NivaarExam PrepOfficial exam papers ↗

04-BS-9 · December 2017

Question 4 of 8: Force Between Two Antiparallel Current-Carrying Wires

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

Notes on this paper

National Exams — December 2017 — 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 for discrete point charges, Gauss's law for cylindrical charge distributions and coaxial capacitors, the Biot–Savart/Ampère force between parallel currents, Biot–Savart on-axis loop fields, Faraday's law for a moving loop in a spatially varying field, and the point (differential) form of Gauss's law for recovering a charge distribution from a given field; Young & Freedman, University Physics with Modern Physics — Snell's law and refracted-ray time-of-flight geometry.

Question 4: Force Between Two Antiparallel Current-Carrying Wires (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 in each wire$I=0.5$ A
Separation $d$1 cm $=0.01$ m
Current directionsantiparallel (east vs. west)
Segment length1 cm $=0.01$ m (on the east-flowing wire)

Find. The magnitude and direction of the force on the 1 cm segment of the east-flowing wire.

west wire, 0.5A → (west) east wire, 0.5A → (east) d=1cm F on east wire (repelled south)
Antiparallel currents repel: the east-flowing wire (south) is pushed away from the west-flowing wire (north), i.e. further south.

Approach. Use the standard force-per-length formula between two long parallel wires, $F/L=\mu_0I_1I_2/(2\pi d)$; antiparallel currents repel, so the force direction is away from the other wire.

  1. Force per unit length. $$\frac{F}{L}=\frac{\mu_0I^2}{2\pi d}=\frac{(4\pi\times10^{-7})(0.5)^2}{2\pi(0.01)}=\boxed{5.000\times10^{-6}\ \text{N/m}}$$
  2. Force on the 1 cm segment. $$F=\frac{F}{L}\times(0.01\ \text{m})=(5.000\times10^{-6})(0.01)$$ $$\boxed{F=5.000\times10^{-8}\ \text{N, directed south (away from the west-flowing wire)}}$$
QuantityResult
Force per unit length$5.000\times10^{-6}$ N/m
Force on the 1 cm segment$5.000\times10^{-8}$ N
Directionsouth — repulsive, away from the (northerly) west-flowing wire