Q127 / 200 · NEET 2010 (Screening) · Physics
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
- A$\frac{\mu_0 q \mathrm{f}}{2 \pi \mathrm{R}}$
- B$\frac{\mu_0 \mathrm{q}}{2 \pi \mathrm{f} R}$
- C$\frac{\mu_0 q}{2 f \mathrm{f}}$
- Correct answerD$\frac{\mu_0 q f}{2 R}$
Solution
The magnetic field at the centre of the circle
\(B=\frac{\mu_0 I}{2 R}=\frac{\mu_0}{2 R}\left(\frac{q}{t}\right)=\frac{\mu_0 q f}{2 R}\)
Hence, the correct answer is option: 'D'.