Answer
6.01 mJ.
Work Step by Step
The power dissipated by a resistor is $P=I^2R$.
The energy dissipated is $E=P\Delta t=I^2R \Delta t$
Calculate the current I, which is created by the induced emf. The emf is given by Faraday’s Law.
$$\epsilon=-\frac{\Delta \Phi_B}{\Delta t}=-\frac{A\Delta B}{\Delta t}$$
$$I=\frac{\epsilon}{R}=-\frac{A\Delta B}{R\Delta t}$$
$$ E=P\Delta t=I^2R \Delta t =\frac{A^2(\Delta B)^2}{R^2(\Delta t)^2}R\Delta t$$
$$ E=\frac{A^2(\Delta B)^2}{R (\Delta t) } $$
$$ E=\frac{((0.240m)^2)^2(0-0.665T)^2}{(6.10\Omega) (0.0400s) }=6.01\times10^{-3}J $$
