JEE PYQ: Electromagnetic Induction Question 19
Question 19 - 2019 (12 Apr Shift 1)
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\pi$ rad s$^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9}$ T, then the charge carried by the ring is close to ($\mu_0 = 4\pi \times 10^{-7}$ N/A$^2$):
(1) $2 \times 10^{-6}$ C
(2) $3 \times 10^{-5}$ C
(3) $4 \times 10^{-5}$ C
(4) $7 \times 10^{-6}$ C
Show Answer
Answer: (2)
Solution
Current $i = \frac{q}{T} = \frac{q\omega}{2\pi}$. Magnetic field $B = \frac{\mu_0 i}{2R} = \frac{\mu_0 q\omega}{4\pi R}$. So $3.8 \times 10^{-9} = \frac{\mu_0}{4\pi} \times \frac{q \times 40\pi}{0.10} = 10^{-7} \times \frac{q \times 40\pi}{0.10}$. Therefore $q = 3 \times 10^{-5}$ C.
BETA
