The Traveling Electromagnetic Wave, Quantitatively — Question 9

Question 9

A perfect-conductor boundary fixes the standing-wave node pattern

In x greater than zero, a vacuum wave incident toward a perfect conductor at x=0 has $E_{iy}=E_0\cos(kx+\omega t)$. The reflected field is $E_{ry}=-E_0\cos(kx-\omega t)$, ensuring tangential E=0 at the surface. Use omega=ck.

1. Find total Ey and Bz by superposition, including the magnetic signs.

2. Locate electric nodes and magnetic nodes, and find Bz at the surface.

Question 9 - Solution

A perfect-conductor boundary fixes the standing-wave node pattern

(a) Method and calculation

$$B_{iz}=-E_{iy}/c,\quad B_{rz}=E_{ry}/c\Rightarrow\boxed{B_z=-\frac{2E_0}{c}\cos kx\cos\omega t}.$$$$\boxed{E_y=-2E_0\sin kx\sin\omega t}.$$

(b) Interpretation and further calculation

$$E\text{ nodes}:x=m\pi/k;\quad B\text{ nodes}:x=(m+1/2)\pi/k.$$$$\boxed{B_z(0,t)=-(2E_0/c)\cos\omega t}.$$

Physical check

The conductor enforces an electric node and a magnetic antinode; the standing wave is not described by E=cB pointwise.