B Field at the Midpoint of a DC Transmission Line: 04-BS-9, May 2014, Q3

Question 3: B Field at the Midpoint of a DC 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
Power delivered $P$	$10^5$ W
Line voltage $V$	$10$ kV
Conductor separation (centre-to-centre)	$0.5$ m

Find. The magnitude and direction of $\vec B$ at the midpoint between the two conductors.

Two-wire DC line: forward current out of the page, return current into the page. Each wire's field at the midpoint points the same way, so they add.

Approach. Find the line current from $P=VI$, then superpose the two long-straight-wire fields at the midpoint — because the currents in a forward/return pair are anti-parallel, their fields at the midpoint add rather than cancel.

Line current. $$I=\frac{P}{V}=\frac{10^5}{10^4}$$ $$I=\boxed{10\ \text{A}}$$
Field of one wire at the midpoint. Distance from each conductor to the midpoint is half the separation, $r=0.25$ m: $$B_1=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi\times10^{-7})(10)}{2\pi(0.25)}$$ $$B_1=8.0\times10^{-6}\ \text{T}$$
Superpose. With the return current anti-parallel to the forward current, the right-hand-rule fields from the two wires point the same way at the midpoint (parallel to the plane containing the two conductors, perpendicular to the line joining them) — they add rather than cancel: $$B=2B_1=\frac{\mu_0 I}{\pi r}=2(8.0\times10^{-6})$$ $$B=\boxed{1.60\times10^{-5}\ \text{T}=16.0\ \mu\text{T}}$$, directed parallel to the plane of the two conductors, perpendicular to the line joining them.

Quantity	Result
Line current $I$	$10$ A
$B$ at midpoint	$1.60\times10^{-5}$ T (16.0 µT)

Concept & Theory

Topic. Superposition of the magnetic field of two long parallel current-carrying wires (a two-wire transmission line).

Key relations. Single infinite wire: $B=\mu_0 I/2\pi r$; power/current/voltage: $P=VI$.

Why this works. A DC line's forward and return currents are equal in magnitude and opposite in direction. By the right-hand rule, at a point exactly between two anti-parallel wires, each wire's field circles the same way past the midpoint, so the two contributions reinforce instead of cancelling — the opposite of what happens at points outside the pair, where the fields subtract.

Common pitfall. Assuming the fields cancel at the midpoint because the currents are "opposite" — that intuition is correct for the fields outside the wire pair, but backwards for the region directly between them.

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Biot–Savart Law, force between current-carrying conductors).

Question 3 of 8: B Field at the Midpoint of a DC Transmission Line

Question 3: B Field at the Midpoint of a DC Transmission Line (20 marks)