$A$ rectangular wire loop of sides $8 \; cm$ and $2 \; cm$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \; T$ directed normal to the loop. Suppose the loop is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of $0.3 \; T$ at the rate of $0.02 \; T \, s^{-1}$. If the cut is joined and the loop has a resistance of $1.6 \; \Omega$,how much power is dissipated by the loop as heat? What is the source of this power?

(N/A) The area of the rectangular loop is $A = 8 \; cm \times 2 \; cm = 16 \; cm^2 = 16 \times 10^{-4} \; m^2$.
The rate of change of the magnetic field is $\frac{dB}{dt} = 0.02 \; T \, s^{-1}$.
The induced electromotive force (emf) $e$ in the loop is given by Faraday's law: $e = \left| \frac{d\phi}{dt} \right| = A \frac{dB}{dt}$.
Substituting the values: $e = (16 \times 10^{-4} \; m^2) \times (0.02 \; T \, s^{-1}) = 0.32 \times 10^{-4} \; V$.
The induced current $i$ in the loop with resistance $R = 1.6 \; \Omega$ is $i = \frac{e}{R} = \frac{0.32 \times 10^{-4} \; V}{1.6 \; \Omega} = 2 \times 10^{-5} \; A$.
The power dissipated as heat is $P = i^2 R = (2 \times 10^{-5} \; A)^2 \times 1.6 \; \Omega = 4 \times 10^{-10} \times 1.6 \; W = 6.4 \times 10^{-10} \; W$.
The source of this power is the external agent (the power supply) that is reducing the current in the electromagnet,thereby changing the magnetic field.