Question 5 of 8: EMF Induced in a Loop by a Plane Wave

Question 5: EMF Induced in a Loop by a Plane Wave (equal value)

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. Plane wave $f=3000$ MHz, $S=2$ W/m$^2$, propagating $60^\circ$ above the horizontal along a north-west bearing, $\mathbf E$ horizontally polarized; sensing loop $N=10$ turns, $A=25$ cm$^2$, lying in the vertical east–west plane (loop normal along north–south).

Find. The RMS EMF induced in the loop.

Approach. Determine the wave's H-field direction from its known propagation direction and (horizontal) E-field direction via $\hat H=\hat k\times\hat E$, project that onto the loop's normal (north–south), then apply Faraday's law $\mathcal E_{rms}=N A\omega B_{\perp,rms}$.

1. Set up axes and find $\hat k$. With $x=$East, $y=$North, $z=$Up, the horizontal NW bearing is $\hat u_{NW}=(-\sin45^\circ,\cos45^\circ,0)$, so with a $60^\circ$ elevation, $$\hat k=\cos60^\circ\,\hat u_{NW}+\sin60^\circ\,\hat z=(-0.354,\,0.354,\,0.866).$$
2. Find $\hat E$. $\mathbf E$ is horizontal and perpendicular to $\hat k$; the only horizontal direction perpendicular to $\hat u_{NW}$ is the NE–SW line, $\hat E=\hat u_{NE}=(\sin45^\circ,\cos45^\circ,0)=(0.707,0.707,0)$ (a horizontal wave can only be transverse to a tilted $\hat k$ along this one line).
3. Find $\hat H=\hat k\times\hat E$. $$\hat H=(-0.612,\ 0.612,\ -0.500)\quad(\text{unit vector}).$$
4. Project onto the loop normal. The loop lies in the vertical E–W plane, so its normal is $\hat n=\hat y$ (north–south): $$\hat H\cdot\hat n=0.612.$$ Only this fraction of $H$ links the loop.
5. Wave's H-field magnitude. From $S=E_0^2/(2\eta_0)=\eta_0 H_0^2/2$, $$H_0=\sqrt{\frac{2S}{\eta_0}}=\sqrt{\frac{2\times2}{377}}=0.1030\ \text{A/m (peak)},\qquad H_{rms}=\frac{H_0}{\sqrt2}=0.0728\ \text{A/m}.$$
6. Induced EMF. $B_{\perp,rms}=\mu_0 H_{rms}(\hat H\cdot\hat n)$ and $\mathcal E_{rms}=N A\omega B_{\perp,rms}$, with $\omega=2\pi f=1.885\times10^{10}$ rad/s: $$B_{\perp,rms}=(4\pi\times10^{-7})(0.0728)(0.612)=5.61\times10^{-8}\ \text{T},$$ $$\mathcal E_{rms}=(10)(25\times10^{-4})(1.885\times10^{10})(5.61\times10^{-8})=\boxed{26.4\ \text{V}}.$$