Question

A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.235 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.40 s.

Required:
a. What average emf is induced in the second coil if it has a diameter of 3.5 cm and N = 7?
b. What is the induced emf if the diameter is 7.0 cm and N = 10?

Answer 1

a) ε = 14.7 μv

b) ε = 21 μv

Step-by-step explanation:

Given the data in the question;

Diameter of solenoid; d = 3 cm

radius will be half of diameter, so, r = 3 cm / 2 = 1.5 cm = 1.5 × 10⁻² m

Number of turns; N = 40 turns per cm = 4000 per turns per meter

Current;
`I` = 0.235 A

change in time Δt = 0.40 sec

Now,

We determine the magnetic field inside the solenoid;

B = μ₀ × N ×
`I`

we substitute

B = ( 4π × 10⁻⁷ Tm/A ) × 4000 × 0.235

B = 1.1881 × 10⁻³ T

Now, Initial flux through the coil is;

∅₁ = NBA = NBπr²

and the final flux

∅₂ = 0

so, the εmf induced ε = -Δ∅/Δt = -( ∅₂ - ∅₁ ) / Δt

= -( 0 - NBπr² ) / Δt

= NBπr² / Δt

a)

for N = 7

ε = [ 7 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 14.7 × 10⁻⁶ v

ε = 14.7 μv

b)

for N = 10

ε = [ 10 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 21 × 10⁻⁶ v

ε = 21 μv