Question 6 of 8: Horizontal magnetic field above a semicircular loop fed by two infinite straight wires

Question 6: Horizontal magnetic field above a semicircular loop fed by two infinite straight wires

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. A horizontal circuit made of one semicircular arc plus two semi-infinite parallel feeders attached to its ends.

Find. The magnitude and direction of the horizontal part of $\mathbf{B}$ at the point $25\ \text{cm}$ vertically above the midpoint of the diameter.

Approach. Integrate Biot–Savart separately over the arc and over each semi-infinite feeder. The field point lies on the symmetry plane of the two feeders, so their horizontal contributions cancel and only the arc leaves a horizontal field.

1. Set the geometry and the sense. Use east–north–up coordinates with the diameter along the east–west line, the arc bulging north, and the feeders running south to infinity from the arc ends at $(\pm R, 0, 0)$. Viewed from above, clockwise means the current runs north up the western feeder, over the crown of the arc from west to east, then south down the eastern feeder. The field point is $P = (0, 0, h)$.
2. Integrate over the arc. Every element of the arc is the same distance $r_0 = \sqrt{R^{2} + h^{2}}$ from $P$. Parameterising by the polar angle and carrying out $d\mathbf{B} = (\mu_0 I/4\pi)\,d\boldsymbol{\ell} \times \hat{r}/r^{2}$ over the half turn, the east–west components cancel by symmetry while the other two survive:$$B_{h,arc} = \frac{\mu_0 I}{4\pi}\,\frac{2Rh}{(R^{2}+h^{2})^{3/2}}, \qquad B_{z,arc} = \frac{\mu_0 I}{4\pi}\,\frac{\pi R^{2}}{(R^{2}+h^{2})^{3/2}}$$Note that a semicircle, unlike a full loop, leaves a transverse component: the two halves that would cancel it are simply not there.
3. Evaluate the arc terms. With $h = R$ the horizontal term collapses to a compact closed form,$$B_{h,arc} = \frac{\mu_0 I}{4\sqrt{2}\,\pi R} = \frac{4\pi \times 10^{-7} \times 2}{4\sqrt{2}\,\pi \times 0.25} = 5.657 \times 10^{-7}\ \text{T} = 0.566\ \mu\text{T}$$directed horizontally along the perpendicular bisector of the diameter and pointing away from the arc (due south, with the arc to the north). The vertical term is $B_{z,arc} = 0.889\ \mu\text{T}$ downward, the downward sense following from the right-hand rule applied to a clockwise circulation.
4. Treat the two feeders. Each is a semi-infinite straight whose perpendicular foot from $P$ falls exactly at its own end, so its field magnitude is half the infinite-wire value, $\mu_0 I/(4\pi r_0) = 0.566\ \mu\text{T}$, resolving into$$\mathbf{B}_{feeder} = \frac{\mu_0 I}{4\pi}\,\frac{(\pm h,\ 0,\ -R)}{R^{2}+h^{2}} \;\Rightarrow\; 0.400\ \mu\text{T horizontal},\ 0.400\ \mu\text{T down each}$$The signs of the horizontal parts are opposite because $P$ lies on the plane midway between the two feeders and they carry opposite currents, so those two contributions cancel exactly. Their vertical parts add.
5. Collect the horizontal answer. Only the arc survives horizontally, so$$\boxed{B_{horizontal} = \frac{\mu_0 I}{4\sqrt{2}\,\pi R} = 0.566\ \mu\text{T}}$$directed horizontally along the perpendicular bisector of the diameter, pointing away from the arc — i.e. due south if the arc bulges north.
6. Cross-check with the full field. The vertical components sum to $0.889 + 2 \times 0.400 = 1.689\ \mu\text{T}$ downward, giving a resultant $|\mathbf{B}| = \sqrt{0.566^{2} + 1.689^{2}} = 1.781\ \mu\text{T}$ dipping $71.5^\circ$ below the horizontal. A direct numerical Biot–Savart integration along the whole conductor path reproduces both the magnitude and the sense of every component, which is the cheapest guard against a handedness slip.