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NEET PHYSICSMOVING CHARGES AND MAGNETISMEasy

Question

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.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMchargeuniformlyspreadradiusrotates

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