Question

Part 1: A resistor of known resistance R is placed within a circular loop of wire of known radius, a, such that the loop is closed. This loop is placed within a sinusoidal magnetic field of a known amplitude B and period T directed along the loop’s axis of symmetry. Calculate the induced emf across the resistor.

Part 2: Now, the situation is the same, but the radius of the loop is also increasing at constant rate x. Calculate the total induced emf across the resistor.

Answer 1

Answer:

Part 1 - ε = (2π²r²B/T)cosωt

Part 2: ε = B√[(2πr{(πr/T)² + x²}]Sin(ωt + Ф) where Ф = tan⁻¹(πr/xT)

Explanation:

Part 1. Since the magnetic field is sinusoidal and has a period of T and amplitude of B, it is of the form B' = Bsinωt where ω = 2π/T

Now, the induced emf in the circular loop is ε = dΦ/dt where Φ = magnetic flux through circular loop of wire. Φ  = AB' where A = area of loop of wire, A = πr² where r = radius of loop of wire.

So,  ε = dΦ/dt

ε = dAB'/dt

= AdB'/dt

= AdBsinωt/dt

= ωABcosωt

=  (2πAB/T)cosωt    

with A = πr²,

ε = (2π²r²B/T)cosωt

Part 2

If the radius of the loop is increasing at a constant rate x, then the induced emf is

ε = dΦ/dt

= dAB'/dt

= AdB'/dt + B'dA/dt

= πr²dBsinωt/dt  + (Bsinωt)dπr²/dt

= (ωπr²)Bcosωt + (Bsinωt)2πrdr/dt

= (ωπr²)Bcosωt + (Bsinωt)2πrx

= (2π²r²B/T)cosωt + (2πrxB)sinωt

Writing this in compound angle form,

ASin(ωt + Ф) = AsinФcosωt + AcosФsinωt

comparing both expressions,

AsinФ = 2π²r²B/T and AcosФ = 2πrxB

(AsinФ)² + (AcosФ)²  = A²

(2π²r²B/T)² + (2πrxB)²  = A²

B²[(2πr{(πr/T)² + x²}]  = A²

B√[(2πr{(πr/T)² + x²}]  = A

tanФ = AsinФ/AcosФ

tanФ = 2π²r²B/T ÷ 2πrxB

tanФ = πr/xT

Ф = tan⁻¹(πr/xT)

ε = B√[(2πr{(πr/T)² + x²}]Sin(ωt + Ф) where Ф = tan⁻¹(πr/xT)