Question 5 of 8: Rotating Direction-Finding Loop in the Earth’s Field

Question 5: Rotating Direction-Finding Loop in the Earth’s 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. A multi-turn vertical loop spins about a vertical axis aboard an aircraft flying in a straight line through the Earth’s field, which points north and dips 45° below the horizontal.

Find. (i) the induced EMF, (ii) whether the aircraft’s translation adds to it, and (iii) the loop orientation at which the EMF vanishes.

Approach. The loop’s normal stays horizontal as it spins about a vertical axis, so only the horizontal component of the Earth’s field is ever linked; differentiate the resulting sinusoidal flux for the EMF, then treat the translation separately using the closed-loop form of the motional term.

1. Resolve the field. With a dip of 45° below horizontal, the component lying in the horizontal plane (and pointing magnetic north) is $$B_{h}=B\cos\delta=\left(10\times10^{-6}\right)\cos45^{\circ}=7.071\ \mu\text{T}$$ The vertical component of equal size is parallel to the spin axis and therefore always lies in the plane of the loop, linking no flux at any moment.
2. Convert the rotation rate. $$\omega=\frac{2\pi(2500)}{60}=261.8\ \text{rad/s},\qquad f=\frac{2500}{60}=41.67\ \text{Hz}$$ The induced signal is therefore a 41.67 Hz sinusoid — the rotation frequency itself, not twice it, because flux linkage varies as the cosine of the azimuth angle and not its square.
3. Write the flux linkage. Letting \(\theta=\omega t\) be the angle between the loop’s (horizontal) normal and magnetic north, $$\Phi(t)=NB_{h}A\cos\omega t$$
4. Differentiate for the EMF. By Faraday’s law, $$e(t)=-\frac{d\Phi}{dt}=NB_{h}A\,\omega\sin\omega t$$ so the peak value is $$e_{\text{peak}}=NB_{h}A\omega=(20)(7.071\times10^{-6})(2.5\times10^{-3})(261.8)=\boxed{92.56\ \mu\text{V}}$$ and, since this is a genuine sinusoid, its rms value is $$e_{\text{rms}}=\frac{e_{\text{peak}}}{\sqrt{2}}=\boxed{65.45\ \mu\text{V}}$$ Both should be quoted: the question says “calculate EMF” without specifying which measure, and the two differ by a factor \(\sqrt{2}\).
5. Part (ii): test the motional contribution. The motional EMF around a closed circuit is $$e_{\text{motional}}=\oint_{C}\left(\mathbf{v}\times\mathbf{B}\right)\cdot d\boldsymbol{\ell}$$ The aircraft flies straight and level, so every element of the loop shares the same velocity \(\mathbf{v}\), and the Earth’s field is uniform over a loop only 5.6 cm across. The vector \(\mathbf{v}\times\mathbf{B}\) is therefore a constant, and the closed line integral of any constant vector vanishes identically because \(\oint d\boldsymbol{\ell}=0\). $$\boxed{e_{\text{motional}}=0:\ \text{the horizontal motion contributes nothing}}$$
6. Confirm that the answer is not merely “small”. The largest possible value of \(|\mathbf{v}\times\mathbf{B}|\) is \((55.56)(10^{-5})=556\ \mu\text{V/m}\), which acting over the loop’s 17.7 cm circumference would look like roughly one hundred microvolts — comparable with the rotational signal, so this is not a case of neglecting a small term. It is an exact cancellation, and the flux picture says the same thing: sliding a rigid loop through a uniform field changes no flux through it. Only a field gradient, a change of loop shape, or rotation can induce an EMF, and the aircraft supplies none of those.
7. Part (iii): locate the nulls. From step 4 the EMF is proportional to \(\sin\theta\), so it vanishes when \(\theta=0^{\circ}\) or \(180^{\circ}\): that is, when the loop’s normal points magnetic north or magnetic south, which means the plane of the loop lies in the vertical east–west plane, perpendicular to \(B_{h}\). $$\boxed{e=0\ \text{when the loop plane is the vertical E--W plane (normal on the N--S line)}}$$ This is precisely the orientation of maximum flux linkage, which is why the rate of change is zero there; the EMF peaks a quarter turn later, when the loop plane contains the magnetic meridian. The nulls occur twice per revolution, and it is this sharp double null — not the broad maximum — that a direction-finding loop is aimed with.