Question 9 of 10: On-Axis Vector Potential, Magnetic Field and Flux of a Current Loop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A3, Electromagnetics — National Exam, May 2019. 3-hour closed-book exam (Casio or Sharp approved calculators only, one 8.5″×11″ aid sheet); any FIVE of the printed questions constitute a complete exam paper and only the first five as they appear in a candidate's answer book are marked, each of equal value, with full justification required for marks. All ten printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — plane waves and boundaries, Gauss's law, resistance and current density, magnetostatics; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients and bounce diagrams; Pozar, Microwave Engineering (4th ed.) — transmission-line theory, rectangular waveguide cutoff; Balanis, Antenna Theory: Analysis and Design (4th ed.) — short-dipole radiation resistance and efficiency.
Question 9: On-Axis Vector Potential, Magnetic Field and Flux of a Current Loop (20 marks)
Given. Circular loop, radius $b=3$ m, in the $xy$-plane centred at the origin, current $I=1$ A in $\hat a_\phi$.
Find. $\mathbf A(0,0,z)$; the magnetic field at $(0,0,0.5\text{ m})$; the flux through the loop's own area.
Approach. Evaluate the Biot–Savart vector-potential integral $\mathbf A=\frac{\mu_0I}{4\pi}\oint\frac{d\mathbf l'}{R}$ directly on the $z$-axis, where by symmetry $R$ is the same for every source point; this shows $\mathbf A\equiv0$ ON the axis exactly. Get $\mathbf H(z)$ instead from the standard on-axis Biot–Savart field integral (equivalent physics, avoiding the technical point that curl needs field values in a NEIGHBOURHOOD of the axis, not just on it). For the flux, evaluate the same on-axis result at $z=0$ (the loop's own plane) and multiply by its area.
Part (a), first result — A is exactly zero on the axis. Parametrize the loop by $\phi'$: source point $b(\cos\phi',\sin\phi',0)$, $d\mathbf l'=b\,d\phi'(-\sin\phi',\cos\phi',0)$. For a field point $(0,0,z)$ on the axis, $R=|(0,0,z)-b(\cos\phi',\sin\phi',0)|=\sqrt{b^2+z^2}$ is the SAME for every $\phi'$ (axial symmetry), so
$$\mathbf A(0,0,z)=\frac{\mu_0I b}{4\pi\sqrt{b^2+z^2}}\int_0^{2\pi}(-\sin\phi',\cos\phi',0)\,d\phi'=\boxed{\mathbf 0}$$
because both $\int_0^{2\pi}\sin\phi'\,d\phi'=0$ and $\int_0^{2\pi}\cos\phi'\,d\phi'=0$ — the azimuthal vector potential vanishes identically at every point of the axis itself (it only becomes nonzero infinitesimally off-axis, which is why $\mathbf B=\nabla\times\mathbf A$ is still nonzero there).
Part (a), field via Biot–Savart. The same integral for $\mathbf H=\frac{I}{4\pi}\oint\frac{d\mathbf l'\times\hat a_R}{R^2}$ has its transverse ($x,y$) components cancel by the same symmetry, leaving only $H_z$:
$$\boxed{H_z(z)=\frac{Ib^2}{2(b^2+z^2)^{3/2}}}\qquad\Rightarrow\qquad H_z(0.5\text{m})=\frac{(1)(3)^2}{2(9+0.25)^{1.5}}=\boxed{0.1600\ \text{A/m}}$$
$$B_z(0.5\text{m})=\mu_0H_z=\boxed{2.01\times10^{-7}\ \text{T}=0.201\ \mu\text{T}.}$$
Part (b) — Flux through the loop's own area. Evaluate the same on-axis expression at $z=0$ (the plane of the loop itself):
$$H_z(0)=\frac{I}{2b}=\frac{1}{2(3)}=0.1667\ \text{A/m},\qquad B_z(0)=\mu_0H_z(0)=2.094\times10^{-7}\ \text{T}.$$
Treating $B_z(0)$ as constant over the loop's area $A=\pi b^2=\pi(3)^2=28.27$ m²:
$$\Phi=B_z(0)\cdot A=\boxed{5.92\times10^{-6}\ \text{Wb}.}$$
On-axis field $B_z(z)$ for the 3 m loop: peaked at the loop's own plane ($z=0$, red dot at $z=0.5$ m evaluation point), falling off as $z^{-3}$ far from the loop.
