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I am working on the below problem: "A region shown below contains a perfect conducting half-space and air. The surface current Ks on the surface of the perfect conductor is Ks= 2 amperes/metre in direction of +ve x-axis amperes per meter. The tangential H field in the air just above the perfect conductor is ?"

enter image description here

Now, since we have enter image description here , here H2 =0 and an12 should be unit vector in direction of -y axis. In that case, the answer comes out to be 2 amperes/metre in direction of +ve x-axis, however, the answer is 2 amperes/metre in direction of +ve z-axis. Can anyone please explain?

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2 Answers 2

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I'm sorry, but I'm not familiar with the equation you've given. In this instance, I'd use this equation: $$ \mathbf { H^{||}_{above} - H^{||}_{below} = K \times \hat n } $$

With $ \mathbf { H^{||}_{below} = 0 , H^{||}_{above} } $ becomes the cross product of the surface current and the normal to the surface. As you've mentioned, it has a magnitude of 2 A/m. The direction is $ \mathbf { \hat x \times \hat y } $ which gives $ \mathbf { \hat z } $.

I hope this helps!

Edit: Fixed a superscript error.

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In your equation, $\vec{n_{12}}$ is pointing upward (+ve y). So, $\vec{H_1}$ should be pointing outside of the screen (which is +ve z) so that $\vec{H_1} \times \vec{a_{12}}=\vec{K}$ is pointing in the +ve x-direction.

An easier way to see this is to consider a loop enclosing the current. And using right-hand rule, the field must point outside of the screen in the air region.

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