Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 sin (ωt) A, where ω = 1000 rad./s. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v = 1 cm/s. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is α/√2 μA. The value of α i $s _________.$
[Resistance of the loop = 10 Ω]

Question

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 sin (ωt) A, where ω = 1000 rad./s. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v = 1 cm/s. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is α/√2 μA. The value of α i $s _________.$

[Resistance of the loop = 10 Ω]

G1 Physics · Accepted Answer

When the loop is fully inside the long solenoid, the changing magnetic field due to the alternating current produces the induced emf. Magnetic field inside the solenoid: B = μ₀nI = μ₀nI₀ sinωt. Loop area: A = πr². Magnetic flux: Φ = AB. Therefore the induced emf is ε = A dB/dt = Aμ₀nI₀ω cosωt. The emf amplitude is ε₀ = πr² μ₀nI₀ω. Hence the current amplitude is i₀ = ε₀/R. The rms current is i(rms) = i₀/√2. Substitute: r = 1 cm = 10⁻² m, μ₀ = 4π×10⁻⁷, n = 500 m⁻¹, I₀ = 10 A, ω = 1000 rad/s, R =…