JEE Main201912 Apr 2019Morning ShiftPhysicsMagnetic Effects of CurrentActual
A thin ring of 10 c m radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 π r a d s - 1 about its axis, perpendicular to its plane. Is the magnetic field its centre is 3.8 × 10 - 9 T , then the charge carried by the ring is close to μ 0 = 4 π × 10 - 7 N / A 2 .
Options
- A4 × 10 - 5 C
- B3 × 10 - 5 C
- C2 × 10 - 6 C
- D7 × 10 - 6 C
Correct answer
B. 3 × 10 - 5 C
Step-by-step solution
i = q t = q 2 π / ω = q ω 2 π B = μ 0 i 2 R = μ 0 q ω 4 π R 4 π × 10 - 7 × q × 40 π 4 π × ( 0.1 ) = 3.8 × 10 - 9 q × 400 π = 3.8 × 10 - 2 q = 3 × 10 - 5 C