Welcome to Electromagnetic Induction. The Magnetic Fields & Forces topic showed that an electric current makes a magnetic field around it, the effect Oersted discovered in 1820. Michael Faraday spent years looking for the reverse, a magnetic field that makes a current, and found it in 1831, though not in the form anyone expected. A magnet lying next to a coil drives no current through it, however strong the magnet is. A current appears only while something changes: while the magnet moves, or while the current in a nearby circuit is switched on or off. The effect is called electromagnetic induction (电磁感应), and the current it produces is an induced current (感应电流). This page finds exactly when an induced current flows, and where it flows. The Lenz's Law page finds which way it goes, the Faraday's Law page finds how large its EMF is, and the Rods on Rails page uses both on the topic's most repeated question.
Induction has had 10 of the 192 questions in the papers so far, at least one in every paper, all of them from Q27 on. Two belong to this page: which action in Faraday's iron-ring experiment gives no current (March 2026, Q35), and a rod sliding on a U-shaped rail, where the question is how the flux through the closed part changes and which pieces of metal carry the current (January 2026, Q42). Both are worked through below. Neither needs a formula; both need the condition for an induced current stated exactly. Their wrong options come from two habits, which this page names as they come up: taking a strong field or a steady current for a changing one, and assuming that every piece of metal in a field carries a current.
A current from a magnet
Connect a coil of wire to a galvanometer and nothing else: there is no battery anywhere in the circuit. A galvanometer (灵敏电流计) is a sensitive current meter with its zero in the middle of the scale, so its pointer can swing either way. It shows a current far too small for an ordinary ammeter, and which way round the circuit that current flows. Now bring up a bar magnet.
Push the magnet's N pole into the coil and the pointer swings to one side while the magnet moves. Stop the magnet inside the coil and the pointer goes back to zero, and it stays there for as long as the magnet rests, however strong the magnet is. Pull the magnet out and the pointer swings to the other side. Push the magnet in faster and the swing is larger; push it in slowly and the swing is small. Finally, hold the magnet still and move the coil onto it instead: the pointer behaves exactly as before. What counts is the motion of the magnet and the coil relative to each other.
A coil joined only to a galvanometer G, with no battery. The pointer moves while the magnet moves into the coil, sits at zero while the magnet rests inside it, however strong the magnet, and swings the other way while the magnet is pulled out. Which way it swings first depends on how the coil is wound; the Lenz's Law page works that out.
The magnet's field is inside the coil the whole time the magnet rests there, so the field on its own cannot be what drives the current. What the three moments differ in is change. While the magnet moves in, the field inside the coil grows. While it rests, the field is steady. While it moves out, the field fades. The current flows only while the field in the coil is changing, and a growing field and a fading field drive it opposite ways.
Try it! · Check-in question
A strong bar magnet rests inside a coil that is connected to a galvanometer. Which of these makes the galvanometer's pointer deflect?
Pulling the coil off the magnet moves the coil and the magnet relative to each other. As the coil moves away, the field inside it fades, and a fading field drives a current, so the answer is C. A changes nothing, and waiting longer does not help: a steady field induces nothing however long it lasts. B moves both things, but together. Nothing moves relative to anything else, so the field inside the coil stays exactly as it was. D has a stronger field, but still a steady one. The size of the field does not matter, only whether it is changing.
What changes is the flux
The Magnetic Induction & Flux page of the Magnetic Fields & Forces topic measured how much field passes through a loop by the magnetic flux, Φ = BS cos θ, in webers (1 Wb = 1 T·m²), where θ is the angle between the field and the normal to the loop; for a loop square-on to the field, Φ = BS. Flux is the quantity that fits every one of Faraday's results. As the magnet comes in, B inside the coil grows, so the flux through each turn grows. While the magnet rests, the flux is steady, whatever its size. As the magnet leaves, the flux falls. Faraday's other experiments, below, point the same way, and together they give the condition on which the rest of this topic is built.
The condition for an induced current. A current is induced in a circuit only while two things are true: (1) the circuit is closed, and (2) the magnetic flux through it is changing. The current flows for as long as the flux changes and stops when the flux stops changing. How large the flux is does not matter, only whether it is changing.
Φ = BS cos θ has three factors, so there are three ways to change the flux.
Change B. Move a magnet towards or away from the circuit, or change the current in a nearby coil: switch it on, switch it off, or move a rheostat's slider.
Change S. Change the area that the field passes through: slide a rod along rails to lengthen or shorten a loop, squash a loop out of shape, or move a loop into or out of a region of field.
Change θ. Turn the loop so that it faces the field more or less squarely. A loop turned from square-on to the field (θ = 0, Φ = BS) to edge-on (θ = 90°, Φ = 0) loses all its flux.
Motion and change are not the same thing. Slide a loop sideways through a uniform field, keeping it inside the field and not turning it, and B, S and θ all stay as they were. The flux is constant, and no current flows, however fast the loop goes. The test is always whether the flux changes, never whether something moves.
Try it! · Check-in question
A closed wire loop lies in a uniform magnetic field, with its plane perpendicular to the field. Which of these does not induce a current in the loop?
Go through Φ = BS cos θ factor by factor. Sliding the loop inside a uniform field leaves B the same, S the same and θ = 0, so the flux is constant and no current is induced, although the loop is moving, so the answer is D. A changes B, so the flux grows. B changes S: a circle encloses the most area for its length of wire, so the narrow oval lets less of the field through and the flux falls. C changes θ: as the loop turns away from square-on, cos θ falls from 1, and so does the flux. Each of the three is a change in flux while it happens, so each induces a current.
A rod that cuts field lines
A second kind of experiment changes the area instead of the field. Lay two straight metal rails side by side, join them at one end through a galvanometer, and place them in a uniform field at right angles to their plane. A metal rod laid across the rails completes a circuit: the rod, the two rails and the meter enclose a rectangle. Slide the rod away from the meter and the pointer deflects. Stop the rod and the pointer returns to zero. Slide it back and the pointer swings the other way. The field never changes here, but the area of the rectangle does. As the rod moves away from the meter, the rectangle gets longer, and with B constant the flux Φ = BS grows with it.
Many books describe the same experiment from the rod's point of view: a current is induced when a conductor that is part of a closed circuit cuts field lines (切割磁感线), that is, moves across them. A rod moving along the field lines, or along its own length, cuts none and gives no current. For a rod on rails the two descriptions always agree, because a rod that sweeps across field lines adds the field it sweeps over to the loop, or takes it away. Of the two, the flux through the closed circuit is the safer test. It also covers circuits in which nothing moves at all, such as the iron ring in the next section.
Try it! · Check-in question
A straight horizontal copper wire is held between the poles of a horseshoe magnet, where the field runs horizontally from the N pole to the S pole, at right angles to the wire. The ends of the wire are connected to a galvanometer by long, flexible leads. Which motion of the wire makes the pointer deflect?
Only moving the wire up carries it across the field lines, so the answer is A. The flux view agrees. The wire, its leads and the meter form a closed circuit, and as the wire rises it changes how much of the magnet's field passes through that circuit. B moves the wire along the field lines, so it cuts none of them and the field through the circuit stays the same. C slides the wire along itself: one piece of wire simply takes the place of another, and nothing about the circuit's flux changes. D is the steady-field case again: the field between the poles is strong, but it is not changing.
Faraday's iron ring
Faraday's first success, in August 1831, used no moving magnet at all. He wound two separate coils on opposite sides of an iron ring. Coil A was in a circuit with a cell and a switch; in the version the March 2026 paper drew, a rheostat sits in that circuit too. Coil B was joined only to a galvanometer. No wire connects the two circuits.
Faraday's iron ring, the circuit of March 2026 Q35 redrawn. Coil A, on the left, is in a circuit with a cell, a switch S and a rheostat. Coil B, on the right, is joined only to a galvanometer G. No wire connects the two circuits: their only link is the ring, which carries coil A's field round through coil B.
Close the switch and the galvanometer's pointer gives a brief kick, then settles back to zero. With the switch held closed, a steady current flows in coil A and the ring is strongly magnetised, yet the pointer stays at zero. Open the switch and the pointer kicks again, the other way. With the switch closed, move the rheostat's slider and the pointer deflects for as long as the slider moves.
The flux explains all four results. The current in coil A magnetises the ring, and the ring carries that field round through coil B, so the flux through coil B rises and falls with coil A's current. Closing the switch sends coil A's current up from zero, so the flux through coil B grows. Opening it lets the current die away, so the flux falls, and the kick reverses. Moving the slider changes the resistance in coil A's circuit, so coil A's current changes, and the flux changes with it. Holding the switch closed gives a steady current and so a steady flux, however large, and a steady flux induces nothing.
Worked example · the action that gives no current (March 2026, Q35)
Figure A shows the coil used in Faraday's electromagnetic induction experiment. The working principle is shown in Figure B. Which of the following does not cause the pointer of the galvanometer to deflect in the experiment? ( ).
A. When opening the switch B. Keeping the switch closed C. When closing the switch D. When adjusting the rheostat
Figure A is a picture of the ring itself; Figure B is the circuit drawn above. The paper's English calls the meter an "ammeter", but the meter is drawn and used as a galvanometer, so the stem is quoted here with that word.
Step 1, the closed circuit: coil B and the galvanometer form a closed circuit the whole time, so the question comes down to whether the flux through coil B changes.
Step 2, the flux: the flux through coil B follows the current in coil A. Opening the switch (A) makes that current fall to zero, closing it (C) makes it rise from zero, and adjusting the rheostat (D) changes it. Each one changes the flux through coil B, and each one deflects the pointer.
Step 3: with the switch kept closed (B), coil A carries a steady current, the flux through coil B is constant, and no current is induced. The answer is B.
Each wrong option is a change that is easy to overlook. A catches students who think switching off is doing nothing: the current in coil A dies away, and that fall in flux kicks the pointer just as the rise did, only the other way. D catches students who think the rheostat belongs to coil A's circuit alone. It does, but it sets coil A's current, and that current sets the flux in the ring. B is the option that feels wrong, because a strong, steady current flows in coil A the whole time the switch is closed. Strong is not enough; the flux has to change.
Exam wording. The papers' English is not consistent about the meter: March 2026 calls it an ammeter in Q35 and a galvanometer (G) in Q45. In both questions it is a centre-zero meter that shows a small current and its direction. In Chinese papers a galvanometer is 灵敏电流计 (or 电流计), a rheostat is 滑动变阻器, electromagnetic induction is 电磁感应, an induced current is 感应电流, magnetic flux is 磁通量, a closed circuit is 闭合电路, and cutting field lines is 切割磁感线.
Try it! · Check-in question
In Faraday's ring experiment the switch has been closed for some time, and the galvanometer reads zero. Which statement is correct?
Moving the slider changes the resistance in coil A's circuit, so coil A's current and the flux through coil B change while it moves, and stop changing when it stops, so the answer is B. A has the reason wrong. The steady current in coil A keeps the ring magnetised the whole time; the pointer is at zero because that magnetisation is not changing. C makes the steady current larger but leaves it steady: the flux is larger and still constant, so there is still no induced current. D forgets that "stops" is a change. The current in coil A falls to zero, the flux through coil B falls with it, and the pointer kicks the opposite way to the kick on closing.
An EMF without a current
The condition has two parts, and an exam option can break either one. The second part, a changing flux, does something even when the first part fails. When the flux through a coil changes, the charges in its wire are pushed along it: the changing flux sets up an induced EMF (感应电动势), just as a battery has an EMF. The EMF & Closed Circuits page of the DC Circuits topic showed that a battery keeps its EMF on an open circuit, where a voltmeter reads it, but drives no current until the circuit is closed. An induced EMF behaves the same way. In a coil whose ends are not joined, the charges are pushed to the two ends, pile up there and stop, leaving a voltage across the gap but no current round the coil. Join the ends, directly or through a meter, and the current flows.
EMF and current. A changing flux always induces an EMF, whether or not the circuit is closed. An induced current needs the circuit to be closed as well. The Faraday's Law page finds the size of the EMF.
Put together, the two parts of the condition give four situations in which no current flows, and the papers' wrong options are built from all four.
No induced current in four situations. A switch held closed: a steady current gives a steady flux (March 2026, Q35). A magnet at rest in a coil, however strong. A loop moving inside a uniform field without turning: B, S and θ all stay the same. An open circuit: there may be an EMF, but there is no current.
Try it! · Check-in question
The two ends of a coil are left unconnected, with a small gap between them. A bar magnet is pushed quickly into the coil. Which statement is correct while the magnet moves?
The flux through the coil is changing, so an EMF is induced, and it appears as a voltage across the gap. The circuit is open, so no current can flow round it, and the answer is A. B ignores the first part of the condition: a current needs a closed path, and the gap breaks it. C goes too far the other way. The open circuit stops the current, not the EMF, just as a battery keeps its EMF with nothing connected. D has the timing backwards: the EMF lasts only while the flux changes, so once the magnet stops there is no EMF either, and joining the ends then would give nothing.
Where the current flows
An induced current, like any current, needs a complete path. It flows round the closed loop whose flux is changing, and nowhere else. A wire that leads to a dead end, such as the end of a rail that is joined to nothing, cannot carry a steady current: charge flowing into it would have nowhere to go. In a circuit made of rails and a rod, then, the first step is to find the closed loop. It is made of the moving rod, the stretches of rail between the rod and whatever joins the rails together, and that joining piece itself. Rails on the far side of the rod belong to no loop and carry no current, however much field passes over them.
The rail of January 2026 Q42, redrawn, with the field into the page everywhere (×). The rod ef, the rails eb and cf and the load bc make the closed loop ebcf, shaded. As ef moves left to the dashed position, the loop gets longer and lets more of the field through. The stretches ae and fd end at a and d, which are joined to nothing.
Worked example · which loop, and which segments (January 2026, Q42)
As shown in the figure, a fixed U-shaped metallic rail abcd with a load connected is placed in a uniform magnetic field perpendicular to the plane of the rails. There is a metal rod ef that can slide freely on the rails. During the process in which the rod ef moves to the left, which of the following statements is correct?
A. There is a current flowing through segment ae
B. The magnetic flux passing through ebcf remains unchanged
C. There is a current flowing through segment bc
D. The magnetic flux passing through ebcf decreases
The printed stem calls the rail "abd", but its figure labels the corners a, b, c and d, so the rail is called abcd here.
Step 1, the closed loop: the rod ef, the rail from e to b, the load bc and the rail from c back to f make a closed loop, ebcf. The stretches ae and fd run from the rod to the open ends a and d, which are joined to nothing.
Step 2, the flux through ebcf: B is uniform and constant, so Φ = BS changes only if the area changes. As ef moves left, eb and cf get longer and the loop's area grows, so the flux through ebcf increases. B and D are false.
Step 3, the current: the flux through a closed loop is changing, so a current flows round ebcf, along the rod, along the rails eb and cf, and through the load bc. C is true.
Step 4, the open end: ae is part of no closed loop, so it carries no current, and A is false. The answer is C.
B is the trap for students who see a uniform, constant field and stop there. Φ = BS has two factors, and here the area is the one that changes. D comes from watching the wrong region. The stretch between a and the rod does get shorter, but a and d are not joined, so that region is not a circuit and its flux drives nothing. A assumes that every piece of metal in the field carries a current. ae sits in the same field as the rest of the rail, but a current cannot flow into a dead end.
Try it! · Check-in question
Two long, straight, parallel rails run from left to right in a uniform magnetic field perpendicular to their plane. The upper rail runs from p on the left to q on the right, and the lower one from s on the left to r on the right. A resistor joins q and r; the ends p and s are open. A metal rod mn lies across the rails, with m on the upper rail and n on the lower one, and is pulled to the right, towards the resistor. Which statement is correct?
The closed loop is mqrn: the rod, the rails from the rod to the resistor, and the resistor. As the rod moves right, towards the resistor, mq and nr get shorter and the loop's area shrinks. With B constant, the flux through mqrn falls, and a changing flux through a closed loop drives a current round it, through the resistor, so the answer is D. A assumes that a moving rod always makes the flux grow. Here the loop is on the side the rod is moving towards, so it shrinks; always look at which side of the rod the closed loop lies. B forgets the area, the trap of January's option B. C puts current in pm, which leads only to the open end p and belongs to no closed loop.
Entering and leaving a field
A loop moving inside a uniform field carries no current. A real field, though, fills only a limited region, and a loop can cross its edges. The flux through the loop is then B times the part of its area that lies inside the field, and that part changes while an edge of the field passes across the loop. Follow a square loop moving at a steady speed straight through a region of field that is wider than the loop. Before the loop reaches the field, no field passes through it: Φ = 0, and nothing happens. While its front side is inside and its back side still outside, more of its area enters the field every moment, so the flux grows and a current flows. Once the whole loop is inside, Φ = BS stays constant and the current stops, although the loop keeps moving at the same speed. While its front side is out and its back side still in, the flux falls and a current flows again, the other way. After the loop has left, Φ = 0 once more.
A square loop moving at a steady speed through a region of field wider than itself. The shaded part of each loop is the part the field passes through. It grows while the loop enters, stays the same while the loop is fully inside, and shrinks while the loop leaves, so a current flows only while the loop crosses an edge of the field.
The rod's-eye view says the same. While the loop crosses an edge, only one of its sides is cutting field lines. Once it is fully inside, both sides cut field lines at the same rate, and their effects cancel round the loop. The Rods on Rails page uses this result for electromagnetic braking, where a frame fixed under a train runs into a region of field.
Three slips the options are built from. Taking a strong or steady field for a changing one: a magnet at rest in a coil, or a switch held closed, induces nothing. Taking motion for change: a loop sliding inside a uniform field, or a coil carried along with its magnet, keeps the same flux. Assuming every piece of metal in the field carries a current: only the closed loop whose flux changes does, and a stretch of rail that ends in an open end carries none.
Try it! · Check-in question
A square wire loop of side 10 cm moves at a steady speed to the right, straight through a region of uniform magnetic field 30 cm wide, and out the other side. The loop's plane stays perpendicular to the field. When is a current induced in the loop?
The flux through the loop is B times the part of its area inside the field. That part grows while the loop enters and shrinks while it leaves, and in between, for 20 cm of travel, the whole loop is inside and the flux is constant. A current flows only while the flux changes, so the answer is C. A takes motion for change; for most of the trip the loop moves without any change in its flux. B picks exactly the stretch when the flux is constant, the one time inside the field when no current flows. D treats the current as something that keeps going once started. An induced current lasts only while the flux keeps changing, so it stops as soon as the loop is fully inside, and it flows again, the other way, while the loop leaves.
Your turn! · Practice questions
Practice · Question 1
A rectangular wire loop lies on a table beside a long straight wire, in the same plane as the wire. In which case is a current induced in the loop?
The field of a straight wire passes through the loop at right angles to the table, and, as the Magnets & Fields of Currents page showed, it gets weaker with distance from the wire. Moving the loop away from the wire takes it into a weaker field, so the flux through it falls and a current is induced, so the answer is B. A moves the loop, but every part of it stays the same distance from the wire, so the field through it, and the flux, stay the same. C is a steady field and a loop at rest: nothing changes. D moves the loop the same way as B, but with no current in the wire there is no field, so the flux is zero before, during and after.
Practice · Question 2
A circular coil, connected to a galvanometer by long, flexible leads, lies flat on a table in a uniform vertical magnetic field. In which case does the galvanometer show a current?
Lying flat, the coil faces the vertical field square-on, θ = 0 and Φ = BS. Tilting it turns its normal away from the field, so cos θ, and with it the flux, falls while it tilts, so the answer is D. A turns the coil, but about an axis along the field, so it stays flat and square-on: B, S and θ are all unchanged. A coil can turn without its flux changing; what matters is whether its angle to the field changes. B moves the coil through a uniform field without turning it, the sliding loop again. C changes nothing at all.
Practice · Question 3
Two coils, P and Q, are wound on the same iron core, and coil Q is connected to a galvanometer. The current in coil P rises steadily from 0 to 2 A during the first 4 s, and then stays at 2 A. When does the galvanometer show a current?
The core carries coil P's field through coil Q, so the flux through Q follows the current in P. It changes while that current rises, during the first 4 s, and is constant afterwards, so the answer is A. The Faraday's Law page shows that a steady rise gives a steady induced current for the whole 4 s. B picks the time when the flux is largest, but also constant, so it induces nothing; that is the steady current of the ring experiment with the switch held closed. C confuses a current that is present with one that is changing. D picks the one moment when the change ends; the current in Q flows during the change, not at its end.
Practice · Question 4
Two metal rods lie across a pair of long, parallel rails in a uniform magnetic field perpendicular to the plane of the rails. The rails are not joined at either end. Both rods are pushed to the right together, always with the same velocity. Which statement is correct?
The only closed loop is made of the two rods and the stretches of rail between them. The rods keep the same distance apart, so the loop keeps the same area and, with B constant, the same flux, and no current flows, so the answer is C. A is true about the rods, but the conclusion does not follow. Each rod cuts field lines and has an EMF, but the two push charge opposite ways round the loop and cancel, just as the two sides of a loop do when it is fully inside a field. The flux test gives the answer in one step. B puts current in rails that lead only to open ends. D looks at the speed, but the rods have the same velocity at every moment, while speeding up as well, so the gap between them never changes.
Practice · Question 5
A flat coil of area 0.04 m², connected to a galvanometer, lies with its plane perpendicular to a uniform magnetic field of 0.5 T. It is turned through 90°, until its plane is parallel to the field. What is the flux through the coil before and after, and is a current induced while it turns?
Use Φ = BS cos θ, with θ measured between the field and the normal to the coil. Before: the plane is perpendicular to the field, so the normal lies along it, θ = 0 and Φ = 0.5 × 0.04 = 0.02 Wb. After: the plane is parallel to the field, so the normal is at 90° to it and Φ = 0. The flux changes while the coil turns, so a current is induced while it turns, and the answer is B. A notices that B and S stay the same and forgets the angle. C is the angle trap from the Magnetic Induction & Flux page: it measures θ from the plane of the coil instead of from its normal, which swaps the two values. D has the timing wrong. Once the coil stops, its flux is constant at zero, so the current flows during the turn and stops with it.
Things to remember
Concept
What to know
Electromagnetic induction
A changing magnetic flux produces a current in a closed circuit, the induced current (感应电流). Faraday found it in 1831. A field alone, however strong, does nothing.
Galvanometer
A sensitive meter with its zero in the middle: it shows a small current and which way it flows. March 2026 calls it an ammeter in Q35 and a galvanometer in Q45.
Magnetic flux
Φ = BS cos θ, with θ between B and the normal to the loop. Φ = BS square-on, 0 edge-on. 1 Wb = 1 T·m². Taught on the Magnetic Induction & Flux page.
The condition
A current flows only while the circuit is closed and the flux through it is changing. It stops when the flux stops changing. The change counts, not the size.
Three ways to change Φ
Change B: move a magnet, or switch, raise or lower a nearby current. Change S: slide a rod, squash a loop, move a loop across a field's edge. Change θ: turn the loop.
Faraday's ring
Closing the switch, opening it and moving the rheostat all change coil A's current, and so the flux through coil B. Holding the switch closed does not. March 2026 Q35: B.
Cutting field lines
A conductor in a closed circuit moving across field lines gives a current; moving along them, or along its own length, gives none. The flux through the closed loop is the safer test.
No current
A switch held closed; a magnet at rest in a coil; a loop moving inside a uniform field without turning; an open circuit.
EMF without current
A changing flux always induces an EMF (感应电动势). A current also needs a closed circuit, so an open coil has a voltage across its ends and no current.
Where it flows
Only round the closed loop whose flux changes. Stretches of rail that end in an open end carry none. January 2026 Q42: the flux through ebcf grows, bc carries current, ae does not: C.
Field edges
A loop moving through a region of field carries a current only while it crosses an edge, entering or leaving. Fully inside a uniform field, its flux is constant and there is no current.
