RMS EMF of a Loop Rotating About a Tilted Horizontal Diameter: 04-BS-9, December 2014, Q5

Question 5: RMS EMF of a Loop Rotating About a Tilted Horizontal Diameter (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
Field $B$ (DC, horizontal, West)	$0.2$ T
Turns $N$	$100$
Radius $r$	$5$ cm $=0.05$ m
Rotation axis	a horizontal diameter, at $30^\circ$ to $B$
Rotation speed	$3600$ RPM

Find. The RMS EMF induced in the loop as it rotates.

Loop rotates about a horizontal diameter set at 30° to the horizontal field B; only the component of B perpendicular to the rotation axis drives a changing flux.

Approach. Resolve the field into a component along the rotation axis (contributes no changing flux, since the loop's normal always stays perpendicular to that axis) and a component perpendicular to it, of magnitude $B\sin30^\circ$ — this is the effective field that sweeps through the loop's normal as it spins, giving a standard sinusoidal generator EMF.

Angular speed. $$\omega=3600\ \text{rev/min}\times\frac{2\pi}{60}$$ $$\omega=\boxed{377.0\ \text{rad/s}}$$
Effective (perpendicular) field component and peak EMF. With loop area $A=\pi r^2=\pi(0.05)^2=7.854\times10^{-3}\ \text{m}^2$: $$\varepsilon_{\text{peak}}=N\,(B\sin30^\circ)\,A\,\omega=(100)(0.2)(7.854\times10^{-3})(377.0)(0.5)$$ $$\varepsilon_{\text{peak}}=\boxed{29.61\ \text{V}}$$
RMS value. For a sinusoidal EMF, $\varepsilon_{\text{rms}}=\varepsilon_{\text{peak}}/\sqrt2$: $$\varepsilon_{\text{rms}}=\frac{29.61}{\sqrt2}$$ $$\varepsilon_{\text{rms}}=\boxed{20.94\ \text{V}}$$

Quantity	Result
$\omega$	$377.0$ rad/s
$\varepsilon_{\text{peak}}$	$29.61$ V
$\varepsilon_{\text{rms}}$	$20.94$ V

Concept & Theory

Topic. Generator EMF for a loop rotating about an axis tilted relative to a uniform field.

Key relations. $\Phi(t)=NBA\sin\theta_{\text{axis}}\cos\omega t$; $\varepsilon_{\text{peak}}=N\,B\sin\theta_{\text{axis}}\,A\,\omega$; $\varepsilon_{\text{rms}}=\varepsilon_{\text{peak}}/\sqrt2$.

Why this works. Decomposing $B$ into components parallel and perpendicular to the rotation axis isolates the only part that can drive a changing flux: the loop's normal always stays perpendicular to the rotation axis, so the parallel component never threads the loop and contributes no flux at all; only the perpendicular component $B\sin30^\circ$ behaves like the field in the textbook "loop rotating about an axis perpendicular to $B$" case.

Common pitfall. Using the full field $B$ (ignoring the $30^\circ$ tilt) or using $\cos30^\circ$ instead of $\sin30^\circ$ — the effective driving field is the component perpendicular to the rotation axis, not the angle to some other reference line.

Source. Young & Freedman, University Physics, Ch. 29 (Faraday's Law, AC generators); Sadiku, Elements of Electromagnetics, Ch. 9.

Question 5 of 8: RMS EMF of a Loop Rotating About a Tilted Horizontal Diameter

Question 5: RMS EMF of a Loop Rotating About a Tilted Horizontal Diameter (20 marks)