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Question: Answered & Verified by Expert
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 in $\mathrm{Wb} \mathrm{m}^{-2}$ at the centre of the ring is
PhysicsElectromagnetic InductionNEETNEET 2010 (Screening)
Options:
  • A $\frac{\mu_0 q \mathrm{f}}{2 \pi \mathrm{R}}$
  • B $\frac{\mu_0 \mathrm{q}}{2 \pi \mathrm{f} R}$
  • C $\frac{\mu_0 q}{2 f \mathrm{f}}$
  • D $\frac{\mu_0 q f}{2 R}$
Solution:
2983 Upvotes Verified Answer
The correct answer is: $\frac{\mu_0 q f}{2 R}$
The magnetic field at the centre of the circle
\(B=\frac{\mu_0 I}{2 R}=\frac{\mu_0}{2 R}\left(\frac{q}{t}\right)=\frac{\mu_0 q f}{2 R}\)

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