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Answer :
The copper loop, when exposed to a changing magnetic field, generates thermal energy at a rate of 89.0 picowatts due to Joule Heating and Faraday's Law of Induction.
This problem is a combination of the physics concepts of Faraday’s Law of electromagnetic induction and the Joule Heating principle.
We want to find the rate at which thermal energy is generated in a copper loop due to a changing magnetic field.
Faraday’s Law states that the electromotive force (EMF) induced in a loop of wire is equal to the negative rate of change of the magnetic field times the area of the loop.
By this law, we have:
EMF = - d(B x A)/dt,
where B is the magnetic field strength, A is the area of the loop, and 'd/dt' denotes the rate of change.
We actually require the magnitude of EMF,
so we ignore the negative sign.
As we are given that the change in the magnetic field is constant, EMF = B x dA/dt = B x π(r^2) = 8.79e-3 T x π(0.565e-2 m)^2 = 8.77e-6 V.
The rate at which thermal energy is generated in a resistor (the Joule
Heating principle) is given by P = (EMF)^2/R, where R is the resistance of the copper loop.
To find the resistance of the wire, we use the resistivity formula R = ρ x (L/A) = 1.69e-8 Ωm x (43.6e-2 m / π(0.565e-2 m)^2), which gives us R = 0.864 Ω.
With these, the rate at which thermal energy is generated can be calculated to be P = (8.77e-6 V)^2 / 0.864 Ω = 8.90e-11 W or 89.0 pW.
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