Question

The instantaneous magnetic flux associated with a closed loop of resistance $10\Omega$ is given by, $\varphi =6t^2-5t+1$, then the magnitude of induced current at $t=0.25\text{s}$ will be -

• A. $0.4\text{A}$
• B. $0.8\text{A}$
• C. $0.6\text{A}$
• D. $0.2\text{A}$

Answer 1

The correct option is D $0.2\text{A}$

According to Faraday's law of electromagnetic induction, emf induced due to the change in magnetic flux is given by,

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

Here, $\varphi =6t^2-5t+1$

So, $E=-\frac{d\varphi }{dt}=-(12t-5)$

Now, $E_{t=0.25}=-12\times 0.25+5=2\text{V}$

$\therefore i=\frac{E}{R}=\frac{2}{10}=0.2\text{A}$