RMS EMF Induced by a Rotating Uniform Field: 04-BS-9, May 2013, Q3

Question 3: RMS EMF Induced by a Rotating Uniform 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.

Quantity	Value
Loop area $A$	100 cm² $=1.0\times10^{-2}$ m²
Turns $N$	10
Field magnitude $B$	0.2 T, at 45° above horizontal
Rotation rate	3600 RPM about the vertical axis

Find. The RMS induced EMF in the fixed loop.

Loop fixed in a vertical plane (horizontal normal $\hat n$); the field vector's horizontal component sweeps around the vertical axis at 3600 RPM.

Approach. Decompose $\vec B$ into a vertical component (always perpendicular to the loop's horizontal normal, so it contributes zero flux at every instant) and a horizontal component that sweeps around the vertical axis; only that horizontal component drives a time-varying flux through the fixed loop, exactly like the field seen by a stationary coil in a rotating-field AC generator.

Horizontal component of $B$. $$B_h = B\cos45^\circ = (0.2)(0.7071) = 0.1414\ \text{T}.$$ The vertical component $B\sin45^\circ$ is parallel to the rotation axis and stays perpendicular to the loop's (horizontal) normal at every instant, so it never contributes flux.
Angular speed. $$\omega = 3600\ \text{rpm}\times\frac{2\pi}{60} = 376.99\ \text{rad/s}.$$
Flux and induced EMF. As $B_h$ sweeps past the loop's normal, $\Phi(t)=NB_hA\cos(\omega t)$, so by Faraday's law $$\text{emf}(t)=-\frac{d\Phi}{dt}=NB_hA\,\omega\sin(\omega t),$$ a sinusoid of peak value $NB_hA\omega$.
Peak and RMS EMF. $$\text{emf}_{\text{peak}} = NB_hA\omega = (10)(0.1414)(0.01)(376.99) = 5.33\ \text{V}$$ $$\text{emf}_{\text{rms}}=\frac{\text{emf}_{\text{peak}}}{\sqrt2}=\boxed{3.77\ \text{V}}$$

Quantity	Result
Horizontal field component $B_h$	0.1414 T
Angular speed $\omega$	377.0 rad/s
Peak EMF	5.33 V
RMS EMF	3.77 V

Concept & Theory

Topic. Faraday's law with a rotating field and a fixed coil (reciprocal of the standard rotating-coil AC generator).

Key relations. $\Phi=N\vec B\cdot\hat n\,A$; Faraday's law $\text{emf}=-d\Phi/dt$; RMS of a sinusoid $=$ peak$/\sqrt2$.

Why this works. Only the component of $\vec B$ along the loop's normal produces flux; because the loop's normal is fixed and horizontal, only the horizontal component of the rotating field ever has a nonzero, time-varying projection onto it — physically this is identical to a generator, just with the roles of "rotating part" and "fixed part" swapped between field and coil.

Common pitfall. Using the full field magnitude $B=0.2$ T instead of only its horizontal component $B_h$ in the EMF formula — the constant vertical component never appears in $\Phi(t)$ at all, so it must be dropped, not merely averaged out.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Maxwell's Equations, Faraday's Law).

Question 3 of 8: RMS EMF Induced by a Rotating Uniform Field

Question 3: RMS EMF Induced by a Rotating Uniform Field (20 marks)