Answer
Thermal energy is generated in the loop at a rate of $~~3.7\times 10^{-6}~W$
Work Step by Step
We can find the radius of the wire:
$r = \frac{0.50~m}{2\pi} = 0.0796~m$
We can find the area of the loop:
$A = \pi~r^2 = (\pi)(0.0796~m)^2 = 0.01989~m^2$
We can find the induced emf:
$\mathscr{E} = -\frac{d\Phi}{dt} = (0.01989~m^2)(10.0\times 10^{-3}~T/s) = 1.989\times 10^{-4}~V$
We can find the resistance of the loop:
$R = \frac{\rho~L}{A} = \frac{(1.68\times 10^{-8}~\Omega~m)(0.50~m)}{(\pi)(5.0\times 10^{-4}~m)^2} = 0.0107~\Omega$
We can find the induced current:
$i = \frac{1.989\times 10^{-4}~V}{0.0107~\Omega} = 0.0186~A$
We can find the rate at which thermal energy is generated in the loop:
$P = i^2~R = (0.0186~A)^2(0.0107~\Omega) = 3.7\times 10^{-6}~W$
Thermal energy is generated in the loop at a rate of $~~3.7\times 10^{-6}~W$
