JEE Main 2019 — Physics solution

JEE Main2019Physics

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 $π$ rad s^–1 about its axis, perpendicular to its plane. If the magnetic field at 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/A²).

A
4 × 10^–5 C
B
3 × 10^–5 C
C
7 × 10^–6 C
D
2 × 10^–6 C

Solution

$B = \frac{μ _{0} i}{2 a} an d \frac{ω q}{2 π} = i ∴ B = \frac{μ _{0}}{2 a} . \frac{ω q}{2 π} B = \frac{10 ^{- 7} \times 40}{0.1} \times q \times π \Rightarrow q = 3 \times 1 0^{- 5} C$

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