Question

Magnetic flux $\varphi$ (in $\text{Weber}$) linked with a closed circuit of resisitance $10\Omega$ varies with time $t$ (in $\text{seconds}$) as $\varphi =5t^2-4t+1$. The induced emf in the circuit at $t=0.2\text{sec}$ is

• A. $0.4\text{V}$
• B. $4\text{V}$
• C. $2\text{V}$
• D. $0.2\text{V}$

Answer 1

The correct option is C $2\text{V}$

Induced emf in circuit is given by,

$E=\frac{d\varphi }{dt}=\frac{d}{dt}(5t^2-4t+1)$

$E=5\times 2t-4+0=(10t-4)\text{V}$

Magnitude of induced emf at $t=0.2\text{sec}$ will be,

$e=|(10\times 0.2-4)|=2\text{V}$

Hence, option $\left(C\right)$ is the correct answer.