Question

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?]

Answer 1

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.