EMF Induced in a Rectangular Loop — Rotational and Transformer Cases: 04-BS-9, Undated paper, Q5

Question 5: EMF Induced in a Rectangular Loop — Rotational and Transformer Cases (15 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
Loop vertices	$A(0,0,0),B(1,0,0),C(1,2,0),D(0,2,0)$ m
Loop area	$A_{\text{loop}}=1\times2=2\ \text{m}^2$
Field direction (a, b)	uniform, $+z$, $B=0.1$ T, $\omega=1$ rad/s
Field (c), (d)	$B(t)=4t^2+2t+3$ T; $B(x,t)=4t^2x^2$ T

Find. Induced EMF for: (a) rotation about $y$; (b) rotation about $z$; (c) stationary, uniform time-varying $B$; (d) stationary, position-and-time-varying $B$.

The loop lies in the xy-plane (normal initially +z, parallel to B). Rotating it about the in-plane y-axis sweeps the normal away from B (part a); rotating it about z spins it about its own normal, which never changes the normal's direction at all (part b).

Approach. Faraday's law $\varepsilon=-d\Phi/dt$ with $\Phi=\int\vec B\cdot d\vec A$. For rigid rotation, track how the loop's unit normal $\hat n(t)$ evolves under each rotation axis; for the stationary cases, differentiate $\Phi(t)=\int B\,dA$ directly (part d requires integrating a position-dependent $B$ over the loop's area first).

Part (a) — rotation about the $y$-axis. The loop's normal starts at $\hat n(0)=\hat z$; rotating about $y$ (which lies in the loop's own plane, perpendicular to the initial normal) sweeps $\hat n(t)=(\sin\omega t,\,0,\,\cos\omega t)$. With $\vec B=B\hat z$: $$\Phi(t)=BA_{\text{loop}}\cos\omega t \implies \varepsilon(t)=BA_{\text{loop}}\omega\sin\omega t$$ $$\boxed{\varepsilon(t)=0.2\sin(t)\ \text{V},\ \text{peak }0.2\ \text{V}}$$ the ordinary AC-generator result: rotation axis perpendicular to $\vec B$ gives sinusoidal EMF.
Part (b) — rotation about the $z$-axis. Here the rotation axis is $\hat z$ — exactly the loop's own initial normal and the field direction. Rotating a vector about an axis it is already parallel to leaves it completely unchanged, so $\hat n(t)=\hat z$ for all $t$: the loop is simply spinning about its own normal, like a coin spinning flat, never tilting toward or away from $\vec B$. $$\Phi(t)=BA_{\text{loop}}\cdot1=\text{constant}\implies$$ $$\boxed{\varepsilon(t)=0\ \text{(identically, for all }t)}$$
Part (c) — stationary, $B(t)=4t^2+2t+3$. With the normal fixed at $\hat z$ (parallel to $\vec B$), $\Phi(t)=A_{\text{loop}}B(t)$: $$\varepsilon(t)=-A_{\text{loop}}\frac{dB}{dt}=-2(8t+2)$$ $$\boxed{\varepsilon(t)=-(16t+4)\ \text{V}}$$ at $t=0$ this is $-4$ V, growing more negative as $t$ increases (the flux is increasing, so by Lenz's law the induced EMF opposes that increase).
Part (d) — stationary, $B(x,t)=4t^2x^2$. Now $B$ varies across the loop's own $1$ m width ($x\in[0,1]$, $y\in[0,2]$), so it must be integrated over the area before differentiating: $$\Phi(t)=\int_0^2\!\!\int_0^1 4t^2x^2\,dx\,dy=4t^2\left[\frac{x^3}{3}\right]_0^1(2)=\frac{8}{3}t^2$$ $$\varepsilon(t)=-\frac{d\Phi}{dt}=-\frac{16}{3}t$$ $$\boxed{\varepsilon(t)=-5.333\,t\ \text{V}}$$

Quantity	Result
(a) rotate about $y$	$\varepsilon(t)=0.2\sin(t)$ V
(b) rotate about $z$	$\varepsilon=0$
(c) $B(t)=4t^2+2t+3$	$\varepsilon(t)=-(16t+4)$ V
(d) $B(x,t)=4t^2x^2$	$\varepsilon(t)=-5.333t$ V

Concept & Theory

Topic. Faraday's law: motional (rotational) and transformer EMF.

Key relations. $\varepsilon=-d\Phi/dt$, $\Phi=\int\vec B\cdot d\vec A$; for rigid rotation, $\Phi(t)=BA\cos\theta(t)$ where $\theta$ is the angle between $\hat n$ and $\vec B$; for a position-dependent field, integrate $B$ over the loop's area before differentiating in time.

Why this works. Only the component of $\vec B$ along the loop's instantaneous normal contributes to flux. Rotating about an axis that lies in the loop's plane continuously changes that angle (generator action); rotating about the axis normal to the loop itself changes nothing about the normal's orientation, so no EMF can be produced no matter how fast the loop spins about that axis.

Common pitfall. Assuming any rotation produces an oscillating EMF — the rotation axis's orientation relative to both the loop's normal and $\vec B$ is what matters, not the mere fact of motion; and forgetting to integrate a spatially-varying field over the loop's area before applying $d/dt$.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Faraday's law, EMF).

Question 5 of 7: EMF Induced in a Rectangular Loop — Rotational and Transformer Cases

Question 5: EMF Induced in a Rectangular Loop — Rotational and Transformer Cases (15 marks)