Magnetic Flux Density at the Midpoint and at the Jumper Centres of a Long Rectangular Current Loop: 04-BS-9, May 2018, Q4

Question 4: Magnetic Flux Density at the Midpoint and at the Jumper Centres of a Long Rectangular Current Loop (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
Loop current $I$	2 A
Wire separation	2 cm $\Rightarrow$ half-separation $a=0.01$ m
Straight wire length	1 km (western wire flows north; eastern wire, the return path, flows south)
End connectors	horizontal semicircular jumpers, radius $a=0.01$ m

Find. $\vec B$ at (i) the loop's midpoint (between the wires, far from both ends) and (ii) the centre of curvature of each semicircular jumper.

Top view of the loop (schematic, not to scale): two 1 km straight wires 2 cm apart, joined by semicircular jumpers of radius 1 cm. Field points: the loop's midpoint (i), and the centre of curvature of each jumper (ii).

Approach. At the midpoint, both straight wires behave as ordinary infinite wires, $B=\mu_0I/(2\pi r)$, since the field point sits ~500 m from either end. At a jumper centre, each straight wire instead acts as a SEMI-INFINITE conductor (it only extends away from that point), contributing half the infinite-wire value, and the semicircular arc itself contributes half of a full loop's centre field, $\mu_0I/(2a)$. All contributions are combined using the right-hand rule.

Part (i) — field at the midpoint from each straight wire. Each wire is $a=0.01$ m from the midpoint and, over 500 m of straight run either side, behaves as an infinite wire: $$B_{\text{each}}=\frac{\mu_0I}{2\pi a}$$ Tracking directions by the right-hand rule (west wire north-flowing, east wire south-flowing) shows both fields point the SAME way — vertically DOWNWARD, perpendicular to the horizontal loop — so they add: $$B_{\text{mid}}=2\times\frac{\mu_0I}{2\pi a}=\frac{\mu_0I}{\pi a}=\frac{(4\pi\times10^{-7})(2)}{\pi(0.01)}$$ $$\boxed{B_{\text{mid}}=8.000\times10^{-5}\ \text{T, directed vertically DOWNWARD}}$$
Part (ii) — the semicircular arc's own field at its centre. A full circular loop of radius $a$ gives $\mu_0I/(2a)$ at its centre; every element of a semicircle sits at the SAME distance $a$ from that centre, so half the current path contributes exactly half the field: $$B_{\text{arc}}=\frac{\mu_0I}{4a}=\frac{(4\pi\times10^{-7})(2)}{4(0.01)}=6.283\times10^{-5}\ \text{T}$$
Part (ii) — straight-wire contribution at a jumper centre. From a jumper's centre, each 1 km wire is now a SEMI-INFINITE conductor (it starts right there and extends only one way), so each contributes HALF the infinite-wire value, and together: $$B_{\text{straight}}=2\times\frac{\mu_0I}{4\pi a}=\frac{\mu_0I}{2\pi a}=4.000\times10^{-5}\ \text{T}$$ — exactly half of $B_{\text{mid}}$, as expected. The same right-hand-rule tracking shows the arc's own field and this straight-wire field also point vertically DOWNWARD, so they add: $$B_{\text{jumper}}=B_{\text{arc}}+B_{\text{straight}}=6.283\times10^{-5}+4.000\times10^{-5}$$ $$\boxed{B_{\text{jumper}}=1.028\times10^{-4}\ \text{T, directed vertically DOWNWARD (identical at both jumpers by symmetry)}}$$

Check: treating the 1 km straight sections as effectively infinite (part i) or semi-infinite (part ii) is justified because 1 km is five orders of magnitude larger than the 1 cm radius — the neglected end-correction is far below the precision retained.

Quantity	Result
(i) Midpoint field $B_{\text{mid}}$	$8.000\times10^{-5}$ T, down
(ii) Arc's own field $B_{\text{arc}}$	$6.283\times10^{-5}$ T, down
(ii) Straight-wire field at jumper $B_{\text{straight}}$	$4.000\times10^{-5}$ T, down
(ii) Total field at each jumper centre $B_{\text{jumper}}$	$1.028\times10^{-4}$ T, down

Concept & Theory

Topic. Superposition of Biot–Savart/Ampère fields from straight and arc-shaped current segments sharing a common radius.

Key relations. Infinite wire $B=\mu_0I/(2\pi r)$; semi-infinite wire (field point at the very end) $B=\mu_0I/(4\pi r)$; circular arc of radius $a$ subtending angle $\theta$ at its centre, $B=\mu_0I\theta/(4\pi a)$ ($\theta=2\pi$ for a full loop, $\theta=\pi$ for a semicircle).

Why this works. The Biot–Savart law is linear, so the field at any point is the vector sum of every segment's independent contribution; this geometry is convenient because the arc and both straight sections all sit exactly $a$ from the two points of interest, and the loop's symmetry keeps every contributing vector aligned along the same vertical axis, turning the vector sum into a plain scalar addition.

Common pitfall. Using the FULL infinite-wire formula for the straight sections at a jumper centre instead of the semi-infinite (half) value, or omitting the arc's own field entirely and reporting only the wire contribution there.

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Biot–Savart law; field of straight and semi-infinite conductors; field at the centre of a circular arc).

Question 4 of 8: Magnetic Flux Density at the Midpoint and at the Jumper Centres of a Long Rectangular Current Loop

Question 4: Magnetic Flux Density at the Midpoint and at the Jumper Centres of a Long Rectangular Current Loop (20 marks)