H Field at the Centre of a Bent Semicircular Loop: 04-BS-9, December 2014, Q1

Question 1: H Field at the Centre of a Bent Semicircular 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.

Check: the source does not state whether the vertical semicircle bulges "up" or "down" from the shared diameter; we take the physically continuous loop — current entering the vertical arc where the horizontal arc leaves it — and resolve the sign by right-hand-rule continuity with the stated clockwise-from-above sense. The reported magnitude is unaffected by that choice; only the exact bisecting direction would flip.

Given.

Quantity	Value
Radius $a$ (both semicircles)	$5$ cm $=0.05$ m
Current $I$	$2$ A
Orientation	one semicircle horizontal, the other vertical, sharing a common diameter and centre; clockwise viewed from above

Find. The magnitude and direction of $\vec{H}$ at the common centre.

Two perpendicular semicircles joined along a common diameter through centre O. Each contributes an H-field along its own plane's normal; the two add as perpendicular vectors.

Approach. Each semicircle contributes half of what a full circular loop of the same current and radius would give at its centre, $H_{\text{loop}}=I/2a$, directed along that semicircle's own plane normal (right-hand rule). The two contributions are perpendicular (horizontal-plane normal is vertical; vertical-plane normal is horizontal), so combine them as perpendicular vectors.

Each semicircle's H contribution (half the full-loop value). $$H_{\text{half}}=\frac{I}{4a}=\frac{2}{4(0.05)}$$ $$H_{\text{half}}=\boxed{10.0\ \text{A/m}}$$ Clockwise-from-above fixes the horizontal semicircle's contribution as $H_{\text{half}}$ pointing straight down; loop continuity (the current must leave the horizontal arc and enter the vertical arc at the same physical point, without a discontinuous jump in sense) then fixes the vertical semicircle's contribution as $H_{\text{half}}$ pointing horizontally, at $90^\circ$ to the first.
Combine the two perpendicular contributions. $$H=\sqrt{H_{\text{half}}^2+H_{\text{half}}^2}=\sqrt2\,H_{\text{half}}$$ $$H=\boxed{14.14\ \text{A/m}}$$, directed at $45^\circ$ between the "straight down" and "horizontal" directions — i.e. bisecting the two semicircles' plane normals.

Quantity	Result
$H_{\text{half}}$ (each semicircle)	$10.0$ A/m
$H$ (resultant)	$14.14$ A/m, at $45^\circ$ between the two plane normals

Concept & Theory

Topic. Biot–Savart field at the centre of a circular arc, expressed as the magnetic field intensity $H$ rather than the flux density $B$.

Key relations. Full circular loop at its centre, $H=I/2a$ (equivalently $B=\mu_0 I/2a$); a semicircular arc contributes exactly half that.

Why this works. $H$ is a purely geometric/current quantity ($H=B/\mu_0$) independent of the medium, so the same half-loop superposition used for $B$ applies unchanged; the two half-loops lie in perpendicular planes, so their $H$ vectors are perpendicular and combine by the Pythagorean rule.

Common pitfall. Reporting the answer in tesla (as $B$) instead of A/m (as $H$) — the question specifically asks for the field intensity vector.

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Biot–Savart Law for circular current loops and arcs).

Question 1 of 8: H Field at the Centre of a Bent Semicircular Loop

Question 1: H Field at the Centre of a Bent Semicircular Loop (20 marks)