Question 3 of 8: Magnetic Flux Density Midway Between Two Parallel Current Loops
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 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 and divergence, Biot–Coulomb law and magnetic field of current loops and coaxial conductors, Faraday's law, Lorentz force, capacitance and field energy, DC circuit power distribution; Young & Freedman, University Physics with Modern Physics — refraction geometry.
Question 3: Magnetic Flux Density Midway Between Two Parallel Current Loops (20 marks)
midpoint of the common axis, $x=1.25$ cm from each loop
Find. The magnitude and direction of $\vec B$ at the centre of the system (the axial midpoint).
Two coaxial loops (5 cm radius, 2.5 cm apart) carrying same-sense current; both contribute equally to $B$ at the axial midpoint.
Approach. Use the on-axis field of a single circular loop for each loop at its own distance from the midpoint ($x=1.25$ cm), then add the two contributions — they reinforce because the currents circulate the same way.
On-axis field of one loop at $x=1.25$ cm from its plane.
$$B_{\text{one}}=\frac{\mu_0 I a^2}{2(a^2+x^2)^{3/2}}=\frac{(4\pi\times10^{-7})(0.1)(0.05)^2}{2\left[(0.05)^2+(0.0125)^2\right]^{3/2}}$$
$$B_{\text{one}}=\boxed{1.147\times10^{-6}\ \text{T}}$$
Add the two loops' contributions. Both loops sit the same distance ($1.25$ cm) from the midpoint and circulate the same way, so their axial fields point the same way and add directly:
$$B_{\text{net}}=2\,B_{\text{one}}=\boxed{2.295\times10^{-6}\ \text{T, along the common axis (sense set by the right-hand rule for the stated current direction)}}$$