Question

The magnetic field through a single loop of wire having radius $12\text{cm}$ and resistance $8.5\Omega$ changes with time as shown in the figure. The magnetic field is perpendicular to the plane of the loop. Plot the induced current as a function of time.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Emf induced due to change in magnetic flux is,

$E=-\frac{d\varphi }{dt}=-\frac{d\left(BA\right)}{dt}=-A\frac{dB}{dt}$

So, the induced current,

$I=\frac{E}{R}=-\frac{A\left(\frac{dB}{dt}\right)}{R}$

For, $0\text{sec}<t<2\text{sec}:$

From given diagram,
$\frac{dB}{dt}=\frac{B_2-B_1}{t_2-t_1}=\frac{1-0}{2-0}=\frac{1}{2}$

$I=-\frac{3.14(0.12{)}^2\times \left[\frac{1}{2}\right]}{8.5}\approx -0.0026\text{A}$

For $2\text{sec}<t<4\text{sec}:$

$I=-\frac{3.14(0.12{)}^2\times \left[\frac{1-1}{4-2}\right]}{8.5}=0\text{A}$

For $4\text{sec}<t<6\text{sec}:$

$I=-\frac{3.14(0.12{)}^2\times \left[\frac{0-1}{6-4}\right]}{8.5}\approx 0.0026\text{A}$

On plotting the value of $I$, we obtain the graph as shown below.


Hence, option $\left(\text{A}\right)$ is correct.