22-Elec-A7 Electromagnetics · May 2013

Question 3 of 8: Rotating Loop — Load Power and RMS Torque

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

Notes on this paper

Paper format. National Examinations, May 2013 — 07-Elec-A7, Electromagnetics. Three hours, closed book; one of two approved calculators (Casio or Sharp) permitted. Eight questions, all of equal value; any five constitute a complete paper and only the first five presented are marked. All eight are solved here, since the set is a study resource. Aids printed on the cover page: ε₀ = 8.85 × 10⁻¹² F/m and μ₀ = 4π × 10⁻⁷ H/m; Question 2 additionally supplies the quarter-wave relation.

Reference texts (22-Elec-A7 Electromagnetics).

D. M. Pozar, Microwave Engineering, 4th ed. — transmission lines, matching, waveguides.
M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. — plane waves, guided waves, radiation.
W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. — field theory and transients on lines.
C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. — short elements, image theory.
F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. — oblique incidence, surface currents.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — rotating-loop induction and torque (Question 3).

Conventions used throughout. The paper itself quotes propagation velocities as $3\times10^{8}\ \text{m/s}$, so free space is taken as $c=3.00\times10^{8}\ \text{m/s}$ and the intrinsic impedance as $\eta_0=\mu_0 c=120\pi=376.99\ \Omega$. Lines are treated as lossless, and "rms" is stated explicitly wherever the exam asks for it.

Question 3: Rotating Loop — Load Power and RMS Torque (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. A ten-turn square loop of 100 cm² area spins about its own vertical axis at 3600 rpm inside a horizontal field of 0.2 T and feeds a 10 Ω resistor; loop inductance and winding resistance are to be ignored.

Given data
Quantity	Symbol	Value
Number of turns	$N$	10
Loop area	$A$	100 cm² = $0.0100\ \text{m}^{2}$
Rotational speed	$n$	3600 rpm
Flux density (horizontal, uniform)	$B$	0.200 T
Load resistance	$R$	10.0 Ω
Loop self-inductance	$L$	neglected (purely resistive circuit)

Find. (i) the average power the resistor absorbs, and (ii) the rms value of the torque that acts on the loop.

Figure 3.1 — The loop rotating about a vertical axis in a horizontal field, with the normal $n$ sweeping through the field direction once per revolution.

Approach. Use Faraday's law to get the sinusoidal EMF from the sweeping flux linkage, divide by the load to obtain the current, then form the magnetic moment and take its cross product with the field to get the instantaneous torque, whose rms value follows from the mean square of $\sin^{2}$.

Convert the speed to angular frequency. $$\omega=\frac{2\pi n}{60}=\frac{2\pi\times3600}{60}=376.99\ \text{rad/s}$$ which is $f=\omega/2\pi=60.0$ Hz — a two-pole 60 Hz alternator.
Apply Faraday's law. With the loop normal at angle $\omega t$ to the field, the flux linked is $N\Phi=NBA\cos\omega t$, so $$e(t)=-N\frac{d\Phi}{dt}=NBA\omega\sin\omega t,\qquad E_{pk}=10\times0.200\times0.0100\times376.99=7.540\ \text{V}$$
Obtain the rms EMF and the current. Because the self-inductance is disregarded, the circuit is purely resistive and the current is in phase with the EMF: $$E_{rms}=\frac{E_{pk}}{\sqrt2}=5.331\ \text{V},\qquad i(t)=\frac{e(t)}{R}=\frac{NBA\omega}{R}\sin\omega t$$
Compute the average load power. $$P_{avg}=\frac{E_{rms}^{2}}{R}=\frac{5.331^{2}}{10.0}=\boxed{2.84\ \text{W}}$$
Build the instantaneous torque. The loop's magnetic moment is $m=NiA$, and the torque on a moment in a uniform field is $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$, whose magnitude is $NiAB\sin\omega t$ because the moment lies along the normal. Substituting the current from Step 3, $$\tau(t)=\frac{(NAB)^{2}\omega}{R}\sin^{2}\omega t,\qquad \tau_{pk}=\frac{(10\times0.0100\times0.200)^{2}\times376.99}{10.0}=0.01508\ \text{N}\cdot\text{m}$$ Note the torque never changes sign: it always opposes the rotation, pulsating at twice the electrical frequency.
Take the rms value of that waveform. For $\tau=\tau_{pk}\sin^{2}\omega t$ the mean square is $\tau_{pk}^{2}\langle\sin^{4}\rangle=\tfrac38\tau_{pk}^{2}$, so $$\tau_{rms}=\tau_{pk}\sqrt{\tfrac{3}{8}}=0.01508\times0.6124=\boxed{9.23\times10^{-3}\ \text{N}\cdot\text{m}}$$
Check against the energy balance. The mean torque is $\tau_{pk}/2=7.540\ \text{mN}\cdot\text{m}$, so the mechanical power the driver must supply is $$P_{mech}=\bar{\tau}\,\omega=7.540\times10^{-3}\times376.99=2.84\ \text{W}$$ identical to the electrical power found in Step 4, as it must be for a lossless loop.

The distinction between the 9.23 mN·m rms value and the 7.54 mN·m average value matters here: only the average torque does net work, whereas the rms value is what a torque-transducer reading would report and what a fatigue calculation on the shaft would use.

Final results — Question 3
Quantity	Value
Angular velocity / electrical frequency	377 rad/s / 60.0 Hz
Peak EMF	7.54 V
RMS EMF	5.33 V
(i) Average power delivered to the load	2.84 W
Peak torque	15.1 mN·m
Average torque	7.54 mN·m
(ii) RMS torque on the loop	9.23 mN·m

Topic. Faraday induction in a rotating coil, and the reaction torque of an elementary alternator.

Key relations. $e=-N\,d\Phi/dt$ with $\Phi=BA\cos\omega t$; $\omega=2\pi n/60$; $P_{avg}=E_{rms}^{2}/R$; magnetic moment $m=NiA$; torque $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$; for $\sin^{2}$ waveforms $\langle\sin^{4}\rangle=3/8$.

Why this works. Rotating the loop makes the linked flux vary sinusoidally, and the derivative in Faraday's law turns that into a sinusoidal EMF whose amplitude is proportional to speed — the reason generator voltage tracks rpm. Neglecting self-inductance keeps the current in phase with the EMF, so both are proportional to $\sin\omega t$; the torque, being the product of the current and the same geometric $\sin\omega t$ factor, is therefore proportional to $\sin^{2}\omega t$. That squared form is why the torque is unidirectional and pulsating rather than alternating, and why its rms and average values differ by the factor $2\sqrt{3/8}=1.22$. The energy check in Step 7 is the surest way to catch an algebra slip, because the average torque times the angular velocity must reproduce the electrical power exactly.

Common pitfall. Quoting the average torque when the question asks for the rms value (or vice versa), and treating $\tau_{rms}$ as $\tau_{pk}/\sqrt2$ — that factor belongs to a sinusoid, not to a $\sin^{2}$ waveform. A second error is to include a reactance from the loop inductance despite the explicit instruction to disregard it, which would introduce a power-factor angle that does not belong here.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 3 (the simple rotating loop); Hayt & Buck, Engineering Electromagnetics, 9th ed., Ch. 9 (motional EMF and magnetic torque).

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