Question 5 of 7: EMF Induced in a Rotating and in a Stationary Rectangular Loop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only, one two-sided aid sheet permitted). Constants given: $\varepsilon_0=8.854\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.602\times10^{-19}$ C. Format: seven questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All seven are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition of point charges, Gauss's law and conductors in electrostatic equilibrium, Ampère's law for cylindrical and sheet current distributions, the force between parallel currents, Faraday's law (rotational and transformer EMF), parallel-plate capacitance and the force between charged plates, and magnetic-circuit (reluctance) analysis of a gapped core.
Question 5: EMF Induced in a Rotating and in a Stationary Rectangular Loop (20 marks)
$(0,0,0),(1,0,0),(1,2,0),(0,2,0)$ m — area $=2$ m$^2$
Field $B$, rotation rate $\omega$ (parts a-c)
$0.1$ T (along $z$), $\omega=1$ rad/s
Time-varying field (part d)
$B(t)=4t+1$ T
Find. Induced EMF for rotation about (a) the $x$-axis, (b) the $y$-axis, (c) the $z$-axis (all at $\omega=1$rad/s), and (d) for the stationary loop in the time-varying field.
Loop ABCD lies in the z=0 plane, area 2 m2. Rotation axes x and y both lie IN the loop's own plane (perpendicular to the initial normal); the z-axis is the loop's own normal direction.
Approach. Track the flux $\Phi(t)=B\cdot A\cdot\cos\theta(t)$, where $\theta$ is the angle between the loop's normal and $\vec B$; EMF $=-d\Phi/dt$. Rotation about any axis lying IN the loop's plane makes $\theta$ sweep at rate $\omega$; rotation about the loop's own normal (here the $z$-axis) never changes $\theta$ at all.
Part (a) — rotation about the $x$-axis. The $x$-axis lies in the loop's plane and is perpendicular to the initial normal $\hat n(0)=\hat z$, so $\hat n(t)$ precesses in the $y$-$z$ plane with $n_z(t)=\cos\omega t$:
$$\Phi(t)=BA\cos\omega t\ \Rightarrow\ \text{EMF}(t)=BA\omega\sin\omega t$$
$$\boxed{\text{EMF}_{\text{peak}}=BA\omega=(0.1)(2)(1)=0.200\ \text{V}}$$
Part (b) — rotation about the $y$-axis. The $y$-axis is also in the loop's plane, perpendicular to $\hat n(0)$; by the identical argument ($n_z(t)=\cos\omega t$ again, this time precessing in the $x$-$z$ plane):
$$\boxed{\text{EMF}_{\text{peak}}=BA\omega=0.200\ \text{V (identical to part a)}}$$
Rotating about EITHER in-plane axis gives the same peak EMF — the formula depends only on $|\omega|$ being perpendicular to $\vec B$, not on which particular in-plane axis is chosen.
Part (c) — rotation about the $z$-axis. The $z$-axis IS the loop's own normal direction, so spinning the loop about its own normal never changes the angle between $\hat n$ and $\vec B$ ($\theta=0$ always):
$$\Phi(t)=BA=\text{constant}\ \Rightarrow\ \boxed{\text{EMF}=0}$$
Part (d) — stationary loop, time-varying field. $\Phi(t)=B(t)A=(4t+1)(2)=8t+2$:
$$\text{EMF}=-\frac{d\Phi}{dt}=-\frac{d}{dt}(4t\cdot A)=-4A$$
$$\boxed{|\text{EMF}|=\left|\frac{dB}{dt}\right|A=(4)(2)=8.00\ \text{V (constant, opposing the increasing flux by Lenz's law)}}$$
