RMS EMF Induced in a Loop by an Oscillating Horizontal Field: 04-BS-9, May 2014, Q5

Question 5: RMS EMF Induced in a Loop by an Oscillating Horizontal 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
Side length $s$	$10$ cm $=0.1$ m
Turns $N$	$70$
Loop plane	vertical, containing the NW–SE horizontal line
Field $B_{\text{rms}}$ (E–W, horizontal)	$0.7$ T
Frequency $f$	$60$ Hz

Find. The RMS EMF induced in the loop.

Top view: loop lies edge-on along the NW–SE line (vertical plane); the loop's horizontal normal is along NE–SW, which is 45° from the E–W field.

Approach. The loop's plane contains the vertical and the NW–SE horizontal direction, so its (horizontal) normal points NE–SW — 45° from the given E–W field. Faraday's law then gives the RMS EMF directly from the RMS rate of change of the flux component along that normal.

Angle between field and loop normal. NE–SW is exactly 45° from E–W, so the flux-linking component of $B$ is $B\cos45^\circ$ at every instant.
Faraday's law for a sinusoidal field. With $B(t)=B_0\cos\omega t$ (so $B_{\text{rms}}=B_0/\sqrt2$) and loop area $A=s^2$: $$\varepsilon(t)=-N\frac{d\Phi}{dt}=NA\cos45^\circ\,\omega B_0\sin\omega t$$ Since $dB/dt$ is itself a sinusoid whose RMS value is $\omega B_{\text{rms}}$, the induced EMF's RMS value follows the same linear relation as the peak values: $$\varepsilon_{\text{rms}}=NA\cos45^\circ\,\omega\,B_{\text{rms}}$$
Substitute the numbers. $A=(0.1)^2=0.01\ \text{m}^2$, $\omega=2\pi(60)=376.99\ \text{rad/s}$: $$\varepsilon_{\text{rms}}=(70)(0.01)(0.70711)(376.99)(0.7)$$ $$\varepsilon_{\text{rms}}=\boxed{130.6\ \text{V}}$$

Quantity	Result
Angle between $B$ and loop normal	$45^\circ$
$\omega$	$377.0$ rad/s
$\varepsilon_{\text{rms}}$	$130.6$ V

Concept & Theory

Topic. Faraday's law for a stationary loop in a time-varying field, with the loop's plane oblique to the field.

Key relations. $\Phi=NBA\cos\theta$; $\varepsilon=-d\Phi/dt$; for a sinusoid, RMS values scale linearly through differentiation, so $\varepsilon_{\text{rms}}=NA\cos\theta\,\omega B_{\text{rms}}$.

Why this works. Only the component of $B$ along the loop's normal links flux through it; a compass-direction geometry problem like this one reduces entirely to finding that one angle (here 45°, from the NW–SE/NE–SW right-angle relationship), after which the induced EMF follows the standard transformer-EMF formula.

Common pitfall. Using the angle between $B$ and the loop's plane (i.e. $\sin\theta$) instead of between $B$ and the loop's normal ($\cos\theta$) — here both give the same numeric factor only because $45^\circ$ is self-complementary; that coincidence would not survive a different angle.

Source. Young & Freedman, University Physics, Ch. 29 (Faraday's Law, motional and transformer EMF).

Question 5 of 8: RMS EMF Induced in a Loop by an Oscillating Horizontal Field

Question 5: RMS EMF Induced in a Loop by an Oscillating Horizontal Field (20 marks)