Q 12-04-225JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
A thin ring of $10$ cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\pi\ \text{rad s}^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8\times10^{-9}$ T, then the charge carried by the ring is close to ($\mu_0 = 4\pi\times10^{-7}\ \text{N/A}^2$)
Answer: (B) $3\times10^{-5}$ C
Equivalent current: $I = \dfrac{q\omega}{2\pi} = 20q$.
$$B = \frac{\mu_0I}{2R} = \frac{4\pi\times10^{-7}\times20q}{0.2} \approx 1.26\times10^{-4}q = 3.8\times10^{-9}$$
$$q \approx 3\times10^{-5}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics