Question

A current sheet

$\overrightarrow{J}=10{\hat{u}}_{y}A/m$ lies on the dielectric interface x = 0 between two dielectric media with

${\epsilon }_{r_1}=1,{\mu }_{r_1}=1$ in Region -1 (x < 0) and ${\epsilon }_{r_2}=2,{\mu }_{r_2}=2$ in Region-2 (x > 0). If the magneitc field in Region-1 at x = $0^{+}$ is ${\overrightarrow{H}}_1=3{\hat{u}}_x+30{\hat{u}}_{y}A/m,$ the magnetic field in Region-2 at $x=0^{+}$ is

• A. ${\overrightarrow{H}}_2=1.5{\hat{u}}_x+30{\hat{u}}_y-10{\hat{u}}_{z}A/m$
• B. ${\overrightarrow{H}}_2=3{\hat{u}}_x+30{\hat{u}}_y-10{\hat{u}}_{z}A/m$
• C. ${\overrightarrow{H}}_2=1.5{\hat{u}}_x+40{\hat{u}}_{y}A/m$
• D. ${\overrightarrow{H}}_2=3{\hat{u}}_x+30{\hat{u}}_y+10{\hat{u}}_{z}A/m$

Answer 1

The correct option is A ${\overrightarrow{H}}_2=1.5{\hat{u}}_x+30{\hat{u}}_y-10{\hat{u}}_{z}A/m$

Ans : (a)

For magnetic field boundary relations are

${\overrightarrow{B}}_{n_1}={{\overrightarrow{B}}_n}_{{}_2}$
$and{\overrightarrow{H}}_{t_1}-{\overrightarrow{H}}_{t_2}=-{\overrightarrow{J}}_s\times {\overrightarrow{a}}_n$
$\Rightarrow B_{x_1}=B_{x_2}$ (x is normal component)
${\mu }_1{H_x}_{{}_1}={\mu }_2{H}_{x_2}$
$\Rightarrow 1\times 3=2\times H_{x_2}$
$H_{x_2}=1.5$
$\Rightarrow {\overrightarrow{H}}_{t_1}-{\overrightarrow{H}}_{t_2}=-10{\hat{u}}_y\times {\hat{u}}_x=+10{\hat{u}}_z$
${\overrightarrow{H}}_{t_2}=H_{t_1}-10{\hat{u}}_z$
${\overrightarrow{H}}_{t_2}=30{\hat{u}}_y-10{\hat{u}}_z$
$\Rightarrow {\overrightarrow{H}}_2=1.5{\hat{u}}_x+30{\hat{u}}_y-10{\hat{u}}_{z}A/m$