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04-BS-9 · December 2017

Question 5 of 8: EMF Induced in a Moving Loop Crossing a Spatially Varying Field

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 5: EMF Induced in a Moving Loop Crossing a Spatially Varying Field (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
Turns $N$10
Loop area $A$4 m$^2$
Velocity $v$ (northward, $=dx/dt$)50 m/s
Field profile$B(x)=B_0e^{-x/a}$, $B_0=10^{-6}$ T, $a=50$ m

Find. The induced voltage (EMF) at the instant the loop is at $x=0$.

x (north) B(x)=B₀e^(−x/a), decaying north loop, x=0, v=50m/s →
The loop moves through a field that decays exponentially with $x$; as it moves north the enclosed flux falls, inducing an EMF even though $B$ is not explicitly time-varying.

Approach. The flux depends on time only through the loop's position, $\Phi(t)=NAB(x(t))$; apply the chain rule $d\Phi/dt=(d\Phi/dx)(dx/dt)$ with $dx/dt=v$.

  1. Rate of change of $B$ with position at $x=0$. $$\frac{dB}{dx}=-\frac{B_0}{a}e^{-x/a}\ \Rightarrow\ \left.\frac{dB}{dx}\right|_{x=0}=-\frac{B_0}{a}=-\frac{10^{-6}}{50}=\boxed{-2.000\times10^{-8}\ \text{T/m}}$$
  2. Induced EMF via the chain rule. $$\varepsilon=-N A\frac{dB}{dx}\Big|_{x=0}\cdot v=-(10)(4)(-2.000\times10^{-8})(50)$$ $$\boxed{\varepsilon=4.000\times10^{-5}\ \text{V}=40.00\ \mu\text{V}}$$
QuantityResult
$dB/dx$ at $x=0$$-2.000\times10^{-8}$ T/m
Induced EMF at $x=0$$4.000\times10^{-5}$ V ($40.00\ \mu$V)