Magnetic Field of a Wire with Radially-Varying Current Density: 17-Phys-A3 Electromagnetics, Undated paper, Q8

Question 8: Magnetic Field of a Wire with Radially-Varying Current Density (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. $\mathbf J(\rho)=\dfrac{12\rho}{a}\hat a_z$ A/m² for $0\le\rho\le a$ (radius $a$ left symbolic — no numeric value is given), $\mathbf J=0$ for $\rho>a$.

Find. Total current $I$; $\mathbf H(\rho)$ for $\rho\ge a$ and for $\rho\le a$.

Approach. Integrate $J$ over the cross-section for $I$; use Ampere's circuital law on a circular path of radius $\rho$, with the enclosed current found by integrating $J$ only out to $\rho$ when $\rho\lt a$.

Part (a) — Total current. $$I=\int_0^aJ(\rho)\,2\pi\rho\,d\rho=\int_0^a\frac{12\rho}{a}(2\pi\rho)\,d\rho=\frac{24\pi}{a}\cdot\frac{a^3}{3}=\boxed{8\pi a^2\ \text{A}.}$$
Part (b) — H outside the wire, $\rho\ge a$. The Amperian loop at any $\rho\ge a$ encloses ALL the current: $$H_\phi(2\pi\rho)=I=8\pi a^2 \Rightarrow \boxed{\mathbf H=\frac{4a^2}{\rho}\,\hat a_\phi\ \text{A/m},\qquad \rho\ge a.}$$
Part (c) — H inside the wire, $\rho\le a$. $$I_{enc}(\rho)=\int_0^\rho\frac{12\rho'}{a}(2\pi\rho')\,d\rho'=\frac{24\pi}{a}\cdot\frac{\rho^3}{3}=\frac{8\pi\rho^3}{a}.$$ $$H_\phi(2\pi\rho)=I_{enc}(\rho) \Rightarrow \boxed{\mathbf H=\frac{4\rho^2}{a}\,\hat a_\phi\ \text{A/m},\qquad 0\le\rho\le a.}$$ Check: at $\rho=a$ both expressions give $\mathbf H=4a\,\hat a_\phi$ — continuous across the wire's surface, as required (no surface current is specified).

$H_\phi$ versus $\rho$ (normalized by $4a$, horizontal axis in units of $\rho/a$): rises as $\rho^2$ inside the wire, then falls as $1/\rho$ outside, peaking at $\rho=a$.

Final results
Quantity	Value
Total current, $I$	$8\pi a^2$ A
$\mathbf H$, $\rho\ge a$	$\dfrac{4a^2}{\rho}\hat a_\phi$ A/m
$\mathbf H$, $\rho\le a$	$\dfrac{4\rho^2}{a}\hat a_\phi$ A/m

Topic: Ampere's circuital law for a nonuniform (radially-varying) current distribution.

Key relations: $\oint\mathbf H\cdot d\mathbf l=I_{enc}$; $I_{enc}(\rho)=\int_0^\rho J(\rho')2\pi\rho'\,d\rho'$ for $\rho\lt a$, $=I_{total}$ for $\rho\ge a$.

Why this works: The problem's azimuthal symmetry means $\mathbf H=H_\phi(\rho)\hat a_\phi$ on any circular Amperian loop, turning the line integral into $H_\phi\cdot2\pi\rho$; the only work is computing how much current a loop of radius $\rho$ actually encloses, which requires integrating $J$ (not just multiplying by area, since $J$ is not uniform).

Common pitfall: Using the TOTAL current in Ampere's law for an INSIDE loop ($\rho\lt a$) — only the current enclosed by that specific loop counts, and here it grows as $\rho^3$ (steeper than $\rho^2$ since $J$ itself grows linearly with $\rho$), not as $\rho^2$.

Source: Sadiku, Elements of Electromagnetics, Ch. 7 (Ampere's circuital law, magnetic field of nonuniform current).

Question 8 of 10: Magnetic Field of a Wire with Radially-Varying Current Density

Question 8: Magnetic Field of a Wire with Radially-Varying Current Density (20 marks)