A 40-turn, 4.0-cm-diameter coil with R=0.40Ω surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The 60 Hz current through the solenoid is I=I0sin (2πft). What is the Io if maximum induced current is. 2 A?

Magnetic field B is produced when a current I Amphere passes through a solenoid. B is parallel to its axis.

B=U N/I I. N is number of turns in the solenoid of lm length

N=200, l = 20cm = 0.2m, I = 1₀sin (2πft) where f is equal to 60Hz

B = 4π * 10⁻⁷ (200/0.2) l₀ sin (2πft) T

=1.256 * 10⁻³ l₀ sin (2πft) Tesla

Area of the coil is πr² = π (1.5cm) ² = 2.25π * 10⁴m²

magnetic flux which is through the coil is given by

Ф = B. A = BA cosФ

ФФ = O since B is in direction of A

A = 40 * π*2.25 * ⁻⁴m² which is the number of turns being 40.

Flux Ф through the coil is,

1.256 * 10⁻³ l₀ sin (2πft) * 9π * * 10⁻³m²

=35.5 * 10⁻⁶ l₀ sin ₀ (2πft) ab

Ф is time-varying emf will be generated in the coil

∈ = dФ/dt

∈ = d/dt [35.5 * 10⁻d l₀ sin (2πft) ab]

∈ = 35.5 * 10⁻⁶ l₀ 2πf cos 2πftV

f = 60Hz

∈∈ 13376.4 * 10⁻⁶ l₀ cos 2πftV

Current I amp shall be induced in the cell of resistance Rohm so

I = E/R

I = 13376.4 * 10⁻⁶ l₀ cos 2πft) V/0.4∩

=33441 * 10⁻⁶ I₀ cos 2πft A

I = 3344q * 10 ⁻⁶ l₀

But I = 0.2A

l₀ = (0.2) (10⁶) / 33441 = 6.0A