Question 3 of 8: EMF Induced in a Rotating Pick-Up Loop

Question 3: EMF Induced in a Rotating Pick-Up Loop (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 horizontally polarised plane wave arrives from below the horizon travelling to the north-west and 45° above horizontal, and a small loop spinning slowly about a vertical axis samples its magnetic field.

Find. (i) the rms voltage induced in the loop, and (ii) the orientation of the loop plane that maximises it.

Approach. Convert the power density to a magnetic field, resolve that field against the loop normal using the wave geometry, and apply Faraday's law — noting that the 1 GHz field varies a billion times faster than the 1 Hz rotation, so the rotation only modulates the coupling.

1. Magnetic field from the power density. For a uniform plane wave in air, $S = \eta_0 H^{2}$ with rms quantities, so$$H_{\text{rms}} = \sqrt{\frac{S}{\eta_0}} = \sqrt{\frac{10^{-3}}{376.8}} = 1.629\times10^{-3}\ \text{A/m},$$and for reference $E_{\text{rms}} = \sqrt{S\eta_0} = 0.614$ V/m.
2. Direction of H. The wave is horizontally polarised, so $\mathbf{E}$ lies horizontally at right angles to the ground track. Since $\hat{\mathbf{H}} = \hat{\mathbf{k}}\times\hat{\mathbf{E}}$, and $\hat{\mathbf{k}}$ makes an angle $\psi = 45^{\circ}$ with the horizontal, $\mathbf{H}$ lies in the vertical plane that contains the ray, tilted below the horizontal, with components$$H_{\text{horiz}} = H\sin\psi,\qquad H_{\text{vert}} = H\cos\psi.$$Its horizontal part therefore lies along the NW–SE line, the ground track of the ray.
3. Flux linked by the loop. The loop spins about its own vertical axis, so its normal $\hat{\mathbf{n}}$ stays horizontal and sweeps the compass. Writing $\phi$ for the angle between $\hat{\mathbf{n}}$ and the NW–SE line, only the horizontal part of H couples:$$\Phi = \mu_0 A\,\mathbf{H}\cdot\hat{\mathbf{n}} = \mu_0 A H \sin\psi\,\cos\phi .$$
4. Faraday's law with two very different time scales. The wave oscillates at $10^{9}$ Hz while $\phi$ advances at only 1 Hz, so the rotation is quasi-static and the induced emf is set by the field oscillation alone:$$V_{\text{rms}}(\phi) = \omega\,\mu_0 A H_{\text{rms}}\sin\psi\,|\cos\phi|,\qquad \omega = 2\pi\times10^{9}\ \text{rad/s}.$$
5. Peak flux and peak reading. Substituting the numbers with $\cos\phi = 1$,$$\Phi_{\text{rms}} = (1.2566\times10^{-6})(2\times10^{-4})(1.629\times10^{-3})(0.7071) = 2.895\times10^{-13}\ \text{Wb},$$$$V_{\max} = (6.2832\times10^{9})(2.895\times10^{-13}),$$$$\boxed{V_{\text{rms,max}} = 1.82\ \text{mV}}$$
6. Value averaged over a revolution. If the meter integrates over the whole 1 s rotation as well as over the carrier, the extra factor $\langle\cos^{2}\phi\rangle = 1/2$ applies, giving$$V_{\text{rms,overall}} = \frac{1.819}{\sqrt{2}},$$$$\boxed{V_{\text{rms,overall}} = 1.29\ \text{mV}}$$Both figures are quoted because the question does not say whether the reading is taken at one orientation or averaged over the spin.
7. (ii) Orientation for maximum output. The output peaks when $\hat{\mathbf{n}}$ is parallel to the horizontal component of H, that is when the loop normal points north-west (or, equivalently, south-east):$$\boxed{\text{loop plane} = \text{the vertical NE--SW plane}}$$The output is zero a quarter turn later, when the loop plane contains the ray's ground track. Note the residual factor $\sin 45^{\circ} = 0.707$: even in the best orientation the horizontal loop normal can only capture the horizontal part of a magnetic field that is itself tilted 45° out of the horizontal plane.