Suppose the loop in the exercise is stationary but the current feeding the electromagnet is gradually reduced so that the field decreases from its initial value 0.3T at the rate $0.02T{s^{ - 1}}$. If the cut is joined and the loop has a resistance of 1.6 ohm, how much power is dissipated by the loop as heat? What is the source of this power?

[Exercise: A rectangular wire loop of sides 8cm and 2cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1cm/s in a direction normal to the (a) longer side (b) shorter side of the loop? For how long does the induced voltage last in each case?]

Hint: According to Faraday’s law of magnetic induction, the magnitude of emf induced in a conducting loop is equal to the rate of change of magnetic flux through the loop.

Formulas used:
Magnetic flux is given by:
$\Phi = \overrightarrow A \cdot \overrightarrow B $ …… (1)
where,
$\Phi $is the magnetic flux.
$\overrightarrow A $is the area vector.
$\overrightarrow B $is the magnetic field vector.
Faraday’s law:
$e = - \dfrac{{d\Phi }}{{dt}}$ …… (2)
where,
e is the emf generate.
Current is given by:
$I = \dfrac{e}{R}$ …… (3)
where,
I is the current through the loop.
R is the resistance of the wire.
Power is given by:
$P = {I^2}R$ …… (4)
where,
P is the power dissipated through the loop.

Complete step by step answer:
Given:
1. Sides of rectangular loop l=8cm, b=2cm.
2. Initial magnetic field $({B_0}) = 0.3T$ .
3. Rate of change of magnetic field $\dfrac{{dB}}{{dt}} = 0.02T/s$.
4. Resistance of wire (R) = 1.6 ohm.

To find:
1. The amount of power dissipated by the loop.
2. The source of this power.

Step 1 of 4:
Find the area of the loop:
$
  \overrightarrow A = l \times b \\
  \overrightarrow A = 8cm \times 2cm \\
  \overrightarrow A = 16c{m^2} \\
  \overrightarrow A = 16 \times {10^{ - 4}}{m^2} \\
 $

Step 2 of 4:
Find the emf using by putting the eq (1) in eq (2):
\[emf = - \dfrac{{d(\overrightarrow {A \cdot } \overrightarrow B )}}{{dt}}\] …… (5)
Since, area vector and direction of magnetic field both lie parallel to each other eq (5) becomes:
\[emf = - \dfrac{{d(A \times ( - B))}}{{dt}}\] …… (6)
Since, area is constant while magnetic field is time varying, eq (6) becomes:
\[emf = A\dfrac{{dB}}{{dt}}\] …… (7)
Put the values of A and $\dfrac{{dB}}{{dt}}$ in eq (7) to find emf:
$
  emf = 16 \times {10^{ - 4}} \times 0.02 \\
  emf = 0.32 \times {10^{ - 4}}V \\
 $

Step 3 of 4:
Put the values of emf and R in eq (3) to find current (I):
$
  I = \dfrac{{0.32 \times {{10}^{ - 4}}}}{{1.6}} \\
  I = 2 \times {10^{ - 5}}A \\
 $

Step 4 of 4:
Put the values of I and R in eq (4) to find power:
\[
  P = {(2 \times {10^{ - 5}})^2} \times 1.6 \\
  P = 6.4 \times {10^{ - 10}}W \\
 \]

Final Answer: The power dissipated through the loop is $6.4 \times {10^{ - 10}}W$. This heat loss is due to the external agent responsible for changing the magnetic field.

Note: In questions where we have to find the power dissipated by a changing magnetic field, go step by step. First calculate flux then emf, then current and finally power.