Question:
A thin ring of radius 'R' metre has charge 'q' coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of 'f' revolution/s. The value of magnetic induction at the centre of the ring in \(\text{Wb/m}^2\) is
( \(μ_0\) = permeability of free space )
A thin ring of radius 'R' metre has charge 'q' coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of 'f' revolution/s. The value of magnetic induction at the centre of the ring in \(\text{Wb/m}^2\) is
( \(μ_0\) = permeability of free space )
( \(μ_0\) = permeability of free space )
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A rotating charge forms a current I = q f.
Updated On: Oct 1, 2026
- \(\frac{μ_0q}{2fR}\)
- \(\frac{μ_0q}{2πR}\)
- \(\frac{μ_0qf}{2R}\)
- \(\frac{μ_0qπ}{2fR}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Charge q on the ring crosses any point \(f\) times each second, so it is equivalent to a current \(I = qf\).
Step 2: Compute
The field at the centre of a circular current loop is \(B = \dfrac{\mu_0I}{2R}\).
\[ B = \frac{\mu_0 q f}{2R} \]
Option (B) uses \(\pi R\) in the denominator, and (A) and (D) put f in the wrong place.
Final Answer:
The magnetic induction is \(\dfrac{\mu_0qf}{2R}\), option (C). \[ \boxed{\frac{\mu_0 q f}{2R}} \]
Charge q on the ring crosses any point \(f\) times each second, so it is equivalent to a current \(I = qf\).
Step 2: Compute
The field at the centre of a circular current loop is \(B = \dfrac{\mu_0I}{2R}\).
\[ B = \frac{\mu_0 q f}{2R} \]
Option (B) uses \(\pi R\) in the denominator, and (A) and (D) put f in the wrong place.
Final Answer:
The magnetic induction is \(\dfrac{\mu_0qf}{2R}\), option (C). \[ \boxed{\frac{\mu_0 q f}{2R}} \]
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