Spinning Loops Lab — which current does each loop make in a half-field?

Answer:  P→3, Q→2, R→1, S→4

1. The idea. Only the part of a loop inside x > 0 carries flux: Φ = B · A_in. As the loop turns at ω = 2π/T, a straight edge of a sector sweeps area r²ω/2 per second as it crosses the boundary (the y axis). So the emf −dΦ/dt, and with it i, is a constant while an edge is crossing, and zero while nothing crosses. Positive i is along the drawn arrow (clockwise): it flows while the flux grows (Lenz).
2. (P) Half-disc → (3). Turning clockwise, its area enters the field steadily for half a turn and leaves steadily for the next half: i is a positive constant, then a negative constant.
3. (Q) Two 60° sectors on the left → (2). The upper sector enters (0 to T/6), nothing crosses (T/6 to T/3), the lower sector enters (T/3 to T/2); then the upper one leaves (T/2 to 2T/3), a gap, and the lower one leaves (5T/6 to T): two positive pulses, then two negative.
4. (R) One 60° sector → (1). It enters (a positive pulse, 0 to T/6), stays inside, leaves (a negative pulse, T/2 to 2T/3), then stays out.
5. (S) Two opposite 60° sectors → (4). One sector is in the field and one is out; they are diametrically opposite, so whenever one crosses out the other crosses in, at the same rate. The wire bends at O without crossing, so both lobes circulate the same way: the total area in the field, and the flux, stay constant, and i = 0. (If the lobes circulated opposite ways, the two changes would add, not cancel, giving pulses that match no graph in List-II.)
6. That is P→3, Q→2, R→1, S→4: option (C). Graph (5), a current growing in time, would need area entering faster and faster, which a turning sector never does.

Understand the concept

Key takeaways

• Flux through a loop in a half-field: Φ = B × (area inside the field).
• A rotating sector's edge sweeps area at r²ω/2: a constant emf while it crosses the boundary.
• No crossing, no current, even with the loop deep in the field.
• Lobes of one loop add their flux changes when they circulate the same way.

Verification: an independent derivation per loop; the overlap of each sector with the half-plane worked out piecewise (and in sympy for the half-disc); Faraday's law applied to each loop exactly as drawn with Green's theorem (flux / B = ∮ max(x, 0) dy, which counts every lobe with its own direction); a control showing a crossing figure-8 would not give zero; every option audited; and the page's own engine, run by tests/verify_p1phyq15.py. The MathonGo solution was read first only for the approach; its explanation of S (opposite-sense lobes cancelling) is not used, but its answer (C) and the official key (C), read last, agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.