Magnetic Flux Density of a Finite-Radius Straight Conductor: 04-BS-9, December 2016, Q4

Question 4: Magnetic Flux Density of a Finite-Radius Straight Conductor (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.

Quantity	Value
Total current $I$	$1$ A, flowing north, uniform current density
Conductor radius $a$	$1$ mm $=10^{-3}$ m
Plot range	$r=0$ to $2$ mm

Find. $B(r)$ for $0\le r\le2$ mm (plot), and the direction of $\vec B$ at points above the axis.

$B(r)$ rises linearly from $0$ to the surface value at $r=a$ (Ampère's law with the enclosed-current fraction $\propto r^2$), then falls as $1/r$ outside the conductor.

Approach. Apply Ampère's law with a circular loop of radius $r$ centred on the axis. Inside the conductor ($r\le a$) the enclosed current is only the fraction of the total carried within radius $r$ (current density uniform, so enclosed current $\propto r^2$); outside ($r>a$) the full current $I$ is always enclosed. The direction at any point follows the right-hand rule for the given (north) current direction.

Inside the conductor ($0\le r\le a$). Enclosed current $I_{\text{enc}}=I(r/a)^2$, so Ampère's law $\oint\vec B\cdot d\vec l=\mu_0I_{\text{enc}}$ gives $$B(r)=\frac{\mu_0I_{\text{enc}}}{2\pi r}=\frac{\mu_0I r}{2\pi a^2}\qquad(0\le r\le a)$$ This is linear in $r$, reaching its maximum at the surface $r=a$: $$B(a)=\frac{\mu_0(1)}{2\pi(10^{-3})}=\boxed{2.000\times10^{-4}\ \text{T}=200\ \mu\text{T (surface, }r=a)}$$
Outside the conductor ($r>a$). The full current is always enclosed, so the field is the standard infinite-wire result: $$B(r)=\frac{\mu_0I}{2\pi r}\qquad(r>a)$$ At the requested endpoint $r=2$ mm: $$B(2\ \text{mm})=\frac{\mu_0(1)}{2\pi(2\times10^{-3})}=\boxed{1.000\times10^{-4}\ \text{T}=100\ \mu\text{T}}$$
Direction above the axis. With current flowing north ($\hat y$ in an East-North-Up frame) and the field point directly above the axis ($\hat z$ direction from the wire), the right-hand rule gives $\vec B\propto\hat I\times\hat r=\hat y\times\hat z=\hat x$: $$\boxed{\vec B\text{ points due EAST at points directly above the axis (for every }r\text{, inside or outside)}}$$

Quantity	Result
$B$ at surface ($r=a=1$mm)	$2.000\times10^{-4}$ T (peak)
$B$ at $r=0.5$mm (inside)	$1.000\times10^{-4}$ T
$B$ at $r=2$mm (outside)	$1.000\times10^{-4}$ T
Direction above the axis	due East, for all $r$ in $[0,2\text{mm}]$

Concept & Theory

Topic. Ampère's law for a finite-radius straight conductor carrying uniform current density — the field both inside and outside the wire.

Key relations. Inside: $B(r)=\mu_0Ir/(2\pi a^2)$ (linear); outside: $B(r)=\mu_0I/(2\pi r)$ (the familiar infinite-wire form); the two expressions agree exactly at $r=a$, so $B(r)$ is continuous with a kink (slope discontinuity) at the surface.

Why this works. Ampère's law only "sees" the current threaded through the loop; for uniform current density the enclosed current scales as the enclosed area ($\propto r^2$) while for $r>a$ the loop always encloses the full wire, which is what produces the linear-then-$1/r$ shape.

Common pitfall. Treating the conductor as a thin filament (ignoring its finite radius) and using $B=\mu_0I/(2\pi r)$ even for points inside the conductor, which wrongly predicts a field that diverges as $r\to0$ instead of the correct linear rise to zero at the axis.

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Ampère's circuital law, field of a finite conductor).

Question 4 of 8: Magnetic Flux Density of a Finite-Radius Straight Conductor

Question 4: Magnetic Flux Density of a Finite-Radius Straight Conductor (20 marks)