RMS EMF of a Loop Rotating in a Uniform Field: 04-BS-9, December 2013, Q4

Question 4: RMS EMF of a Loop Rotating in a Uniform Field (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Number of turns $N$	$10$
Loop area $A$	$10$ cm² $=1\times10^{-3}$ m²
Field $B$ (horizontal, uniform)	$10^{-5}$ T (of the order of the Earth's field)
Rotation rate	$3600$ RPM about the vertical diameter

Find. The RMS voltage (EMF) induced in the loop.

Loop of N turns rotating about a vertical diameter in a horizontal uniform field B.

Approach. Faraday's law for a loop rotating at constant angular speed $\omega$ in a uniform field: the flux is sinusoidal, so the induced EMF is sinusoidal with a peak value $N B A\omega$, and RMS $=$ peak$/\sqrt2$.

Angular speed. $$\omega=3600\ \text{RPM}=\frac{3600\times2\pi}{60}\ \text{rad/s}$$ $$\omega=\boxed{377.0\ \text{rad/s}}$$
Flux and peak EMF. With the loop normal at angle $\theta=\omega t$ to $B$, $\Phi(t)=BA\cos\omega t$, so $\varepsilon(t)=-N\,d\Phi/dt=NBA\omega\sin\omega t$: $$\varepsilon_{\text{peak}}=NBA\omega=(10)(10^{-5})(1\times10^{-3})(377.0)$$ $$\varepsilon_{\text{peak}}=\boxed{3.770\times10^{-5}\ \text{V}}$$
RMS value. For a sinusoid, RMS $=$ peak$/\sqrt2$: $$\varepsilon_{\text{rms}}=\frac{\varepsilon_{\text{peak}}}{\sqrt2}=\frac{3.770\times10^{-5}}{1.4142}$$ $$\varepsilon_{\text{rms}}=\boxed{2.666\times10^{-5}\ \text{V}=26.7\ \mu\text{V}}$$

Quantity	Result
$\omega$	$377.0$ rad/s
$\varepsilon_{\text{peak}}$	$3.770\times10^{-5}$ V
$\varepsilon_{\text{rms}}$	$2.666\times10^{-5}$ V $=26.7\ \mu$V

Concept & Theory

Topic. Faraday's law applied to an AC generator (rotating-loop EMF).

Key relations. $\Phi=BA\cos\omega t$; $\varepsilon=-N\,d\Phi/dt=NBA\omega\sin\omega t$; $\varepsilon_{\text{rms}}=\varepsilon_{\text{peak}}/\sqrt2$ for any pure sinusoid.

Why this works. As the loop turns, the flux linking it varies sinusoidally between $+BA$ and $-BA$; the rate of change of a sinusoid is itself sinusoidal (a cosine differentiates to a sine), so the EMF waveform is sinusoidal at the same angular frequency as the mechanical rotation — this is the operating principle of an AC alternator.

Common pitfall. Reporting the peak EMF when RMS is asked for (or vice versa); forgetting to convert RPM to rad/s before using it in $\omega=d\theta/dt$.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Faraday's Law, motional/rotational EMF); Young & Freedman, University Physics, Ch. 29 (AC generators).

Question 4 of 8: RMS EMF of a Loop Rotating in a Uniform Field

Question 4: RMS EMF of a Loop Rotating in a Uniform Field (20 marks)