Question

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?

Answer 1

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