{\mu_0 i}{2R} = \frac{\mu_0 q f}{2R}$.

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