Question

Magnetic flux $\varphi$ (in Webers) linked with a closed circuit of resisitance $100\Omega$ varies with time $t$ (in seconds) as $\varphi =5t^2-4t+1$. The induced current in the circuit at $t=0.2\text{sec}$ is

• A. $5\text{mA}$
• B. $10\text{mA}$
• C. $20\text{mA}$
• D. $1\text{A}$

Answer 1

The correct option is C $20\text{mA}$

Given,

$R=100\Omega$

$\varphi =5t^2-4t+1$

$t=0.2\text{Sec}$

We know that emf induced in a circuit is given by,

$E_{ind}=\left|\begin{array}{c}\frac{d\varphi }{dt}\end{array}\right|$

$\Rightarrow E_{ind}=\left|\begin{array}{c}\frac{d}{dt}(5t^2-4t+1)\end{array}\right|$

$E_{ind}=\left|\begin{array}{c}10t-4\end{array}\right|$

$\therefore i_{ind}=\frac{E_{ind}}{R}=\frac{\left|\begin{array}{c}10t-4\end{array}\right|}{100}$

${\left(i_{ind}\right)}_{t=0.2}=\frac{\left|\begin{array}{c}10\times 0.2-4\end{array}\right|}{100}$

$\therefore i_{ind}=20\text{mA}$

Hence ($C$) is the correct answer.

Why this Question? Note: The value of induced current in a closed path can always be calculated by using, $i=\frac{E_{ind}}{R}$ Where, $E_{ind}=\left|\begin{array}{c}\frac{d\varphi }{dt}\end{array}\right|$