Current Density Producing a Triangular H(x) Profile: 04-BS-9, May 2018, Q5

Question 5: Current Density Producing a Triangular H(x) Profile (20 marks)

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 piecewise-linear "tent" profile $H_z(x)$ (only a $z$-component, depending only on $x$): rising linearly from 0 at $x=-w$ to a peak $H_0$ at $x=0$, then falling linearly back to 0 at $x=+w$, and exactly zero for $|x|>w$, with $H_0=5\times10^{-3}$ A/m and $w=1\times10^{-6}$ m.

Find. The current density $\vec J(x)$ producing this static $H$ field (the exam's own "curl" aid is Ampère's law in differential form, $\nabla\times\vec H=\vec J$, valid here since the field is static).

Triangular H(x) profile (peak H₀ at x=0, zero at x=±w) and the two uniform, oppositely-directed volume-current slabs it implies via ∇×H=J.

Approach. With $\vec H=(0,0,H_z(x))$ depending only on $x$, the curl reduces to a single non-zero component: $\nabla\times\vec H=-\dfrac{\partial H_z}{\partial x}\hat y$. Differentiate each piece of $H_z(x)$ and read off $J_y$.

General reduction of the curl. Since $H_z$ depends only on $x$ (no $y$ or $z$ dependence) and $H_x=H_y=0$ everywhere: $$\nabla\times\vec H=\left(\frac{\partial H_z}{\partial y}-0,\ 0-\frac{\partial H_z}{\partial x},\ 0-0\right)=\left(0,\ -\frac{\partial H_z}{\partial x},\ 0\right)=\vec J$$ so only $J_y=-\partial H_z/\partial x$ can be non-zero.
Differentiate each branch. For $-w<x<0$: $H_z=H_0(1+x/w)\Rightarrow \partial H_z/\partial x=H_0/w\Rightarrow J_y=-H_0/w$. For $0<x<w$: $H_z=H_0(1-x/w)\Rightarrow\partial H_z/\partial x=-H_0/w\Rightarrow J_y=+H_0/w$. For $|x|>w$: $H_z=0\Rightarrow J_y=0$.
Substitute the numbers. $H_0/w=(5\times10^{-3})/(1\times10^{-6})=5.000\times10^{3}$ A/m$^2$: $$\boxed{J_y=\begin{cases}-5.000\times10^{3}\ \text{A/m}^2, & -w<x<0\\ +5.000\times10^{3}\ \text{A/m}^2, & 0<x<w\\ 0, & |x|>w\end{cases}}$$

Region	Current density $J_y$
$-w<x<0$	$-5.000\times10^{3}$ A/m$^2$
$0<x<w$	$+5.000\times10^{3}$ A/m$^2$
$\|x\|>w$	$0$

Concept & Theory

Topic. Recovering a static volume current distribution from a given H field, using the differential (point) form of Ampère's law.

Key relations. $\nabla\times\vec H=\vec J$ (magnetostatics, $\partial\vec D/\partial t=0$); for $\vec H=(0,0,H_z(x))$ this collapses to $J_y=-\partial H_z/\partial x$.

Why this works. The differential form is a LOCAL statement: it relates the field's spatial rate of change at a point directly to the current density there, with no need to know the field elsewhere. That is exactly why a piecewise-LINEAR $H(x)$ (finite slope, no discontinuous jump) corresponds to a piecewise-CONSTANT (finite) current density rather than an infinite sheet current — a sharp KINK in $H$ is a step in $J$, whereas a true JUMP in $H$ would require a delta-function surface current instead.

Common pitfall. Assuming $H$ depends on $y$ or $z$ (it doesn't — the curl reduces to one term only), or mis-signing the two slopes: the "tent" shape means the two half-widths carry current in OPPOSITE directions, a back-to-back pair of slabs rather than one uniform slab.

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Ampère's circuital law, differential form, curl of $\vec H$).

Question 5 of 8: Current Density Producing a Triangular H(x) Profile

Question 5: Current Density Producing a Triangular H(x) Profile (20 marks)