The area of the elastic conductive coil increases at a constant speed = 0.5 m ^ 2 / s. The coil is in a uniform magnetic field

The area of the elastic conductive coil increases at a constant speed = 0.5 m ^ 2 / s. The coil is in a uniform magnetic field, the induction modulus of which is B = 0.1 T. Determine the EMF of self-induction at times t0 = 0 and t1 = 3 s /, if the lines of induction of the magnetic field are perpendicular to the plane of the loop?

ΔS / Δt = 0.5 m2 / s.
B = 0.1 T.
t0 = 0 s.
t1 = 3 s.
∠β = 90 °.
EMF -?
According to Faraday’s law of electromagnetic induction, the electromotive force of induction is directly proportional to the rate of change in the magnetic flux: EMF = – ΔF / Δt.
The “-” sign shows the direction of the current in the circuit.
The change in magnetic flux ΔФ is expressed by the formula: ΔФ = Ф1 – Ф0 = B * S1 * cosα – B * S0 * cosα = B * cosα * ΔS.
∠α = 90 ° – ∠β = 0 °.
cos0 ° = 1.
ΔФ = B * ΔS.
EMF = ΔF / Δt = B * ΔS / Δt.
EMF = 0.1 T * 0.5 m2 / s = 0.05 V.
Answer: EMF = 0.05 V appears in the circuit.