(a) Use Ampère's law (8.6.4) to prove that the current density \(J(z)\) in the conductor of Fig. \(8.6 .2\) is related to the net magnetic field just outside the conductor by $$ B_{\text {outeide }}=\mu_{0} \int_{0}^{\infty} J(z) d z=\mu_{0} K, $$ where the integral symbolized by \(K\) has the dimensions of a surface current density, namely, amperes per meter. \((b)\) Consider an artificial model whereby the surface current \(K\) is distributed uniformly in the skin layer \(0

Ampère's law states that the circulation of the magnetic field around a closed loop is equal to the product of the permeability \(\mu_0\) and the total current passing through the surface enclosed by the loop. For this exercise, we need to relate the current density \(J(z)\) to the net magnetic field just outside the conductor, \(B_\text{outside}\). We can write Ampère's law as: $$ \oint \vec{B} \cdot d\vec{l} = \mu_0 \int \vec{J} \cdot d\vec{A} $$ Consider a rectangular loop just outside the conductor of width \(dA\). The magnetic field is tangential to the conductor and has constant magnitude over the surface. Thus, the circulation of the magnetic field around the loop is equal to \(B_\text{outside}\times dA\). On the other hand, the total current passing through the surface enclosed by the loop is \(\int_0^\infty J(z) dz\). Substituting these expressions into Ampère's law, we get: $$ B_\text{outside} \times dA = \mu_0 \int_{0}^{\infty} J(z) dz $$ Dividing by \(dA\), we have: $$ B_\text{outside} = \mu_{0} \int_{0}^{\infty} J(z) d z = \mu_{0} K $$ where \(K\) has the dimensions of surface current density.

Let's consider the simplified model with a uniform surface current \(K\) distributed in the skin layer \(0 < z < \delta\), such that the current density is \(J_\text{skin} = \frac{K}{\delta}\). To find the time-average power per unit area, we use the following expression: $$ \left(\frac{\overline{J_\text{skin}^2}}{g}\right) \delta $$ Substituting \(J_\text{skin} = \frac{K}{\delta}\), we get: $$ \frac{\overline{K^2}}{g \delta} = \frac{\overline{B_\text{outside}^2}}{\mu_0^2 g \delta} $$ This result matches with the time-average power per unit area found in Prob. \(8.6.6 b\). Now, the resistance ratio of a round wire for high-frequency currents to direct current can be found using the ratio of effective areas: $$ \frac{R_\text{ac}}{R_\text{dc}} = \frac{2 \delta}{a} $$ where \(a\) is the radius of the wire. This concludes the exercise.