RMS EMF of a Coil Rotating About a Vertical Diameter in a Tilted Field: 04-BS-9, May 2017, Q4

Question 4: RMS EMF of a Coil Rotating About a Vertical Diameter in a Tilted 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
Turns $N$	$10$
Radius $a$	$5$cm $=0.05$m
Rotation speed	$10000$ RPM about a vertical diameter
Field $B$	$10^{-5}$ T, $45^\circ$ above horizontal

Find. The RMS induced EMF.

The loop's normal always stays horizontal, sweeping a full circle as it rotates about the vertical diameter; only $B$'s horizontal component threads a time-varying flux.

Approach. The rotation axis (a vertical diameter, lying IN the loop's plane) is vertical, so the loop's normal — always perpendicular to its own plane — stays horizontal and sweeps a full circle as the loop turns. Only $B$'s horizontal component ($B\cos45^\circ$) ever has a nonzero dot product with this horizontal normal; the vertical component of $B$ produces zero flux change and contributes no EMF.

Isolate the flux-driving component of $B$ and write the flux. With normal $\hat n(t)=(\cos\omega t,\sin\omega t,0)$ and $\vec B=(B\cos45^\circ,0,B\sin45^\circ)$ (arbitrary horizontal azimuth for $B$'s in-plane part, taken along $x$): $$\Phi(t)=NAB\cos45^\circ\cos\omega t$$
Angular speed and peak EMF. $$\omega=10000\ \text{RPM}=\frac{10000\times2\pi}{60}=\boxed{1047.20\ \text{rad/s}}$$ $$\varepsilon_{\text{peak}}=-\frac{d\Phi}{dt}=NAB\cos45^\circ\,\omega=(10)(\pi(0.05)^2)(10^{-5})(0.70711)(1047.20)$$ $$\varepsilon_{\text{peak}}=\boxed{5.816\times10^{-4}\ \text{V}}$$
RMS value. For a sinusoidal EMF, $\varepsilon_{\text{rms}}=\varepsilon_{\text{peak}}/\sqrt2$: $$\boxed{\varepsilon_{\text{rms}}=4.112\times10^{-4}\ \text{V}=411.2\ \mu\text{V}}$$

Quantity	Result
Angular speed $\omega$	$1047.20$ rad/s
Peak EMF	$5.816\times10^{-4}$ V
RMS EMF	$4.112\times10^{-4}$ V ($411.2\ \mu$V)

Check: the source gives only the vertical-plane tilt of $B$ ("$45^\circ$ up") and not its horizontal azimuth relative to the rotation axis's zero-position; this is immaterial to the RMS result because the loop's normal sweeps ALL horizontal azimuths once per revolution, so the peak flux always occurs when the normal aligns with whatever horizontal direction $B$ happens to point — only the magnitude $B\cos45^\circ$ of the horizontal component matters, not its azimuth.

Concept & Theory

Topic. Faraday's law for a coil rotating at constant angular speed about an axis lying in its own plane (the basic AC generator), with a magnetic field that is not purely perpendicular to the rotation axis.

Key relations. $\Phi(t)=NAB_{\perp}\cos\omega t$ where $B_{\perp}$ is the component of $B$ perpendicular to the rotation axis; $\varepsilon_{\text{peak}}=NAB_{\perp}\omega$; $\varepsilon_{\text{rms}}=\varepsilon_{\text{peak}}/\sqrt2$.

Why this works. Only the field component PERPENDICULAR to the rotation axis produces a time-varying flux, because the normal vector (which is always perpendicular to that axis) can only ever "see" that component as it sweeps around; the component of $B$ PARALLEL to the axis (here, vertical) is always perpendicular to the horizontal normal, contributing zero flux at all times and hence zero EMF — exactly analogous to the earlier translational motional-EMF case where only the field component perpendicular to both velocity and rod contributed.

Common pitfall. Using the FULL field magnitude $B$ (instead of only $B\cos45^\circ$) in the EMF formula, or converting RPM to rad/s incorrectly (RPM$\times2\pi/60$, not RPM$\times\pi/60$ or RPM$\times2\pi$).

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Faraday's law, rotating-loop / generator EMF).

Question 4 of 8: RMS EMF of a Coil Rotating About a Vertical Diameter in a Tilted Field

Question 4: RMS EMF of a Coil Rotating About a Vertical Diameter in a Tilted Field (20 marks)