A wire loop with a radius of 1 cm, having a resistance of 1 mOhm, is penetrated by a uniform magnetic field

A wire loop with a radius of 1 cm, having a resistance of 1 mOhm, is penetrated by a uniform magnetic field, the induction lines of which are perpendicular to the plane of the loop. The magnetic field induction changes smoothly at a rate of 0.01 T / s. How much heat will be released in a loop in 1 minute?

r = 1 cm = 0.01 m.

R = 1 mΩ = 0.001 Ω.

ΔВ / t = 0.01 T / s.

T = 1 min = 60 s.

∠α = 0 °.

Q -?

The released amount of heat Q is expressed by the Joule-Lenz law: Q = EMF2 * T / R, where EMF is the electromotive force of induction, T is time, R is the resistance of the wire loop.

When the magnetic induction changes in the wire loop, an induction current arises.

EMF = ΔФ / t – Faraday’s law.

The change in the magnetic flux ΔФ in the loop is expressed by the formula: ΔФ = Δ (B * S) * cosα.

Since the area of ​​the loop S does not change, then ΔФ = S * cosα * ΔB.

EMF = S * cosα * ΔB / t.

The area of ​​the loop S is expressed by the formula: S = P * r ^ 2.

EMF = P * r ^ 2 * cosα * ΔB / t.

EMF = 3.14 * (0.01 m) ^ 2 * cos 0 ° * 0.01 T / s = 3.14 * 10 ^ -6 V.

Q = (3.14 * 10 ^ -6 V) 2 * 60 s / 0.001 Ohm = 6 * 10 ^ -7 J.

Answer: Q = 6 * 10 ^ -7 J of thermal energy is released in the wire loop.