ARCHETYPE: Loop Entering & Leaving a Field

Swipe a card and you run this archetype in reverse: the stripe's magnetized segments stream past a fixed pickup coil, each flux reversal writing a voltage blip, and the read head's electronics decode the pulse train — boundaries, not plateaus, carry all the information. Airport security wands and buried-cable locators pulse a field and listen for the induced response of conducting loops crossing it. And in particle-accelerator beamlines, "flux gates" and stripline monitors read the passage of charge bunches as exactly these entry/exit pulse pairs — the flat-top triptych, drawn a billion times a second.

Set the stage precisely, because every mark in this gym lives in the geometry. A rectangular loop — edge length $L$ perpendicular to the motion, width $b$ along it — slides at constant velocity $v$ along the $x$-axis. Ahead lies a slab of uniform field $B$, pointing into the page, occupying a region of width $d$ wider than the loop ($d > b$). Outside the slab: no field. Five acts. Approach: no flux, nothing happens. Entry: the leading edge is inside, the trailing edge still out; the overlap — and the flux — grows. Full immersion: both edges inside; flux at its maximum $BLb$, frozen there. Exit: the leading edge has escaped, the trailing edge is still in; the overlap shrinks. Gone: flux zero again.

The instinct this figure must build: hunt for the active edge. During entry the leading edge is a rod of length $L$ sweeping through field $B$ at speed $v$ — a motional-emf source exactly as in Motional EMF, wired to the rest of the loop. During full immersion both edges are such rods, generating equal emfs that chase each other around the loop and cancel: no net push, no current, silence. During exit only the trailing edge is active, and the push has flipped.

Write the flux as a function of position $x$ of the leading edge, taking $x = 0$ where it meets the boundary. During entry, the overlap width is $x$, so $\Phi = BLx$ — a straight ramp. Fully inside, $\Phi = BLb$ — the plateau. During exit (leading edge past the far boundary, overlap shrinking), the ramp runs back down to zero. Trapezium. The emf follows as the slope of that trapezium swept at speed $v$. On the entry ramp the flux grows at rate

where: $L$ — the active edge's length; $v$ — the drag speed. This is Motional EMF's law wearing gym clothes, and The Rod on Rails already turned it into a system: current, force, power. Here the new content is when it switches on and off — the emf is a pulse of height $BLv$ lasting exactly the time $b/v$ it takes one edge to cross, then nothing, then an equal and opposite pulse on the way out. Close the loop with total resistance $R$ and the current during either crossing is

where: the direction flips between entry and exit — Lenz's Law & Energy Conservation applied twice: entering, the flux (into the page) grows, so the induced current circulates anticlockwise to oppose it; leaving, the flux dies, and the current flips to clockwise to prop it up. And the active edge, carrying current $I$ in field $B$, feels a force $BIL$ which — both times — points against the motion:

where: the magnitude is The Rod on Rails' drag formula, cited not re-derived. What this gym adds is the profile: force pulses coincide with emf pulses, and both pulses drag. The field never pushes the loop onward; leaving is resisted just as entering was. If that feels wrong, it is Lenz's honesty: any help would be free energy.

Here is the whole story as one figure — the triptych every graph MCQ is sampled from:

Read the three panels against each other until the correspondences are automatic: ramps make pulses; the plateau makes silence; the emf pulses are mirror images, but the drag pulses are twins.

Two totals summarize a complete transit, and they make a beautifully treacherous pair.

The heat ledger. Your hand must supply force $F = B^2L^2v/R$ over a distance $b$ during entry, and again over $b$ during exit — the immersion stretch is free. At constant speed all of that work becomes circuit heat:

where: the factor $2$ counts the two crossings. Note $W \propto v$: dragging faster costs more energy per transit, not less — the pulses are taller by more than they are briefer. (Cross-check by power: $P = \varepsilon^2/R$ during each pulse, times total pulse time $2b/v$, gives the same $W$.)

The charge ledger. Each crossing moves charge $q = \Delta\Phi/R = BLb/R$ — the speed-blind endpoint law from Flux-Change Numericals, applied here per boundary. But the two crossings drive current in opposite directions, so the net charge for a full transit is zero, even while the heat is decidedly not. Heat adds; signed charge cancels. Exam setters exploit exactly this contrast.

Circles instead of rectangles. A circular loop crossing the same slab has no constant active edge: the chord inside the boundary grows from zero to the diameter and back. The flux–position curve becomes S-shaped, and the emf pulse becomes a rounded hump instead of a rectangle. The graph-reading rule survives untouched — emf is still the slope — only the shapes soften. MCQs love offering you the rectangle answer for the circle stem, and vice versa.

Loop wider than the region ($b > d$). Now the loop can never be fully immersed; the flux plateau sits at $BLd$ (the region is the limiting window), and it lasts while the slab lies wholly inside the loop. The emf is again zero on that plateau — this time it is the region that is fully swallowed. The lesson generalizes: silence whenever the overlap stops changing, whoever swallows whom.

The falling magnet's two blips. Drop a bar magnet through a horizontal coil and watch the emf on a scope: a hump as it approaches (flux rising), a zero-crossing near the coil's plane (flux at maximum — the moment of greatest field and total silence), then an opposite hump as it recedes. Because the magnet accelerates, the second blip is taller and narrower than the first — but the two blips enclose equal areas, since area is flux change and the flux rises exactly as far as it falls. That equal-areas read, with the charge bookkeeping cited from Flux-Change Numericals, is a classic AP and JEE graph question — and it is precisely the trace André Blondel's oscillograph first made visible.

Board level 1 — one full pass, all four quantities (AP-2/board pattern)

A rectangular loop with edge $L = 0.20$ m perpendicular to its motion and width $b = 0.10$ m is dragged at $v = 2.0$ m/s into a region of uniform field $B = 0.50$ T (into the page). Loop resistance $R = 0.80\ \Omega$. For the entry phase find (a) the emf, (b) the current, (c) the applied force needed to keep $v$ constant, and (d) how long the phase lasts.

Step 1. Emf while the leading edge crosses: $\varepsilon = BLv = 0.50 \times 0.20 \times 2.0 = \boxed{0.20\ \mathrm{V}}$

Step 2. Current: $I = \varepsilon/R = 0.20/0.80 = \boxed{0.25\ \mathrm{A}}$ — anticlockwise, opposing the growing into-page flux (Lenz, stated in words).

Step 3. Force: the field drags with $F = BIL = 0.50 \times 0.25 \times 0.20 = 0.025$ N, so your hand supplies $\boxed{0.025\ \mathrm{N}\ (= 25\ \mathrm{mN})}$ forward to hold $v$ steady.

Step 4. Duration: the edge crosses in $\Delta t = b/v = 0.10/2.0 = \boxed{0.050\ \mathrm{s}}$. Sanity: power in $= Fv = 0.05$ W; power out $= \varepsilon^2/R = 0.04/0.8 = 0.05$ W ✓ — the rails ledger balancing, as The Rod on Rails demands.

Board level 2 — reading a flux–time graph (graph-literacy pattern)

The flux through a single loop rises linearly from $0$ to $8.0$ mWb in $0.05$ s, stays constant for $0.10$ s, then falls linearly to zero in $0.05$ s. Sketch the emf and give its value in each interval.

Step 1. Rising segment: $\varepsilon = \dfrac{\Delta\Phi}{\Delta t} = \dfrac{8.0\times10^{-3}}{0.05} = \boxed{0.16\ \mathrm{V}}$ (one direction).

Step 2. Flat segment: slope zero $\Rightarrow \boxed{\varepsilon = 0}$ — regardless of the $8.0$ mWb sitting there.

Step 3. Falling segment: same slope magnitude, opposite sign: $\boxed{0.16\ \mathrm{V}}$, reversed direction. The sketch is two opposite rectangles separated by silence — the flat-top signature.

Step 4. Bonus mark that boards love: the area of each rectangle is $0.16 \times 0.05 = 8.0$ mV·s $= 8.0$ mWb $= \Delta\Phi$ ✓.

Competitive level 1 — the transit's heat and net charge (JEE-Main pattern)

The loop of Board 1 makes one complete transit of a wide field region (fully in, then fully out). Find (a) the total heat dissipated and (b) the net charge that circulates. Then explain in one line why the answers differ so violently in spirit.

Step 1. Heat: drag acts only during the two crossings, $F = B^2L^2v/R = 0.025$ N over distance $b$ each time:

Step 2. Charge per crossing: $q = \dfrac{BLb}{R} = \dfrac{0.50 \times 0.20 \times 0.10}{0.80} = 12.5$ mC — in on entry, out on exit, opposite senses.

Step 3. Net charge: $\boxed{0}$ (magnitude per crossing $12.5$ mC).

Step 4. The line: heat adds squares (direction-blind), charge adds signed flows — the flux returns to its starting value, so the signed sum must vanish while the thermal bill stands. Timing note: two minutes. The only computation is $F$ once and $q$ once; everything else is the two-crossings bookkeeping. If you find yourself integrating anything, you have missed the constant-pulse structure.

Competitive level 2 — double the speed, audit everything (JEE-Advanced flavour)

The same loop repeats the full transit at $v = 4.0$ m/s. For the faster run, find the emf pulse height, the drag force, the total heat, the charge per crossing, and the pulse duration — and state which of the five changed.

Step 1. Emf: $\varepsilon = BLv = 0.50\times0.20\times4.0 = \boxed{0.40\ \mathrm{V}}$ — doubled.

Step 2. Drag: $F = B^2L^2v/R = \dfrac{0.25\times0.04\times4.0}{0.80} = \boxed{0.050\ \mathrm{N}}$ — doubled.

Step 3. Heat: $W = 2Fb = 2\times0.050\times0.10 = \boxed{10\ \mathrm{mJ}}$ — doubled ($W \propto v$: faster transits cost more, not less).

Step 4. Charge per crossing: $q = BLb/R = \boxed{12.5\ \mathrm{mC}}$ — unchanged (endpoints only).

Step 5. Pulse duration: $b/v = 0.10/4.0 = \boxed{25\ \mathrm{ms}}$ — halved. Summary: rates ($\varepsilon, F$) and totals-of-squares ($W$) scaled with $v$; the endpoint total ($q$) ignored it; the clock compressed. Timing note: under two minutes as pure ratio work from Competitive 1's anchors — double, double, double, same, half. Recomputing each from scratch is the slow lane.

Emf whenever flux exists. The plateau is silent. Any option showing nonzero emf while the loop is fully immersed is dead on arrival — this single read eliminates half the graph choices.

"The field pushes it out as it leaves." Never. Drag opposes motion during both crossings — leaving is resisted exactly as entering was (Lenz's Law & Energy Conservation; free rides are free energy). Force graphs showing a sign flip between entry and exit are wrong; it is the emf that flips, not the drag's opposition.

Confusing the two flips. Between entry and exit: current direction flips, emf sign flips, drag direction (relative to the page) flips so as to keep opposing $v$ — three statements, one physics. Sort them before bubbling.

Charge scaled with speed. $q = BLb/R$ per crossing, endpoints only — Flux-Change Numericals' speed-blindness, which examiners re-test here dressed in position language. And net transit charge is zero; "total charge" versus "charge during entry" is a wording read.

Rectangle answers for circle stems. A circular loop's emf profile is a rounded hump, not a flat top — no constant active edge. Match shapes to geometry before matching numbers.

Area ignored. The area under each emf pulse equals $\Delta\Phi$ (and equals $qR$). Graph questions that give you no formulas are usually asking for exactly this read.

Which constants? None — the gym runs on givens. Keep $L$ (perpendicular edge) and $b$ (along-motion width) straight: swapping them poisons $\Phi_{max}$ and $q$ but not $\varepsilon$, which is how setters catch the half-understanding.

Texts. NCERT Physics Class 12, Ch. 6 (motional emf sections and exemplar graph problems) · HC Verma Vol. 2, Ch. 38 (rectangular-loop exercises) · AP Physics 2 released free-response items on loops and field regions.

Provenance. All numbers recomputed for this article: Board 1 gives $0.20$ V, $0.25$ A, $25$ mN, $50$ ms with the $0.05$ W power balance checked both routes; Board 2 gives $0.16$ V / $0$ / $0.16$ V and the $8.0$ mWb area check; Competitive 1 gives $5.0$ mJ and $\pm12.5$ mC (net zero); Competitive 2 gives $0.40$ V, $0.050$ N, $10$ mJ, $12.5$ mC, $25$ ms.

Prerequisites. The Rod on Rails · Flux-Change Numericals — Pulls, Flips & Ramps. Leads to. Lenz Without Tears — Signs in EMI (the sign discipline these graphs keep testing) · The AC Generator (rotation instead of translation) · Motional EMF (UG deep dive).