We now apply Ampere's circuit law to determine H for some symmetrical current distributions as we did for Gauss's law. We will consider an infinite line current, an infinite current sheet, and an infinitely long coaxial transmission line.

A. Infinite Line Current

Consider an infinitely long filamentary current I along the z-axis as in Figure 1. To determine H at an observation point P, we allow a closed path pass through P. This path, on which Ampere's law is to be applied, is known as an Amperian path (analogous to the term Gaussian surface). We choose a concentric circle as the Amperian path in view of eq. (1),

$\mathbf{H}=\frac{I}{2 \pi \rho} \mathbf{a}_{\phi}$ ...(1)

which shows that H is constant provided $\rho$ is constant. Since this path encloses the whole current I, according to Ampere's law

$I=\int H_{\phi} \mathbf{a}_{\phi} \cdot \rho d \phi \mathbf{a}_{\phi}=H_{\phi} \int \rho d \phi=H_{\phi} \cdot 2 \pi \rho$

or

$\mathbf{H}=\frac{l}{2 \pi \rho} \mathbf{a}_{\phi}$

B. Infinite Sheet of Current

Consider an infinite current sheet in the z = 0 plane. If the sheet has a uniform current density $K = K_{y}a_{y}$ A/m as shown in Figure 2, applying Ampere's law to the rectangular closed path (Amperian path) gives

$\oint \mathbf{H} \cdot d \mathbf{l}=I_{\mathrm{enc}}=K_{y} b$ ...(2)

To evaluate the integral, we first need to have an idea of what H is like. To achieve this, we regard the infinite sheet as comprising of filaments; dH above or below the sheet due to a pair of filamentary currents can be found using eqs. (1) and (3). As evident in Figure 2(b), the resultant dH has only an x-component. Also, H on one side of the sheet is the negative of that on the other side. Due to the infinite extent of the sheet, the sheet can be regarded as consisting of such filamentary pairs so that the characteristics of H for a pair are he same for the infinite current sheets, that is,

$\mathbf{a}_{\phi}=\mathbf{a}_{\ell} \times \mathbf{a}_{\rho}$ ...(3)

$\mathbf{H}=\left\{\begin{array}{ll}{H_{0} \mathbf{a}_{x}} & {z\gt0} \\ {-H_{\mathrm{o}} \mathbf{a}_{x}} & {z\lt0}\end{array}\right.$...(4)

where Ho is yet to be determined. Evaluating the line integral of H in eq. (4) along the closed path in Figure 2(a) gives

$\begin{aligned} \oint \mathbf{H} \cdot d \mathbf{l} &=\left(\int_{1}^{2}+\int_{2}^{3}+\int_{3}^{4}+\int_{4}^{1}\right) \mathbf{H} \cdot d \mathbf{l} \\ &=0(-a)+\left(-H_{\mathrm{o}}\right)(-b)+0(a)+H_{\mathrm{o}}(b) \\ &=2 H_{\mathrm{o}} b \end{aligned}$ ...(5)

From eqs. 2 and 5, we obtain $H_{\mathrm{o}}=\frac{1}{2} K_{y}$ . Substituting $H_{\mathrm{o}}$ in eq. 4 gives

$\mathbf{H}=\left\{\begin{array}{ll}{\frac{1}{2} K_{y} \mathbf{a}_{x},} & {z\gt0} \\ {-\frac{1}{2} K_{y} \mathbf{a}_{x}} & {z\lt0}\end{array}\right.$

In general, for an infinite sheet of current density K A/m,

$\mathbf{H}=\frac{1}{2} \mathbf{K} \times \mathbf{a}_{n}$

where $a_n$ is a unit normal vector directed from the current sheet to the point of interest.

C. Infinitely Long Coaxial Transmission Line

Consider an infinitely long transmission line consisting of two concentric cylinders having their axes along the z-axis. The cross section of the line is shown in Figure 3, where the z-axis is out of the page. The inner conductor has radius a and carries current I while the outer conductor has inner radius b and thickness t and carries return current -I. We want to determine H everywhere assuming that current is uniformly distributed in both conductors. Since the current distribution is symmetrical, we apply Ampere's law along the Amperian path for each of the four possible regions: $0 \leq \rho \leq a, a \leq \rho \leq b, b \leq \rho \leq b+t$ and $\rho \geq b+t$

For region $0 \leq \rho \leq a,$ we apply Ampere's law to path $L_{1}$ , giving

$\oint_{L_{1}} \mathbf{H} \cdot d \mathbf{l}=I_{\mathrm{enc}}=\int \mathbf{J} \cdot d \mathbf{S}$ ...(6)

since the current is uniformly distributed over the cross section,

$$ \mathbf{J}=\frac{1}{\pi a^{2}} \mathbf{a}_{z}, \quad d \mathbf{S}=\rho d \phi d \rho \mathbf{a}_{z} $$

$I_{\mathrm{enc}}=\int \mathbf{J} \cdot d \mathbf{S}=\frac{I}{\pi a^{2}} \iint \rho d \phi d \rho=\frac{I}{\pi a^{2}} \pi \rho^{2}=\frac{I \rho^{2}}{a^{2}}$

Hence eq. 6 becomes

$H_{\phi}=\frac{I \rho}{2 \pi a^{2}}$