A region shown below contains a perfect conducting half-space and
A region shown below contains a perfect conducting half-space and air. The surface current \(\overrightarrow {{K_S}} \) on the surface of the perfect conductor is \(\overrightarrow {{K_S}} = \hat x2\) amperes per meter. The tangential \(\vec H_t\) field in the air just above the perfect conductor is
A. (x̂ + ẑ) 2 amperes per meter
B. x̂ 2 amperes per meter
C. –ẑ 2 amperes per meter
D. ẑ 2 amperes per meter
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
The boundary condition for the tangential magnetic field is given by:
\({H_{t1}} - {H_{t2}} = \overrightarrow {{K_s}} \times {\hat a_n}\)
Also, the tangential component of the magnetic field inside a perfect conductor is zero.
Calculation:
Given:
The boundary condition for tangential magnetic field is:
\({H_{{t_1}}} - {H_{{t_2}}} = \overrightarrow {{K_S}} \times {\hat a_n}\)
The tangential component of the magnetic field inside a perfect conductor is zero.
i.e. \({H_{{t_2}}} = 0\)
\({H_{{t_1}}} = \overrightarrow {{K_S}} \times {\hat a_n} = 2\hat x \times \hat y\)
\({H_{{t_1}}} = 2\hat z\;A/m\)
