Question 5 of 8: Magnetic Flux Density at the Centre of a Bent Semicircular Loop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition, Gauss's law, Biot–Savart law, magnetic field of current loops and coaxial conductors, Faraday's law, capacitance and field energy, plane-wave propagation; Young & Freedman, University Physics with Modern Physics — reflection geometry.
Question 5: Magnetic Flux Density at the Centre of a Bent Semicircular Loop (20 marks)
one semicircle horizontal, the other vertical (E–W plane, upper half-space), joined at a common E–W diameter
Sense
clockwise, viewed from above
Find. The magnitude and direction of $\vec B$ at the common centre.
Bent loop: horizontal semicircle (blue) joined to a vertical, upward-bulging E–W semicircle (red) at a common diameter. Each contributes half a full loop's centre field, along its own plane's normal; the two are perpendicular.
Approach. Each semicircular arc contributes exactly half of a full circular loop's centre field, directed along its own plane's normal. The horizontal arc's normal is vertical; the vertical arc's normal is horizontal (north–south) — perpendicular to each other — so the two contributions combine as perpendicular vectors (Pythagoras), not by simple addition.
Each semicircle's contribution (half of a full loop's centre field).
$$B_{\text{half}}=\frac{\mu_0 I}{4a}=\frac{(4\pi\times10^{-7})(2)}{4(0.10)}$$
$$B_{\text{half}}=\boxed{6.283\times10^{-6}\ \text{T}}$$
Direction of the horizontal semicircle's contribution. Viewed from above, the current circulates clockwise; by the right-hand rule this drives $\vec B$ straight down (into the ground) at the centre, magnitude $B_{\text{half}}$.
Direction of the vertical semicircle's contribution. Its plane's normal is the north–south line, so this contribution is horizontal, magnitude $B_{\text{half}}$, pointing either north or south. The "clockwise viewed from above" statement fixes the down-pointing component (step 2) but does not by itself say which side (north or south) the horizontal semicircle bulges toward — and that unstated detail is exactly what fixes whether the vertical arc's current (and hence its field) points north or south (see check callout below).
Combine as perpendicular vectors.
$$B_{\text{net}}=\sqrt{B_{\text{half}}^2+B_{\text{half}}^2}=\sqrt2\,B_{\text{half}}$$
$$B_{\text{net}}=\boxed{8.886\times10^{-6}\ \text{T, at }45^\circ\text{ from straight down, tilted toward the N–S direction}}$$
Quantity
Result
Each semicircle's contribution $B_{\text{half}}$
$6.283\times10^{-6}$ T
$B_{\text{net}}$ at the common centre
$8.886\times10^{-6}$ T, $45^\circ$ off vertical (down), tilted N or S
Check: the source fixes the vertical component of $\vec B$ unambiguously (down, via the stated "clockwise viewed from above"), but does not state whether the horizontal semicircle bulges north or south of the common E–W diameter — a detail that flips the sense (north vs. south) of the horizontal component, and hence the exact tilt of the resultant, without changing its magnitude. Both sign choices are shown consistent with the stated circulation by continuity of current at the joint; the magnitude $8.886\times10^{-6}$ T at a $45^\circ$ tilt from vertical is unaffected either way.
