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Electromagnetic Induction — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
Suppose the loop in Exercise $6.4$ 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 $0.3T$ at the rate of $0.02T s^{–1}$. 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?
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Answer

Sides of the rectangular loop are $8\ cm$ and $2\ cm.$
Hence, area of the rectangular wire loop,
$A =$ length $ \times$ width
$= 8 \times 2 = 16\ cm^2$
$= 16 \times 10^{-4}m^2$
Initial value of the magnetic field $, B^r = 0.3T$
Rate of decrease of the magnetic field, $\frac{\text{dB}}{\text{dt}}=0.02\text{T/s}$
$\text{Emf}$ developed in the loop is given as:
$\text{e}=\frac{\text{d}\phi}{\text{dt}}$
$\text{d}\phi=$ Change in flux through the loop area
$= AB$
$\therefore\ \text{e}=\frac{\text{d(AB)}}{\text{dt}}=\frac{\text{AdB}}{\text{dt}}$
=$ 16 \times 10^{-4} \times 0.02 = 0.32$
$\therefore 10^{-4}V$
Resistance of the loop, $\text{R}=1.6\Omega$
The current induced in the loop is given as:
$\text{i}=\frac{\text{r}}{\text{R}}$
$=\frac{0.32\times10^{-4}}{1.6}=2\times10^{-5}\text{A}$
Power dissipated in the loop in the form of heat is given as:
$P = i^2R$
$= (2 \times 10^{-5})^2 \times 1.6$
$= 6.4 \times 10^{-10} W$
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

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