Magnetic Flux Density at the Center of a Three-Quarter-Circle Current Loop: 04-BS-9, December 2018, Q7

Question 7: Magnetic Flux Density at the Center of a Three-Quarter-Circle 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
Radius of each quarter circle $R$	$10\ \text{cm}=0.1$ m
Arcs	one each in the $+xy$, $+yz$, $+zx$ quadrant planes, all centered at the origin

Find. The magnitude of $\vec B$ at the origin (the common center of the three arcs).

Oblique (cabinet) projection of the three mutually perpendicular quarter-circle arcs, radius R, meeting at the origin. Each arc's field contribution is normal to its own plane; the resultant B lies along the body diagonal x=y=z.

Approach. Each quarter-circle arc contributes a field along the axis normal to its own plane, with magnitude equal to one quarter of a full loop's center field ($\mu_0I/2R$); with a consistent current sense around the loop, the three contributions are mutually perpendicular and add as a vector sum.

Field of one quarter-circle arc. A full loop gives $B_{\text{full}}=\mu_0I/(2R)$ at its center; an arc subtending $\theta=\pi/2$ contributes proportionally: $$B_{\text{arc}}=\frac{\mu_0I}{2R}\times\frac{1}{4}=\frac{\mu_0I}{8R}$$ Substituting $\mu_0=4\pi\times10^{-7}$ H/m, $I=2$ A, $R=0.1$ m: $$B_{\text{arc}}=\frac{(4\pi\times10^{-7})(2)}{8(0.1)}=3.142\times10^{-6}\ \text{T}$$
Direction of each arc's contribution. With current circulating $+x\to+y\to+z\to+x$ around the closed loop, the right-hand rule gives: the $xy$-arc contributes along $+z$, the $yz$-arc along $+x$, the $zx$-arc along $+y$ — three mutually perpendicular, equal-magnitude vectors.
Vector sum. Three equal, mutually perpendicular components combine by the Pythagorean rule: $$\vec B=B_{\text{arc}}(\hat x+\hat y+\hat z)\ \Rightarrow\ |\vec B|=\sqrt3\,B_{\text{arc}}$$ $$\boxed{|\vec B|=\sqrt3(3.142\times10^{-6})=5.441\times10^{-6}\ \text{T}}$$ directed along the body diagonal $x=y=z$ of the positive octant.

Quantity	Result
Field from each arc, $B_{\text{arc}}$	$3.142\times10^{-6}$ T
Resultant $\|\vec B\|$	$5.441\times10^{-6}$ T, along $x=y=z$

Concept & Theory

Topic. Biot–Savart field of a circular arc; vector superposition of mutually perpendicular field contributions.

Key relations. Full-loop center field $B=\mu_0I/2R$; arc of angle $\theta$: $B=\mu_0I\theta/(4\pi R)$; Pythagorean vector sum of perpendicular components.

Why this works. The Biot–Savart law is linear in current element, so the loop's total field is exactly the sum of the three arcs' individual fields. Because the arcs lie in mutually perpendicular coordinate planes, each arc's field (normal to its own plane) is automatically mutually perpendicular to the other two, reducing the sum to a simple three-term Pythagorean combination.

Common pitfall. Adding the three magnitudes as scalars ($3B_{\text{arc}}$) instead of as perpendicular vectors ($\sqrt3\,B_{\text{arc}}$), or getting one arc's direction backwards from an inconsistent current sense around the loop.

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Biot–Savart law, arc segments).

Question 7 of 8: Magnetic Flux Density at the Center of a Three-Quarter-Circle Current Loop

Question 7: Magnetic Flux Density at the Center of a Three-Quarter-Circle Current Loop (20 marks)