Magnetic Fields Around a Three-Conductor Coaxial Transmission Line: 04-BS-9, December 2015, Q6

Question 6: Magnetic Fields Around a Three-Conductor Coaxial Transmission Line (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
Cylinder radii	$5$ mm, $10$ mm, $15$ mm (thin, coaxial)
Current on $5$ mm cylinder	$2$ A (outgoing)
Return current on $10$ mm and $15$ mm cylinders	$1$ A each (opposite sense)

Find. $B(r)$, magnitude and sense of circulation, in each of the three free-space regions.

Cross-section of the three coaxial cylinders (5, 10, 15 mm radii). Region I: between 5–10 mm; Region II: between 10–15 mm; Region III: outside 15 mm.

Approach. Apply Ampere's law on a circular path of radius $r$ centred on the axis in each region; the enclosed current is the algebraic sum of all conductor currents inside that radius, giving $B(r)=\mu_0 I_{\text{enc}}/(2\pi r)$.

Region I, $5\text{ mm}<r<10\text{ mm}$: only the inner (5 mm) conductor is enclosed. $$I_{\text{enc,I}}=+2\ \text{A}\qquad B_{\text{I}}(r)=\frac{\mu_0(2)}{2\pi r}=\frac{2\times10^{-7}\times2}{r}$$ At the two boundaries: $B_{\text{I}}(5\text{mm})=\boxed{8.0\times10^{-5}\ \text{T}}$, $B_{\text{I}}(10\text{mm})=\boxed{4.0\times10^{-5}\ \text{T}}$, circulating in the sense given by the right-hand rule about the $2$ A current.
Region II, $10\text{ mm}<r<15\text{ mm}$: the 5 mm conductor plus the 10 mm return are enclosed. $$I_{\text{enc,II}}=2-1=+1\ \text{A}\qquad B_{\text{II}}(r)=\frac{\mu_0(1)}{2\pi r}=\frac{2\times10^{-7}\times1}{r}$$ At the two boundaries: $B_{\text{II}}(10\text{mm})=\boxed{2.0\times10^{-5}\ \text{T}}$, $B_{\text{II}}(15\text{mm})=\boxed{1.33\times10^{-5}\ \text{T}}$ — the net enclosed current is still in the same sense as the 5 mm conductor's (just reduced by 1 A), so region II circulates in the same sense as region I.
Region III, $r>15\text{ mm}$: all three conductors are enclosed. $$I_{\text{enc,III}}=2-1-1=0\ \text{A}\implies B_{\text{III}}=\boxed{0}$$

Region	$I_{\text{enc}}$	$B(r)$	Sense
I ($5$–$10$ mm)	$+2$ A	$2\times10^{-7}(2)/r$: $80\to40\ \mu$T	about the $2$A current (RH rule)
II ($10$–$15$ mm)	$+1$ A	$2\times10^{-7}(1)/r$: $20\to13.3\ \mu$T	same as region I
III ($>15$ mm)	$0$	$0$	none (no net enclosed current)

Concept & Theory

Topic. Ampere's law for a multi-conductor coaxial line.

Key relations. $\oint \vec B\cdot d\vec l=\mu_0 I_{\text{enc}}\Rightarrow B(r)=\dfrac{\mu_0 I_{\text{enc}}}{2\pi r}$, with $I_{\text{enc}}$ the algebraic sum of every conductor current inside radius $r$.

Why this works. Ampere's law only sees the NET current threading the Amperian loop; adding a return conductor inside the loop's radius reduces $I_{\text{enc}}$ by that conductor's current (with its sign), so the field steps down each time a new return conductor is crossed, and vanishes entirely once every conductor's current has cancelled — exactly the coaxial-cable shielding principle.

Common pitfall. Treating the return currents as adding to, rather than subtracting from, the enclosed current; or forgetting that region III's field is zero not because the conductors are far away, but because the ENTIRE current returns before that radius (net enclosed current is identically zero for any $r$ beyond the outermost conductor).

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Ampere's circuital law, coaxial conductors).

Question 6 of 8: Magnetic Fields Around a Three-Conductor Coaxial Transmission Line

Question 6: Magnetic Fields Around a Three-Conductor Coaxial Transmission Line (20 marks)