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NEET PHYSICSEasy

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 :

A

\mu ₀qf / 2R

B

\mu ₀q / 2fR

C

\mu ₀q / 2\pi fR

D

\mu ₀qf / 2\pi R

Step-by-Step Solution

  1. Current Equivalent: A charge qq rotating with frequency ff constitutes an equivalent electric current II. Since current is the rate of flow of charge (I=q/tI = q/t) and the time period of rotation is T=1/fT = 1/f, the effective current is given by: I=qT=qfI = \frac{q}{T} = qf .
  2. Magnetic Field Formula: The magnetic field (BB) at the center of a circular current-carrying loop of radius RR is given by the formula: B=μ0I2RB = \frac{\mu_0 I}{2R} .
  3. Substitution: Substituting the expression for equivalent current (I=qfI = qf) into the magnetic field formula: B=μ0(qf)2RB = \frac{\mu_0 (qf)}{2R} This matches Option 1.
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