*9. A small loop of area 6.8 mm2 is placed inside a long solenoid that has 854 turns/cm and carries a sinusoidally varying current i of amplitude 1.28 A and angular frequency 212 rad/s . The central axes of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop

E = 1.98*10^-4 V

Explanation:

Area = 6.8mm² = 6.8*10^-6m²

Turns = 854N/mm = 85400N/m

I = 1.28A

w = 212 rad/s

From Faraday's law of induction, the induced EMF = -dΦᵦ / dt

where Φᵦ = magnetic flux through the loop and its given by Φᵦ = BA

B = μ₀ni is the field produced by the solenoid and A is the area of the loop.

B = 4π*10⁻⁷ * 85400 * 1.28*sin(212t)

B = 0.1373 * sin(212t)

Φᵦ = B*A = 0.1373 * 6.8*10^-6 * sin(212t)

Φᵦ = 9.3364*10^-7 * sin212t

The induce EMF E = -dΦᵦ / dt = -(9.3364*10^-7) * d(sin212t) / dt

After differentiating, sin212t = cos212

E = -(1.98*10⁴) cos 212t

The magnitude of the induced EMF |E| = 1.98*10^-4 V

magnitude and direction