Q101 / 150 · MHT CET 2024 (03 May Shift 2) · Physics
A thin ring of radius ' $R$ ' carries a uniformly distributed charge. The ring rotates at constant speed ' $N$ ' r.p.s. about its axis perpendicular to the plane. If ' $B$ ' is the magnetic field at the centre, the charge on the ring is ( $\mu_0=$ permeability of free space)
- A$\frac{\mu_0 \mathrm{~N}}{2 \mathrm{RB}}$
- B$\frac{\mathrm{RB}}{2 \mu_0^{\prime} \mathrm{N}}$
- C$\frac{\mu_0 \mathrm{~N}}{\mathrm{RB}}$
- Correct answerD$\frac{2 R B}{\mu_0 \mathrm{~N}}$
Solution
A thin uniformly charged rotating ring acts like a current carrying coil.
Magnetic field at the centre of current carrying coil,
$B=\frac{\mu_0 I}{2 R}$
Current is given by, $\mathrm{I}=\mathrm{qN} \quad$ where N is revolutions per second. $\begin{array}{ll} \therefore & B=\frac{\mu_0 q N}{2 R} \\ \therefore & q=\frac{2 R B}{\mu_0 N} \end{array}$ ...[From(i)]
Current is given by, $\mathrm{I}=\mathrm{qN} \quad$ where N is revolutions per second. $\begin{array}{ll} \therefore & B=\frac{\mu_0 q N}{2 R} \\ \therefore & q=\frac{2 R B}{\mu_0 N} \end{array}$ ...[From(i)]
Hence, the correct answer is option: 'D'.