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

Question 4: RMS EMF of a Loop Rotating in a Uniform DC Magnetic 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.

Find. (i) The RMS induced voltage; (ii) the loop orientation at which the induced voltage is maximum.

Approach. Convert the rotation rate to angular frequency, compute the peak EMF from Faraday's law for a loop of fixed area rotating in a uniform field, then convert peak to RMS; the position of maximum EMF follows directly from where $d\Phi/dt$ is largest.

1. Angular frequency of rotation. $$f=\frac{10^4\ \text{RPM}}{60}=166.7\ \text{Hz}\qquad \omega=2\pi f=\boxed{1047\ \text{rad/s}}$$
2. Peak EMF, $\varepsilon(t)=N B A\omega\sin(\omega t)$. With $A=\ell^2=(0.10)^2=0.01\ \text{m}^2$: $$\varepsilon_{\text{peak}}=N B A\omega=(10)(10^{-5})(0.01)(1047)$$ $$\varepsilon_{\text{peak}}=\boxed{1.047\times10^{-3}\ \text{V}}$$
3. RMS voltage. $$\varepsilon_{\text{rms}}=\frac{\varepsilon_{\text{peak}}}{\sqrt2}=\frac{1.047\times10^{-3}}{\sqrt2}$$ $$\varepsilon_{\text{rms}}=\boxed{7.405\times10^{-4}\ \text{V}}$$
4. Position of maximum induced voltage. $\varepsilon(t)\propto\sin(\omega t)$ is largest when $\cos(\omega t)=0$, i.e. when the flux $\Phi=BA\cos(\omega t)$ is instantaneously zero — this occurs exactly when the loop's plane contains the field direction (loop plane oriented north–south, normal pointing east–west), not when the plane faces the field: $$\boxed{\text{Maximum EMF when the loop's plane is parallel to }B\text{ (contains the N–S line)}}$$