Induced Voltage Across a Gapped Loop in a Collapsing Field: 04-BS-9, December 2015, Q7

Question 7: Induced Voltage Across a Gapped Loop in a Collapsing 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 diameter (radius $r$)	$10$ cm ($r=0.05$ m)
Loop plane / field	vertical E–W plane; $B$ horizontal, pointing north (along the loop's normal)
$B$: initial $\to$ final	$0.2$ T $\to$ $0$ in $0.1$ s (uniform rate)
Gap	$1$ mm, at the bottom of the loop

Find. Magnitude and polarity of the voltage across the gap.

Vertical E–W loop viewed facing north; the field points along the loop's normal (north) and is uniformly reduced to zero. Gap sits at the bottom of the loop.

Approach. Apply Faraday's law to the (nearly) closed loop; since the gap carries no current (open circuit), the entire induced EMF appears as an open-circuit voltage across it, with polarity fixed by Lenz's law.

Loop area. $$A=\pi r^2=\pi(0.05)^2$$ $$A=\boxed{7.854\times10^{-3}\ \text{m}^2}$$
Rate of change of the field. $$\frac{dB}{dt}=\frac{0-0.2}{0.1}=\boxed{-2\ \text{T/s}}$$
Induced EMF magnitude. $$|\text{EMF}|=A\left|\frac{dB}{dt}\right|=(7.854\times10^{-3})(2)$$ $$|\text{EMF}|=\boxed{15.71\ \text{mV}}$$
Polarity, by Lenz's law. The northward flux is decreasing, so the induced current tries to maintain it — by the right-hand rule this drives (hypothetical) current from east to west across the bottom of the loop. Current arriving at the gap from the east side and unable to cross makes the east terminal the positive (higher-potential) side, the west terminal negative.

Quantity	Result
Loop area $A$	$7.854\times10^{-3}$ m$^2$
$dB/dt$	$-2$ T/s
Voltage across the 1 mm gap	$15.71$ mV, east terminal positive

Concept & Theory

Topic. Faraday's law of induction in a nearly-closed (gapped) loop.

Key relations. $\text{EMF}=-d\Phi/dt=-A\,dB/dt$ for a fixed-area loop with $B$ aligned along the loop's normal; Lenz's law for polarity.

Why this works. With no current able to flow across the open gap, there is no resistive drop anywhere in the wire, so the entire induced EMF must appear across the gap — exactly as if an ideal battery equal to the EMF sat there. Lenz's law identifies which way the (blocked) current would try to flow, and the terminal that current is driven INTO becomes the higher-potential (positive) one.

Common pitfall. Getting the correct current SENSE from Lenz's law but then mislabelling the polarity — the terminal the hypothetical current flows toward (and piles up against) is positive, not negative.

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

Question 7 of 8: Induced Voltage Across a Gapped Loop in a Collapsing Field

Question 7: Induced Voltage Across a Gapped Loop in a Collapsing Field (20 marks)