Question

Magnetic flux linked with a stationary loop resistance $R$ varies with respect to time during the time period $T$ as follows: $\varphi =at(T-t)$ The amount of heat generated in the loop during that time (inductance of the coil is negligible) is?

Answer 1

Step 1: Given Data

Given equation: $\varphi =at(T-t)$

Resistance $=R$

Time period $=T$

Step 2: Faraday's Law of Electromagnetic Induction

1. Induced emf in a circuit is proportional to the rate of change of magnetic flux linked with the circuit.
2. The direction of induced emf is such that it tries to oppose the cause of its generation, i.e., the variation of magnetic flux producing it.

If $\varphi$ be the magnetic flux linked with a circuit at any time $t$ then the laws of electromagnetic induction can be expressed mathematically as

$\epsilon =-\frac{d\varphi }{dt}$, $\left(1\right)$

Where $\epsilon$ is the induced emf, here, it is important to point out that $\epsilon$ does not depend on how the flux $\varphi$ is changed.

Step 3: Calculate the emf

From the question we have,

$\varphi =at(T-t)$

We know that emf is given as,

$\epsilon =-\frac{d\varphi }{dt}$

Upon substituting the expression for $\varphi$ the emf is given as

$\epsilon =-\frac{d}{dt}\left\{at\left(T-t\right)\right\}$

$=-\frac{d}{dt}\left(atT-a{t}^2\right)$

$=-aT+2at$

$=a\left(2t-T\right)$

Step 4: Calculate the heat generated

We know that the expression of power is

$P=\frac{{\epsilon }^2}{R}$

$=\frac{a^2{\left(2t-T\right)}^2}{R}$

Therefore, heat generated,

$H={\int }_0^{T}Pdt$

$={\int }_0^T\frac{a^2{\left(2t-T\right)}^2}{R}dt$

$=\frac{a^2}{R}{\int }_0^T\left(4t^2-4t+T^2\right)dt$

$=\frac{4a^2}{R}{\int }_0^T{t}^{2}dt-\frac{4a^2}{R}{\int }_0^{T}tdt+\frac{T{a}^2}{R}{\int }_0^{T}dt$

$=\frac{4a^2}{R}{\left[\frac{t^3}{3}\right]}_0^T-\frac{4a^2}{R}{\left[\frac{t^2}{2}\right]}_0^T+\frac{a^2}{R}{\left[t\right]}_0^T$

$=\frac{4a^2{T}^3}{3R}-\frac{2a^2{T}^2}{R}+\frac{a^{2}T}{R}$

$=\frac{a^{2}T}{R}\left(\frac{4T^2}{3}-2T+1\right)$

Hence, the heat generated in the loop is $\frac{a^{2}T}{R}\left(\frac{4T^2}{3}-2T+1\right)$.