Question 10: Electromagnetic Induction

Question

The magnetic flux $\phi$ through a stationary loop of wire having a resistance R varies with time as $\phi=4 t^2+3 t$. The average emf and total charge flowing in the loop in the time interval $t=0$ to $t=	au$ respectively are

Accepted Answer

First, recall that the induced emf is given by FaradayÃ¢ÂÂs law
$$\mathcal{E}=-\frac{d\phi}{dt}\,. $$
Here
$$\phi(t)=4t^2+3t\quad\Longrightarrow\quad \frac{d\phi}{dt}=8t+3\,, $$
so in magnitude
$$|\mathcal{E}|=8t+3\,. $$
Average emf (magnitude) over $0$ to $	au$:
$      \langle\mathcal{E}angle      =\frac{1}{	au}\int_{0}^{	au}(8t+3)\,dt      =\frac{1}{	au}\Bigl[4t^2+3t\Bigr]_{0}^{	au}      =4	au+3\,.    $
Total charge $Q$ that flows in that time (using $I=\mathcal{E}/R$, $dQ=I\,dt$):
$      Q=\frac{1}{R}\int_{0}^{	au}(8t+3)\,dt      =\frac{1}{R}\bigl[4	au^2+3	au\bigr]\,.    $
Hence the correct choice is Option D:
Average emf = $4	au+3$,
Total charge = $\dfrac{4	au^2+3	au}{R}$.

Answer

$	au+3, \quad \frac{4 	au^2+3 	au}{R}$

Answer

$3 	au+4, \quad \frac{4 	au^2+3 	au}{R}$

Answer

$4 	au+3, \quad \frac{4 r+3}{R}$

Answer

$4 	au+3, \quad \frac{4 	au^2+3 	au}{R}$

Electromagnetic Induction

The magnetic flux \phi through a stationary loop of wire having a resistance R varies with time as \phi=4 t^2+3 t. The average emf and total charge flowing in the loop in the time interval t=0 to t=\tau respectively are

Choose the correct answer: