## A circular loop in the plane of the paper lies in a 0.75 T magnetic field pointing into the paper. The loop’s diameter is changed from 20.0

Question

A circular loop in the plane of the paper lies in a 0.75 T magnetic field pointing into the paper. The loop’s diameter is changed from 20.0 cm to 6.0 cm in 0.50 s.

a) Determine the direction of the induced current and justify your answer.

b) Determine the magnitude of the average induced emf.

c) If the coil resistance is 2.5 Ω, what is the average induced current?

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3 years 2021-08-31T13:34:42+00:00 1 Answers 78 views 0

a) The direction of the Induced current is clockwise and point outside the paper

b) Magnitude of the average induced emf = 107.5 mV

c) Average induced current = 43 mA

Explanation:

a) According to Lenz’s law “the direction of the current induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field.”

The direction of the Induced current is clockwise and point outside the paper

b) Magnitude of the average induced emf

B = 0.75 T

The magnitude of the induced emf is given by:

ε=-N(ΔΦ/Δt)

Initial diameter, d₁ = 20 cm = 0.2 m

Initial Radius , r₁ = 0.2/2 = 0.1 m

Initial Area, A₁ = πr₁² = π * (0.1)² = 0.01π m²

Final diameter, d₂ = 6 cm = 0.06 m

Final Radius , r₂ = 0.06/2 = 0.03 m

Final Area = A₂ = πr₂² = π * (0.03)² = 0.0009π m²

Since Φ=BA

Φ₁=BA₁ = 0.75 * 0.01π = 0.0236 Wb

Φ₂=BA₂ = 0.75 * 0.0009π = 0.0021 Wb

ε = -N(ΔΦ/Δt)

ΔΦ = Φ₂ – Φ₁ =  0.0021  – 0.0236 = -0.0215 Wb

Δt = 20.0cm = 0.2 m

Since it is a loop, N = 1

ε = -(-0.0215)/0.2

ε = 0.1075 V

ε = 107.5 mV

c) Average induced current

ε = IR

R = 2.5 Ω

I = ε /R

I = 0.1075/2.5

I = 0.043 A

I = 43 mA