Question 3 of 8: Rotating loop in an inclined field: best axis and rms EMF

Question 3: Rotating loop in an inclined field: best axis and rms EMF (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. A 10-turn coil of 10 cm$^2$ area spins at 3600 rev/min about one of its own diameters, in a uniform DC field of 10$^{-5}$ T directed north and dipping 30$^\circ$ below the horizontal.

Find. The orientation of the rotation axis that maximises the induced EMF, and the rms value of the EMF in that orientation.

Approach. Write the flux linkage as the projection of B onto the rotating loop normal, isolate the factor that depends on the axis orientation, maximise it, and then divide the resulting sinusoidal peak by $\sqrt{2}$.

1. Angular speed. $$\omega = 2\pi n = 2\pi\left(\frac{3600}{60}\right) = 2\pi(60) = 376.99\ \text{rad/s}$$ so the generated waveform is a 60 Hz sinusoid.
2. Flux linkage of a loop spinning about a diameter. Let $\hat{a}$ be the unit vector along the rotation axis and $\alpha$ the angle between $\hat{a}$ and B. The loop normal $\hat{n}$ is perpendicular to $\hat{a}$ and sweeps a full circle in the plane normal to the axis. Only the component of B lying in that plane, of magnitude $B\sin\alpha$, can ever thread the loop: $$\Lambda(t) = N B A \sin\alpha \,\cos\omega t$$
3. Apply Faraday’s law. $$e(t) = -\frac{d\Lambda}{dt} = N B A \omega \sin\alpha \,\sin\omega t \;\Longrightarrow\; E_{\text{pk}} = N B A \omega \sin\alpha$$ The axis orientation enters only through $\sin\alpha$.
4. (i) Best orientation. $\sin\alpha$ is maximum at $\alpha = 90^\circ$, so $$\boxed{\text{the rotation axis must be perpendicular to }\mathbf{B}}$$ B lies in the north–south vertical plane, tilted 30$^\circ$ below the horizontal, so every axis in the plane normal to it qualifies. The two easiest to set up are a horizontal east–west axis, and an axis in the north–south vertical plane inclined 60$^\circ$ above the horizontal (i.e. 30$^\circ$ from vertical, tilted toward the north). Any linear combination of those two works equally well; the horizontal east–west axis is the practical choice.
5. Peak EMF in that orientation. With $\sin\alpha = 1$, $$E_{\text{pk}} = N B A\,\omega = (10)(1.0\times10^{-5})(1.0\times10^{-3})(376.99)$$$$E_{\text{pk}} = 3.770\times10^{-5}\ \text{V} = 37.70\ \mu\text{V}$$
6. (ii) RMS value. The waveform is a pure sinusoid, so $$E_{\text{rms}} = \frac{E_{\text{pk}}}{\sqrt{2}} = \frac{3.770\times10^{-5}}{1.4142} = \boxed{2.67\times10^{-5}\ \text{V} = 26.7\ \mu\text{V}}$$

The result is deliberately tiny: a geomagnetic-scale field, a postage-stamp loop area and only ten turns cannot produce more than tens of microvolts even at 3600 rev/min. That is exactly why practical rotating-coil magnetometers use thousands of turns and integrate over many cycles. Note that the EMF here is a true sinusoid, so the $1/\sqrt{2}$ factor is legitimate — that is not the case for the torque or power of the same loop, which vary as $\sin^2\omega t$.