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The electric and magnetic fields of a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity $\varepsilon_r$ and relative permeability $\mu_r=1$ are given by\n\n$\mathbf{E}=E_p e^{j(\omega t-280\pi z)}\hat{u}_z\ \text{V/m}$\n\n$\mathbf{H}=3 e^{j(\omega t-280\pi z)}\hat{u}_x\ \text{A/m}$\n\nAssuming the speed of light in free space to be $3\times10^8$ m/s and the intrinsic impedance of free space to be $120\pi$, the relative permittivity $\varepsilon_r$ of the medium and the electric field amplitude $E_p$ are

1. $\varepsilon_r=3,\ E_p=120\pi$
2. $\varepsilon_r=3,\ E_p=360\pi$
3. $\varepsilon_r=9,\ E_p=360\pi$
4. $\varepsilon_r=9,\ E_p=120\pi$

Correct answer: $\varepsilon_r=9,\ E_p=120\pi$

Solution

The phase constant is $\beta=280\pi$ rad/m, so the wavelength is $\lambda=2\pi/\beta=1/140$ m. With $f=14$ GHz, the wave speed is $v=f\lambda=10^8$ m/s, giving $\sqrt{\varepsilon_r\mu_r}=c/v=3$, so $\varepsilon_r=9$ since $\mu_r=1$. Also, the intrinsic impedance is $\eta=\eta_0/\sqrt{\varepsilon_r}=120\pi/3=40\pi$, and since $|H|=3$ A/m, $E_p=\eta|H|=40\pi\times3=120\pi$ V/m.

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