You are designing an ac generator with a maximum emf of $8.0 \mathrm {~V}$. If the generator coil has 200 turns, each with a cross-sectional area of $0.030 \mathrm {~m} ^ { 2 }$, what should be the frequency of the generator in a magnetic field of $0.030 \mathrm {~T}$ ?
A) $44 \mathrm {~Hz}$
B) $8.0 \mathrm {~Hz}$
C) $7.1 \mathrm {~Hz}$
D) $7.5 \mathrm {~Hz}$

Question

Quizplus · Accepted Answer

The maximum emf ($E_{max}$) generated by an AC generator can be calculated using the formula $E_{max} = NAB\omega$, where $N$ is the number of turns, $A$ is the cross-sectional area, $B$ is the magnetic field strength, and $\omega$ is the angular frequency ($\omega = 2\pi f$, with $f$ being the frequency). Rearranging for $f$, we get $f = \frac{E_{max}}{2\pi NAB}$. Substituting the given values: $f = \frac{8.0}{2\pi \times 200 \times 0.030 \times 0.030} \approx 7.1 \mathrm{Hz}$.

You Are Designing an Ac Generator with a Maximum Emf 8.0 V8.0 \mathrm {~V}8.0 V

You Are Designing an Ac Generator with a Maximum Emf $8.0 \mathrm {~V}$