Time-Average Torque on a Synchronously Rotating AC Loop: 04-BS-9, December 2014, Q8

Question 8: Time-Average Torque on a Synchronously Rotating AC Loop (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.

Quantity	Value
Side length $s$	$10$ cm $=0.1$ m
Turns $N$	$20$
Current $I_{\text{rms}}$, frequency	$2$ A RMS, $60$ Hz
Field $B$ (DC, horizontal)	$0.2$ T
Mechanical rotation speed (about the loop's vertical axis)	$3600$ RPM $=60$ rev/s
Phase condition	$I=0$ when the loop plane is $\perp$ to $B$

Find. The time-averaged torque exerted by the field on the loop.

Square loop spinning about its vertical axis at exactly the electrical frequency (60 Hz), with current phased to zero when the plane is perpendicular to B — the synchronous-motor condition.

Approach. Write the loop's magnetic moment as $\vec m(t)=NI(t)A\,\hat n(t)$, with $\hat n(t)$ rotating mechanically at $\Omega=2\pi(60)\ \text{rad/s}$ and $I(t)$ oscillating electrically at the same $\omega=2\pi(60)\ \text{rad/s}$ — the mechanical and electrical frequencies are numerically identical (both 60 Hz), which is the key feature of this problem. Compute $\vec T=\vec m\times\vec B$ and average over one cycle.

Set up the rotating normal and phased current. Let $\hat n(t)=(\cos\Omega t,\ \sin\Omega t,\ 0)$ with $\vec B=B\hat x$, so the loop plane is $\perp B$ (i.e. $\hat n\parallel\hat x$) at $t=0$. The stated phase condition ($I=0$ exactly then) gives $$I(t)=I_0\sin(\omega t),\qquad I_0=I_{\text{rms}}\sqrt2$$ with $\omega=\Omega=2\pi(60)=377.0\ \text{rad/s}$ (electrical and mechanical frequencies match — a synchronous condition).
Instantaneous torque. $\hat n(t)\times\hat x=(0,0,-\sin\Omega t)$, so $$\vec T(t)=NI(t)A\,\hat n(t)\times\vec B=-NI_0\sin(\omega t)\,A\,B\,\sin(\Omega t)\,\hat z=-NI_0AB\sin^2(\omega t)\,\hat z$$ (using $\omega=\Omega$). The torque is entirely along the (fixed) rotation axis and never changes sign, because $\sin^2(\omega t)\geq0$ at all times.
Time average. Since $\langle\sin^2(\omega t)\rangle=\tfrac12$ over a full cycle: $$\langle T\rangle=\frac12 NI_0AB=\frac{N I_{\text{rms}}\sqrt2\,AB}{2}=\frac{NI_{\text{rms}}AB}{\sqrt2}$$ With $A=s^2=(0.1)^2=0.01\ \text{m}^2$: $$\langle T\rangle=\frac{(20)(2)(0.01)(0.2)}{\sqrt2}$$ $$\langle T\rangle=\boxed{0.0566\ \text{N}\cdot\text{m}}$$, directed steadily about the rotation axis (a constant, non-oscillating driving/braking torque).

Quantity	Result
$I_0$ (peak)	$2.828$ A
$\langle T\rangle$	$0.0566$ N·m

Concept & Theory

Topic. Time-average torque on a current loop whose mechanical rotation frequency matches its electrical current frequency — the operating principle of a single-phase synchronous motor.

Key relations. $\vec T=\vec m\times\vec B=NIA\,\hat n\times\vec B$; time average of $\sin^2$ over a full cycle is $\tfrac12$.

Why this works. If the mechanical and electrical frequencies did not match, the torque would contain a term proportional to $\sin(\omega t)\sin(\Omega t)$ with $\omega\neq\Omega$, which averages to exactly zero over any long interval — no net torque, no useful motor action. Here the problem deliberately sets $\Omega=\omega=2\pi(60)$ rad/s and phases the current so the product becomes $\sin^2(\omega t)$, which has a strictly positive time average of $\tfrac12$: this is precisely why real synchronous motors must run at (or very near) their electrical supply frequency to produce steady torque.

Common pitfall. Assuming any rotating AC loop in a DC field averages to zero torque — that is true only when the mechanical and electrical frequencies differ; matched frequencies with the right phase give a genuine non-zero average, which is the whole point of this question.

Source. Young & Freedman, University Physics, Ch. 27 (torque on a current loop, DC motors) and Ch. 31 (AC circuits, synchronous machines); Sadiku, Elements of Electromagnetics, Ch. 8.

Question 8 of 8: Time-Average Torque on a Synchronously Rotating AC Loop

Question 8: Time-Average Torque on a Synchronously Rotating AC Loop (20 marks)