Question

The number of turns in the coil of AC generator is $5000$, and the area of the coil is $0.25{\text{m}}^2$. The coil is rotated at the rate of $100\left(\text{cycles}/\text{sec}\right)$ in a magnetic field of $0.2{\text{Wb m}}^{-2}$. The peak value of emf generated is nearly

• A. $766\text{kV}$
• B. $440\text{kV}$
• C. $157\text{kV}$
• D. $210\text{kV}$

Answer 1

The correct option is C $157\text{kV}$

Given:
$N=5000;A=0.25{\text{m}}^2;B=0.2{\text{Wb m}}^{-2}$
$f=100\left(\text{cycles}/\text{sec}\right)$

For a loop having $N$ turns, the total induced emf across the two ends of the loop is,

$\left|\begin{array}{c}E\end{array}\right|=NBA\omega \sin \omega t$

For peak value, $\sin \omega t=1$

${\left|\begin{array}{c}E\end{array}\right|}_{max}=\omega NBA=\left(2\pi f\right)\left(5000\right)\left(0.2\right)\left(0.25\right)$

${\left|\begin{array}{c}E\end{array}\right|}_{max}=\frac{2\times 3.14\times 100\times 5000\times 50}{1000}=157000$

$\therefore {\left|\begin{array}{c}E\end{array}\right|}_{max}=157\text{kV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.