Question

Two coils have a mutual inductance $0.005H$. The current changes in the first coil according to the equation $I=I_0\sin \left(\omega t\right)$, where $I_0=10A$and $\omega =10\pi \frac{rad}{sec}$. The maximum value of emf in the second coil is?

Answer 1

Step 1: Given that

Two coils having mutual inductance i.e. $M=0.005H$

The alternating current is given by, $I=I_0\sin \left(\omega t\right)$

The peak value of the current, $I_0=10A$

And the angular frequency of AC, $\omega =10\pi \frac{rad}{sec}$

Step 2: Formula used

Now the value of emf ($\epsilon$) in the secondary coil can be given by,

$\epsilon =M\frac{dI}{dt}$

So,

$\begin{array}{rcl}\epsilon & =& M\frac{dI}{dt}\\ & =& M\times \frac{d}{dt}\left(I_0\sin \left(\omega t\right)\right)\\ & =& M\times I_0\cos \left(\omega t\right)\times \omega ........\left(1\right)\end{array}$

Step 3: To find the maximum value of emf in the second coil.

Now for the maximum value of emf, $\cos \left(\omega t\right)=1$.

So from equation (1), the maximum value of emf can be given by,

$$\begin{gathered} {\epsilon }_{max}=M\times I_0\times \omega \\ =(0.005)\times \left(10\right)\times \left(10\pi \right) \\ =0.5\pi V \end{gathered}$$

Therefore, the maximum value of emf in the second coil is $0.5\pi V$.