Question 8 of 10: Ampère's law for a wire with graded current density

Question 8: Ampère's law for a wire with graded current density

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A cylindrical conductor of radius $a$ on the $z$-axis whose axial current density grows linearly with radius, $\mathbf{J} = (12\rho/a)\mathbf{a}_z$ A/m$^2$.

Find. The total current in the wire, and $\mathbf{H}$ both outside ($\rho \ge a$) and inside ($0 \le \rho \le a$).

Approach. Integrate the graded density over an annular element to get the enclosed current, then apply Ampère's circuital law on a circle of radius $\rho$, where cylindrical symmetry makes $\mathbf{H}$ purely azimuthal and constant around the path.

1. (a) Integrate the density over the cross-section. The density varies with $\rho$ only, so use the annular element $dS = 2\pi\rho\,d\rho$: $$I = \int_S \vec{J}\cdot d\vec{S} = \int_0^a \frac{12\rho}{a}(2\pi\rho)\,d\rho = \frac{24\pi}{a}\int_0^a \rho^2\,d\rho = \frac{24\pi}{a}\cdot\frac{a^3}{3}$$ $$\boxed{I = 8\pi a^2\ \text{A}}$$ Note that a uniform density of 12 A/m$^2$ would have given $12\pi a^2$ A; the linear grading reduces the total to two thirds of that, because the weak inner region occupies the small-area core.
2. Set up Ampère's circuital law. Cylindrical symmetry with an axial current forces $\vec{H} = H_\phi(\rho)\hat{a}_\phi$: there is no $\phi$ or $z$ dependence and no radial or axial component. On a circular path of radius $\rho$ centred on the axis, $\vec{H}$ is everywhere parallel to $d\vec{l}$ and of constant magnitude, so $$\oint\vec{H}\cdot d\vec{l} = H_\phi(2\pi\rho) = I_{enc}(\rho) \;\Rightarrow\; H_\phi = \frac{I_{enc}(\rho)}{2\pi\rho}$$
3. (b) Field outside the wire, $\rho \ge a$. Any path beyond the surface encloses the whole current, so $I_{enc} = 8\pi a^2$ and $$\boxed{\vec{H} = \frac{8\pi a^2}{2\pi\rho}\hat{a}_\phi = \frac{4a^2}{\rho}\hat{a}_\phi\ \text{A/m}, \qquad \rho \ge a}$$ This is the ordinary $1/\rho$ decay of a filament carrying $8\pi a^2$ A: outside a cylindrically symmetric conductor, the internal distribution is invisible.
4. (c) Field inside the wire, $0 \le \rho \le a$. Repeat the area integral only out to $\rho$: $$I_{enc}(\rho) = \int_0^\rho \frac{12r}{a}(2\pi r)\,dr = \frac{24\pi}{a}\cdot\frac{\rho^3}{3} = \frac{8\pi\rho^3}{a}$$ which correctly returns $8\pi a^2$ at $\rho = a$. Dividing by $2\pi\rho$, $$\boxed{\vec{H} = \frac{4\rho^2}{a}\hat{a}_\phi\ \text{A/m}, \qquad 0 \le \rho \le a}$$ The field rises as $\rho^2$ — faster than the linear rise of a uniform-density wire — because the current is pushed toward the outside.
5. Check continuity at the surface and the limits. Both expressions give $H_\phi(a) = 4a$ A/m, so the field is continuous across the boundary as it must be (no surface current is present). It also vanishes on the axis, as symmetry demands. For an illustrative $a = 1$ cm: $I = 2.513$ mA, $H_\phi(a) = 0.0400$ A/m, $H_\phi(a/2) = 0.0100$ A/m and $H_\phi(2a) = 0.0200$ A/m — the peak sits exactly at the surface.