The expression for E^{\to}_{r} is

Correct Answer is : 0.23E02(a^x+a^z)e−jπ×104(x−z)32V∕m

0.23{E_{0}}/{\sqrt{2}}(a^{\^}_{x} + a^{\^}_{z})e^{−j {\pi \times 10^{4}(x−z)}/{3\sqrt{2}}}V\/ m

− {E_{0}}/{\sqrt{2}}(a^{\^}_{x} + a^{\^}_{z})e^{−j\frac{\pi \times 10^{4}z}{3}}V\/ m

0.44{E_{0}}/{\sqrt{2}}(a^{\^}_{x} + a^{\^}_{z})e^{−j {\pi \times 10^{4}(x−z)}/{3\sqrt{2}}}V\/ m

{E_{0}}/{\sqrt{2}}(a^{\^}_{x} + a^{\^}_{z})e^{−j\frac{\pi \times 10^{4}z}{3}}V\/ m

GATE Electronics and Communications (EC) 2013 Solved Paper

Question : 63 of 65

Marks: +1, -0

The expression for

E↖{→}_{r}

0.23 $\;{E_{0}}/{{√{2}}}(a↖{\^}_{x} + a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}(x−z)}/{3{√{2}}}}V\/ m$
$−\;{E_{0}}/{{√{2}}}(a↖{\^}_{x} + a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}z}/{3}}V\/ m$
0.44 $\;{E_{0}}/{{√{2}}}(a↖{\^}_{x} + a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}(x−z)}/{3{√{2}}}}V\/ m$
$\;{E_{0}}/{{√{2}}}(a↖{\^}_{x} + a↖{\^}_{z})e^{−j\;{π\;\;×\;10^{4}z}/{3}}V\/ m$

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