GATE EC 1993 Electromagnetics Solution Q 5

GATE EC 1993 Electromagnetics Solution Q 5

A plane wave is incident normally on a perfect conductor as shown in figure. Here E_x^{i\;},\;H_y^i\;and\;\;\overrightarrow P^i are electric field, magnetic field and pointing vector, respectively, for the incident wave. The reflected wave should have

GATE EC 1993 Electromagnetics Solution img3 - Grad Plus

(a) E_x^r=-E_x^i

(b) H_y^r=-H_y^i

(c) \overrightarrow P^r=-\overrightarrow P^i

(d) E_x^r=E_x^i

Ans : (a) and (c)

Explanation

According to the Boundary Conditions for perfect conductor 1) tangential components of electric field are equal and opposite to have Et equal to 0 outside the surface and 2) tangential component of magnetic field intensity are continuous.

E_x^i=-E_x^r\;\;\;\;and\;\;\;\;H_y^i=\;H_y^r \overrightarrow P=\overrightarrow E\times\overrightarrow H

So the direction for reflected wave is reversed as-

-E_x^r\times H_y^r=-P_z^r

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