B Field at the Centre of a Bent Semicircular Loop: 04-BS-9, May 2014, Q4

Question 4: B 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 magnitude reported here 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 $\vec B$ at the common centre.

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

Approach. Each semicircle contributes half the field a full circular loop of the same current and radius would give, directed along that semicircle's own plane normal (by the right-hand rule for its own sense of circulation). The two contributions are perpendicular (horizontal-plane normal is vertical; vertical-plane normal is horizontal), so combine them as perpendicular vectors.

Field from one full circular loop, for reference. $$B_{\text{loop}}=\frac{\mu_0 I}{2a}=\frac{(4\pi\times10^{-7})(2)}{2(0.05)}=2.513\times10^{-5}\ \text{T}$$
Each semicircle's contribution (half the full-loop value). $$B_{\text{half}}=\frac{\mu_0 I}{4a}=\frac{2.513\times10^{-5}}{2}$$ $$B_{\text{half}}=\boxed{1.257\times10^{-5}\ \text{T}}$$ Clockwise-from-above fixes the horizontal semicircle's contribution as $B_{\text{half}}$ 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 $B_{\text{half}}$ pointing horizontally, at $90^\circ$ to the first.
Combine the two perpendicular contributions. $$B=\sqrt{B_{\text{half}}^2+B_{\text{half}}^2}=\sqrt2\,B_{\text{half}}$$ $$B=\boxed{1.78\times10^{-5}\ \text{T}}$$, directed at $45^\circ$ between the "straight down" and "horizontal" directions — i.e. bisecting the two semicircles' plane normals.

Quantity	Result
$B_{\text{half}}$ (each semicircle)	$1.257\times10^{-5}$ T
$B$ (resultant)	$1.78\times10^{-5}$ T, at $45^\circ$ between the two plane normals

Concept & Theory

Topic. Biot–Savart field at the centre of a circular arc; superposition of perpendicular field contributions.

Key relations. Full circular loop at its centre, $B=\mu_0 I/2a$; a semicircular arc contributes exactly half that, along the same axis its own plane would use if closed into a full loop.

Why this works. The Biot–Savart contribution from each element of the arc depends only on that element's own position and current direction, so a half-loop simply contributes half of what the corresponding full loop would — and because the two half-loops here lie in perpendicular planes, their field vectors are perpendicular and combine by the Pythagorean rule, not by simple addition of magnitudes.

Common pitfall. Adding the two contributions' magnitudes directly (giving $2B_{\text{half}}$) instead of combining them as perpendicular vectors (giving $\sqrt2\,B_{\text{half}}$).

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

Question 4 of 8: B Field at the Centre of a Bent Semicircular Loop

Question 4: B Field at the Centre of a Bent Semicircular Loop (20 marks)