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04-BS-9 · December 2018

Question 3 of 8: Magnetic Flux Density from a Finite-Thickness Sub-Surface Current Layer

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2018 — 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 — Gauss's law and boundary conditions for graded (spatially-varying) dielectrics, magnetic-circuit (reluctance) analysis of a partially-filled long solenoid, superposition of infinite current sheets and Ampère's law, Coulomb's-law equilibrium of collinear point charges, Faraday's law and motional EMF, the Biot–Savart law for arc segments, and the Lorentz force in a velocity selector; Young & Freedman, University Physics with Modern Physics — Snell's law and plane-wave reflection/refraction at a dielectric interface.

Question 3: Magnetic Flux Density from a Finite-Thickness Sub-Surface Current Layer (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.

QuantityValue
Current layer thickness $t$ (surface to zero-density depth)$1\times10^{-5}$ m
Total linear (sheet) current $K$$0.05$ A/m, direction north
Current-density profileuniform horizontally, decreasing with depth from max at $z=0$ to $0$ at $z=t$

Find. $B$ at the surface ($z=0$) and at depth $t$ (the bottom of the current layer) — magnitude and direction, for both.

surface, z=0 z=t=1e-5 m ⊗⊗⊗⊗ current layer (into page = north) B(0), east B(t), west
Cross-section (current flows north, into the page). The 1×10⁻⁵ m current layer sits just below the metal surface. B reverses direction across the layer: east just above the surface, west just below the layer's lower edge, equal magnitude.

Approach. Model the layer as a stack of infinitesimally thin infinite current sheets; by the translational symmetry of an infinite sheet, Ampère's law shows the field on either outer face of the whole layer depends only on the TOTAL enclosed sheet current, and reverses direction from one face to the other.

  1. Field just above the surface (all current lies below this point). Every thin sheet in the layer contributes with the same sign at $z=0$, so the sheets add exactly like a single infinite sheet of the total current $K$: $$B(0)=\frac{\mu_0K}{2}=\frac{(4\pi\times10^{-7})(0.05)}{2}$$ $$\boxed{B(0)=3.142\times10^{-8}\ \text{T, pointing EAST}}$$ (right-hand rule: current north, field on the shallow side of the sheet points east).
  2. Field just below the layer, at depth $t$ (all current now lies above this point). Every sheet now contributes with the OPPOSITE sign, giving the same magnitude but reversed direction: $$B(t)=\frac{\mu_0K}{2}$$ $$\boxed{B(t)=3.142\times10^{-8}\ \text{T, pointing WEST}}$$
QuantityResult
$B$ at the surface, $z=0$$3.142\times10^{-8}$ T, east
$B$ at depth $t=1\times10^{-5}$ m$3.142\times10^{-8}$ T, west

Stated assumption. The paper names no return current, so this answer treats the layer as the only current present (magnetostatics, with the metal's $\mu_r\approx1$). The two values are then equal in magnitude and opposite in direction. If the layer were instead a skin-effect current with its field confined above the layer, Ampère's law would give $\mu_0K=6.28\times10^{-8}$ T at the surface and zero below. A candidate who adopts that model should say so explicitly.