Field at the centre of two orthogonal semicircles: 22-Elec-A7 Electromagnetics, May 2017, Q6

Question 6: Field at the centre of two orthogonal semicircles (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. A single closed loop made of two half-turns of the same radius, set at right angles to one another and joined along a north–south diameter, with the sense of circulation specified as seen from above.

Given data
Quantity	Symbol	Value
Loop current	$I$	$2$ A
Semicircle diameter	$D$	$5$ cm
Semicircle radius	$R$	$2.5$ cm $= 0.025$ m
Orientation of the shared diameter	—	north–south, horizontal
Planes of the two halves	—	one horizontal, one vertical
Sense of circulation	—	clockwise viewed from above
Field point	$P$	midpoint of the shared diameter (the common centre)

Find. The magnitude and the direction of $\mathbf{B}$ at $P$.

Figure 6.1 — the two half-turns share their centre P, so each contributes a field along its own normal: one vertical, one horizontal along east-west.

Approach. Use the Biot–Savart result for a circular arc about its own centre; each half-turn gives half of a full loop's field along its own normal, and because the two normals are perpendicular the resultant is the square root of the sum of squares.

Field of one semicircle at its centre. Every element of a circular arc is perpendicular to its radius vector at distance $R$, so $\mathbf{B} = \dfrac{\mu_0 I}{4\pi R^{2}}\displaystyle\oint |d\boldsymbol{\ell}|$ along the arc normal, giving $B = \mu_0 I \theta/(4\pi R)$ for an arc of angle $\theta$. For a half-turn ($\theta = \pi$) that is exactly half a full loop: $$B_{\text{semi}} = \frac{\mu_0 I}{4R} = \frac{(4\pi\times 10^{-7})(2)}{4(0.025)} = \boxed{25.1\ \mu\text{T}} .$$ Both halves have the same $I$ and the same $R$, so both contribute this same magnitude.
Direction from the horizontal half-turn. Its normal is vertical. The circulation is clockwise seen from above, so the right-hand rule puts its field downward: $$\mathbf{B}_1 = 25.1\ \mu\text{T}\ (\text{vertically down}) .$$
Direction from the vertical half-turn. That half lies in the vertical plane containing the shared north–south diameter, so its normal is horizontal and points east–west: $$\mathbf{B}_2 = 25.1\ \mu\text{T}\ (\text{horizontal, along east--west}) .$$ Which of east or west it is depends on which side of the diameter each half-turn occupies, and the paper does not say; taking the horizontal half to lie east of the diameter and the vertical half above it, the current runs north→east→south along the horizontal arc and returns south→up→north over the vertical one, which puts $\mathbf{B}_2$ to the west. An explicit Biot–Savart integration of that path reproduces both components exactly.
Combine the two perpendicular contributions. Since $\mathbf{B}_1 \perp \mathbf{B}_2$ and the magnitudes are equal, $$|\mathbf{B}| = \sqrt{B_1^{2}+B_2^{2}} = \sqrt{2}\,\frac{\mu_0 I}{4R} = \boxed{35.5\ \mu\text{T}} ,$$ directed at $\arctan(B_1/B_2) = 45^\circ$ below the horizontal, in the vertical east–west plane through $P$ (that is, downward and to the west for the configuration described above).
Sanity check on the size. A full 5 cm loop carrying 2 A would give $\mu_0 I/2R = 50.3\ \mu$T at its centre; two orthogonal halves give $\sqrt{2}/2 = 0.707$ of that, i.e. $35.5\ \mu$T, so the answer is a sensible fraction of the closed-loop value and roughly the strength of the earth's own field.

Check: the paper fixes the orientation of the two planes and the sense of circulation, but not which half-plane each semicircle occupies (the horizontal half may lie east or west of the diameter, the vertical half above or below it). All four admissible loops give the same magnitude $35.5\ \mu$T and the same $45^\circ$ downward tilt in the vertical east–west plane; two of them point west of vertical and two east of vertical. The solution quotes the east-half/upper-half case (field down and to the west) and notes the mirror alternative, per the paper's note 1 inviting a clear statement of assumptions.

Final results
Quantity asked	Result
Field from each semicircle	$\mu_0 I/4R = 25.1\ \mu$T
Contribution of the horizontal half	$25.1\ \mu$T, vertically downward
Contribution of the vertical half	$25.1\ \mu$T, horizontal, east–west
Resultant magnitude	$\sqrt{2}\,\mu_0 I/4R = 35.5\ \mu$T
Resultant direction	$45^\circ$ below horizontal in the vertical E–W plane (down and to the west for the stated configuration)
Full-loop comparison	$\mu_0 I/2R = 50.3\ \mu$T (the resultant is $0.707$ of this)

Topic. Biot–Savart field of circular arcs and vector superposition of orthogonal contributions.

Key relations. $d\mathbf{B} = \dfrac{\mu_0 I}{4\pi}\dfrac{d\boldsymbol{\ell}\times\hat{r}}{r^{2}}$; arc of angle $\theta$ at its own centre: $B = \mu_0 I\theta/(4\pi R)$; full loop $\mu_0 I/2R$; half-turn $\mu_0 I/4R$; right-hand rule fixes the sense.

Why this works. At the centre of an arc every current element is perpendicular to its own radius vector and every $d\boldsymbol{\ell}\times\hat{r}$ points the same way, so the integral collapses to arc length over $R^{2}$ — no angular integration survives. The two straight parts of this loop (the shared diameter) contribute nothing at $P$ because they are collinear with the field point, so only the two arcs matter. Their normals are perpendicular by construction, so the components add in quadrature and the resultant tilt is exactly $45^\circ$ whenever the two arcs have equal radius and carry the same current.

Common pitfall. Using the full-loop formula $\mu_0 I/2R$ for a half-turn (a factor of two), or adding the two contributions arithmetically to get $50.3\ \mu$T instead of combining them as perpendicular vectors. It is also easy to forget that the straight diameter contributes zero and to look for a third term.

Source. Sadiku, Elements of Electromagnetics 7th ed. ch. 7 (Biot–Savart law, circular loops); Hayt & Buck, Engineering Electromagnetics 9th ed. ch. 7.

Question 6 of 8: Field at the centre of two orthogonal semicircles

Question 6: Field at the centre of two orthogonal semicircles (20 marks)