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+### Statement
+
+$12.1.21.$ [Insert the problem statement]
+
+### Solution
+
+1. Configuration and fields
+
+We take the z-axis perpendicular to the wall, with the metal occupying z < 0 and the vacuum in z > 0. The wave is incident from the right.
+
+Incident wave (traveling in the -z direction):
+
+$\mathbf{E}_i = E \cos(\omega t + kz)\,\hat{\mathbf{x}}, \qquad
+\mathbf{B}_i = \frac{E}{c} \cos(\omega t + kz)\,\hat{\mathbf{y}} \quad (\text{SI})$
+
+In CGS: $\mathbf{B}_i = E \cos(\omega t + kz)\,\hat{\mathbf{y}} (same units as E)$.
+
+Reflected wave (from a perfect conductor):
+
+$\mathbf{E}_r = -E \cos(\omega t - kz)\,\hat{\mathbf{x}}, \qquad
+\mathbf{B}_r = \frac{E}{c} \cos(\omega t - kz)\,\hat{\mathbf{y}} \quad (\text{SI})$.
+
+In CGS:$ \mathbf{B}_r = E \cos(\omega t - kz)\,\hat{\mathbf{y}}$.
+
+At the surface (z = 0):
+
+$\mathbf{E}_{\text{total}} = E \cos\omega t - E \cos\omega t = 0 \quad (\text{node, electric field is zero})$.
+
+$\mathbf{B}_{\text{total}} = \frac{E}{c}\cos\omega t + \frac{E}{c}\cos\omega t = \frac{2E}{c}\cos\omega t\,\hat{\mathbf{y}} \quad (\text{SI}), \qquad
+\mathbf{B}_{\text{total}} = 2E\cos\omega t\,\hat{\mathbf{y}} \quad (\text{CGS})$
+
+At the surface of a perfect conductor, the magnetic field has a discontinuity related to the surface current $\mathbf{K}$
+
+In SI:
+
+$\mathbf{n} \times (\mathbf{H}_{\text{vacuum}} - \mathbf{H}_{\text{metal}}) = \mathbf{K}$
+
+The normal vector \mathbf{n} points from the metal into the vacuum:
+$\mathbf{n} = \hat{\mathbf{z}}$.
+
+Inside the metal
+
+$\mathbf{H} = 0$.
+
+In vacuum
+
+$\mathbf{H} = \mathbf{B}/\mu_0$.
+
+$\mathbf{H}_{\text{vacuum}} = \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}}$.
+
+Then:
+
+$\mathbf{K} = \hat{\mathbf{z}} \times \left( \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}} \right) = -\frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{x}}.$
+
+The amplitude (linear current density) is:
+
+$\boxed{i = \frac{2E}{\mu_0 c} = 2\varepsilon_0 c E \quad (\text{SI}).}$
+
+In CGS:
+
+$\mathbf{n} \times (\mathbf{B}_{\text{vacuum}} - \mathbf{B}_{\text{metal}}) = \frac{4\pi}{c}\mathbf{K}$
+
+With
+$\mathbf{B}_{\text{vacuum}} = 2E\cos\omega t\,\hat{\mathbf{y}}, \mathbf{B}_{\text{metal}} = 0, \mathbf{n} = \hat{\mathbf{z}}$:
+
+$\hat{\mathbf{z}} \times (2E\cos\omega t\,\hat{\mathbf{y}}) = -\frac{4\pi}{c}\mathbf{K} \quad\Rightarrow\quad \mathbf{K} = -\frac{cE}{2\pi}\cos\omega t\,\hat{\mathbf{x}}$
+
+Amplitude:
+
+$\boxed{i = \frac{cE}{2\pi} \quad (\text{CGS}).}$
+
+The average pressure on a perfect reflector is$ p = 2I/c$ where I is the average incident intensity.
+
+In SI:
+Average intensity: $I = \frac{1}{2}\varepsilon_0 c E^2$
+
+$p = \frac{2I}{c} = \varepsilon_0 E^2$
+
+$\boxed{p = \varepsilon_0 E^2 \quad (\text{SI}).}$
+
+In CGS:
+Average intensity:
+$ I = \frac{c}{8\pi} E^2$
+
+p = \frac{2I}{c} = \frac{E^2}{4\pi}.
+
+$\boxed{p = \frac{E^2}{4\pi} \quad (\text{CGS}).}$
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]