GATE EC 2011 Electromagnetics Solution Q 6

GATE EC 2011 Electromagnetics Solution Q 6

A current sheet

\overrightarrow J=10{\widehat u}<em>y\;A/m

lies on the dielectric interface x = 0 between two dielectric media with

\varepsilon{r_1}=1,\;\mu_{r_1}=1\;in\;region-1\;\left(x<0\right)

and \varepsilon_{r_2}=2,\;\mu_{r_2}=2\;in\;region- 2\;\left(x>0\right) .
If the magnetic field in region-1 at x=0^- is {\overrightarrow H}_1=3{\widehat u}_x+30{\widehat u}_y\;A/m, the magnetic field in Region-2 at x=0^+ is

(a) {\overrightarrow H}_2=1.5{\widehat u}_x+30{\widehat u}_y-10{\widehat u}_z\;A/m

(b) {\overrightarrow H}_2=3{\widehat u}_x+30{\widehat u}_y-10{\widehat u}_z\;A/m

(c) {\overrightarrow H}_2=1.5{\widehat u}_x+40{\widehat u}_y\;A/m

(d) {\overrightarrow H}_2=3{\widehat u}_x+30{\widehat u}_y+10{\widehat u}_z\;A/m

Ans : (a)

Explanation

GATE EC 2011 Electromagnetics Solution img1 - Grad Plus

We know the  magnetic boundary conditions,

{\overrightarrow B}<em>{n_1}={\overrightarrow B}</em>{n_2} {\overrightarrow H}_{t_1}-{{\overrightarrow H}_t}_2=-{\overrightarrow J}_s\times{\overrightarrow a}_n

In our question, the normal component of B is x component.

\therefore B_{x_1}=B_{x_2} \mu_1H_{x1}=\mu_2H_{x2} \begin{array}{l}\therefore1\times3=2\times H_{x2}\\\therefore H_{x2}=1.5\end{array}

In this question tangential component of H is y component.

\begin{array}{l}{\overrightarrow H}<em>{t_1}-{\overrightarrow H}</em>{t_2}=-10{\widehat u}<em>y\times{\widehat u}_x\\\therefore{\overrightarrow H}</em>{t_1}-{\overrightarrow H}_{t_2}=+10{\widehat u}_z\end{array} {\overrightarrow H}<em>{t_2}=H</em>{t_1}-10{\widehat u}_z {\overrightarrow H}_{t_2}=30{\widehat u}_y-10{\widehat u}_z {\overrightarrow H}_2=1.5{\widehat u}_x+30{\widehat u}_y-10{\widehat u}_z\;A/m

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