GATE EC 2011 Electromagnetics Solution Q 4

GATE EC 2011 Electromagnetics Solution Q 4

The electric and magnetic fields for 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

\overrightarrow E=E_pe^{j\left(\omega t-280\pi y\right)}{\widehat u}_zV/m \overrightarrow H=3e^{j\left(\omega t-280\pi y\right)}{\widehat u}_xA/m

Assuming the speed of light in free space to be 3\times10^8m/s ,intrinsic impedance of free space to be 120\pi , the relative permittivity \varepsilon_r , of the medium and the electric field amplitude E_r are

a) \varepsilon_r=3,\;E_p=120\pi

b)\varepsilon_r=3,\;E_p=360\pi

c)\varepsilon_r=9,\;E_p=360\pi

d)\varepsilon_r=9,\;E_p=120\pi

Ans-(d)

Explanation

\begin{array}{l}\overrightarrow E=E_pe^{j\left(\omega t-280\pi y\right)}{\widehat u}_z\;V/m\\\overrightarrow H=3\;e^{j\left(\omega t-280\pi y\right)}\;{\widehat u}_x\;A/m\\c=3\times10^8\;m/s\end{array}

Comparing the given equation with standard form we have,

\beta=280\pi=\frac{2\pi}\lambda \begin{array}{l}\therefore\;\;\lambda=\frac1{140}meter\\\;\;\;\;\;v=f\lambda\\\;\;\;\;\;\;\;=14\times10^9\times\frac1{140}m/sec\end{array} v=1\times10^8m/sec v=\frac C{\sqrt{\in_r\mu_r}} 1\times10^8=\frac{3\times10^8}{\sqrt{1\times\in_r}} \therefore\;\in_r=9 \frac{E_P}{H_P}=\eta=\sqrt{\frac\mu\in}=\sqrt{\frac{\mu_0\times1}{\in_0\times9}}=\frac13=\sqrt{\frac{\mu_0}{\in_0}} \Rightarrow\frac{E_P}3=\frac13\times120\pi \therefore\;E_P=120\pi

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