Question

Flux;$\varphi$ (in Weber's) in a closed circuit of resistance $10ohm$ varies with time $t$ (in seconds) according to the equation $\varphi =6t^2-5t+1$. What is the magnitude of the induced current in $0.25$ seconds?

Answer 1

Step 1: Given

Resistance, $R=10ohm$

Time $t=0.25s$

Equation according to which flux $\varphi$is varying, $\varphi =6t^2-5t+1$

Step 2: Find the emf

The derivative of flux gives us the emf.

Hence,

$$\begin{gathered} emf=\frac{d\varphi }{dt} \\ =\frac{d(6t^2-5t+1)}{dt} \\ =12t-5 \end{gathered}$$

Substituting the time,

$$\begin{gathered} \left|emf\right|=\left|12(0.25)-5\right| \\ =\left|-2\right|=2 \end{gathered}$$

Step 3: Find the magnitude of induced current

Using Ohm's law,

$$\begin{gathered} I=\frac{emf}{R} \\ =\frac{2}{10} \\ =0.2A \end{gathered}$$

Hence, the magnitude of current induced is $0.2A$