Electromagnetic Induction
Changing magnetic flux makes emf.
Faraday's & Lenz's Laws
Changing flux, induced emf, and the direction it takes.
Currents make magnetic fields. Faraday asked the reverse question, and found that a changing magnetic field makes a current. This discovery powers every generator on Earth.
Faraday’s experiments
Put a bar magnet on the axis of a wire loop connected to a galvanometer. Held still, nothing happens. Push the magnet toward the loop and the needle swings; pull it away and it swings the other way. The current lasts only while the magnet moves.
Faraday found the rule behind this and many similar experiments:
Φ is the magnetic flux, measured in webers (1 Wb = 1 T m²). It can change because B changes, because the loop’s area changes, or because the angle between them changes. The emf is the induced emf; in a loop of resistance R it drives an induced current:
A 200-turn coil of area 50 cm² lies perpendicular to a field that rises steadily from 0.10 T to 0.40 T in 0.15 s. Find the induced emf.
Which way does the current go?
Using the sign in Faraday’s law:
- Put an arrow on the loop to choose a positive sense for current (any choice).
- Curl your right-hand fingers along the arrow: the thumb gives the positive normal.
- Find Φ with that normal. If Φ is increasing, ℰ is negative: the current flows against your arrow. If Φ is decreasing, it flows along it.
Lenz’s law
A quicker way. The induced current makes its own magnetic field, which can strengthen or weaken the original flux. Nature always chooses:
- An N pole pushed toward a loop raises the flux; the induced current makes a field pointing back at the magnet, so the near face of the loop acts as an N pole and repels it.
- A sliding wire that shrinks a circuit in a field into the page lowers the flux; the induced current makes field into the page, i.e. it runs clockwise.
- A loop beside a straight wire whose current is increasing: the flux through the loop grows, so the induced current circulates to make field opposite to the wire’s.
A circular loop lies perpendicular to a uniform 0.050 T field. Its radius shrinks at 2.0 mm/s. Find the emf when the radius is 5.0 cm.
A square frame is made from 80 cm of copper wire of diameter 1.0 mm (ρ = 1.7 × 10⁻⁸ Ω m) and placed perpendicular to a field changing at 0.050 T/s. Find the induced current.
- 1ℰ = −dΦ/dt (38.1), Φ = ∫B⃗·dS⃗ in webers; i = ℰ/R (38.2); N turns multiply the emf.
- 2Flux can change through B, area or orientation.
- 3Lenz's law: the induced current opposes the change causing it (energy conservation).
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Videos, simulations & further reading (9)
Motional EMF
A moving conductor becomes a battery.
Where does an induced emf actually come from? When a conductor moves through a field, the answer is simply the magnetic force on its moving electrons.
Two ways to change the flux
Flux can change because (a) the field is steady but the loop (or part of it) moves, (b) the loop is still but the field changes, or both. The mechanism is different in the two cases. This lesson covers (a), motional emf.
A rod moving through a field
A rod PQ of length l moves at velocity v⃗ through a field B⃗, with rod, velocity and field all mutually perpendicular. Its free electrons share the velocity v⃗, so each feels qv⃗ × B⃗ along the rod. Electrons pile up at one end (P) and leave positive charge at the other (Q). The resulting electric field E inside the rod pushes back. Charge stops building up when the two balance: qE = qvB, i.e. E = vB.
Join P and Q by a resistor and current flows. The magnetic force keeps driving electrons back up the rod, maintaining the potential difference. With external resistance R and rod resistance r: i = vBl/(R + r).
Faraday’s law agrees: if x is the length of the circuit in the field, Φ = Blx and |dΦ/dt| = Bl dx/dt = vBl.
Forces and energy
Once current i flows in the rod, the field pushes on it with ilB, against its motion (Lenz again). To keep it moving at constant v something must push with force F = ilB = B²l²v/R. That force delivers power Fv = B²l²v²/R, and the circuit produces heat i²R = (vBl/R)²R = B²l²v²/R. Exactly equal: your mechanical work becomes the heat in the circuit.
A 25 cm rod slides at 4.0 m/s across rails in a 0.60 T field; the total resistance of the circuit is 2.0 Ω. Find the emf, current, the force needed to keep it moving and the power supplied.
A rectangular loop of total resistance R, with its side MN (length l) still inside a field B and the rest outside, is pulled out at constant speed v. Find the current, the magnetic force, the pulling force and the powers.
- 1Motional emf comes from the magnetic force on the conductor's moving free electrons: ℰ = vBl (38.3).
- 2The rod acts as a battery; i = vBl/(R + r).
- 3Induced current makes a retarding force ilB; mechanical power Fv equals the heat i²R.
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Induced Electric Field & Eddy Currents
A changing B makes a curly E.
If the loop isn’t moving, the magnetic force can’t start its electrons going. Something else must: an electric field produced by the changing magnetic field itself.
The induced electric field
Hold a loop still in a steady field, then let the field change. A current starts. But the electrons were at rest, and a magnetic field cannot push charges at rest, so it must be an electric field that sets them moving. It is not made by charges (Coulomb’s law has nothing to say about it); it is made by the changing magnetic field.
- It is non-electrostatic and non-conservative: no potential can be defined for it.
- Its field lines are closed loops, with no beginning or end.
- It exists whenever B changes, loop or no loop. A conducting loop merely lets its free electrons respond, giving a current.
The work per unit charge round a loop is the emf, so Faraday’s law becomes a statement about the field:
A uniform field fills a long cylindrical region of radius 5.0 cm and changes at 0.20 T/s. Find the induced electric field 3.0 cm and 10 cm from the axis.
Eddy currents
A solid metal plate moving into a field has no single wire to guide the current, but every closed path drawn on it has a changing flux. Currents swirl round many such paths at once: eddy currents.
They heat the plate, and that energy comes out of its kinetic energy: the plate slows down. This is electromagnetic damping. Cutting slots in the plate breaks up the current paths and greatly reduces it.
- 1A changing B creates an induced electric field: ∮E⃗·dl⃗ = −dΦ/dt (38.4).
- 2It is non-conservative, has closed field lines, and exists without any conductor.
- 3Eddy currents in bulk conductors cause heating and electromagnetic damping; slots or laminations reduce them.
Check yourself
Self-Induction
Coils resist changes in their own current.
A coil carrying a current threads its own magnetic flux. Change the current and the coil fights back with an emf of its own: self-induction.
Self-inductance
The field of a current, and so the flux through the loop carrying it, is proportional to the current:
The unit is the weber per ampere, equivalently volt-second per ampere, named the henry (H). For a coil of N turns, add the emf of every turn: ℰ = −N d(∫B⃗·dS⃗)/dt. By Lenz’s law the self-induced emf opposes the change: it pushes against a rising current and tries to keep a falling one going.
An average emf of 0.30 V appears in a coil while its current is reversed from 3.0 A one way to 3.0 A the other way in 0.10 s. Find L.
Inductance of a long solenoid
A solenoid of radius r with n turns per length carries i: B = μ₀ni inside, so each turn links μ₀niπr². A length l has nl turns, giving a total emf −(nl)(μ₀nπr²) di/dt. Comparing with −L di/dt:
Find the self-inductance of a solenoid 30 cm long and 2.0 cm in radius with 2000 turns per metre.
A coil of thick wire has negligible resistance but appreciable inductance: an ideal inductor, drawn as a row of loops.
Inductance resists sudden change
A 12 V cell drives current through an inductor and a rheostat. At one instant the rheostat reads 8 Ω and is being increased. Is the current more or less than 1.5 A at that instant?
- 1Φ = Li (38.5); ℰ = −L di/dt (38.6); unit henry = Wb/A = V s/A.
- 2Long solenoid: L = μ₀n²πr²l (38.7).
- 3Self-induced emf opposes changes in current (Lenz).
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LR Circuits & Magnetic Energy
Inductors give circuits inertia and store energy.
An inductor gives a circuit inertia. Current can’t jump on or off; it rises and falls exponentially, with a time constant L/R, and the inductor stores energy in its magnetic field.
Growth of current
An inductor L, a resistor R and a cell ℰ are joined in series and the switch is closed at t = 0. As the current rises, the inductor’s emf −L di/dt opposes it. The loop law gives
L/R has the dimensions of time: the time constant. After one time constant the current is i₀(1 − 1/e) = 63 % of its final value. In principle the final value is reached only as t → ∞; in practice a few time constants is enough. A small τ means a steep rise.
A 50 mH inductor, a 25 Ω resistor and a 5.0 V cell are in series. Find the time constant, the final current, and the time to reach 90 % of it.
Decay of current
With a steady current i₀ flowing, the cell is replaced by a plain wire (a special switch closes the new path before opening the old one, so the circuit is never broken). Now the inductor’s emf is the only one: −L di/dt = Ri, so
In the circuit above, after a long time the cell is shorted out and removed. Find the current 5.0 ms later.
Energy stored in an inductor
Multiply the growth equation by i dt and integrate from 0 to t:
The work done by the cell equals the Joule heat in R plus a remaining term: the energy stored in the inductor’s magnetic field.
How much energy is stored in an 80 mH inductor carrying 1.5 A?
Energy density of a magnetic field
For a long solenoid (radius r, length l, n turns per length, no magnetic material), L = μ₀n²πr²l and B = μ₀ni, so U = ½Li² = (B²/2μ₀)πr²l. Dividing by the volume πr²l:
What is the energy density of a 0.50 T field in air?
- 1Growth: i = (ℰ/R)(1 − e^(−t/τ)) (38.8); decay: i = i₀e^(−t/τ) (38.11); τ = L/R.
- 2Energy stored: U = ½Li² (38.12).
- 3Magnetic energy density u = B²/2μ₀ (38.13).
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Mutual Induction & the Induction Coil
One circuit's changing current drives another.
A changing current in one circuit can induce an emf in a completely separate circuit nearby. That coupling is the heart of transformers, induction coils and wireless chargers.
Mutual inductance
Current i in one circuit produces a field whose flux through a second circuit is proportional to i:
For circuits with several turns, add the flux through every turn before writing Φ = Mi.
A solenoid of radius 1.5 cm and 15 turns/cm sits inside a longer solenoid with 8 turns/cm. Find their mutual inductance over a 40 cm length.
A long solenoid of radius 1.5 cm with 80 turns/cm is surrounded by a 50-turn coil of total resistance 10 Ω connected to a galvanometer. The solenoid current is reversed from 4.0 A one way to 4.0 A the other. How much charge passes through the galvanometer?
The induction coil
Ruhmkorff’s induction coil makes a very high emf from a low-voltage cell. A primary coil T is wound on a laminated soft-iron core; a secondary coil S of many turns is wound over it and ends in two rods G₁ and G₂ with an adjustable gap. The primary circuit has a cell and an automatic interrupter: a springy metal strip A carrying a soft-iron hammer H, touching a screw D. A capacitor C is connected across the interrupter.
How it works
- Make and break. Current through the primary magnetizes the core, which pulls the hammer and breaks the contact. The current stops, the core lets go, the strip springs back and the contact closes again, over and over.
- Slow rise, fast fall. On make, the primary’s self-inductance makes the current grow slowly. On break, it collapses very quickly.
- Secondary emf. The secondary emf −M di/dt is therefore small on make and very large on break (opposite in sign), enough to throw sparks across the gap: about 50 000 V from a 12 V cell, enough to run a discharge tube.
- The capacitor. On break, the primary’s own induced emf would spark across the interrupter, keeping current flowing, slowing its fall and pitting the contacts. The capacitor absorbs that surge instead, so the current drops faster; it then discharges back through the primary, swinging the current almost to −i. The change in flux nearly doubles and so does the secondary emf.
- 1Φ₂ = Mi₁ (38.14); ℰ₂ = −M di₁/dt (38.15); M is symmetric between the two circuits.
- 2Coaxial solenoids: M = μ₀n₁n₂πr₁²l.
- 3Induction coil: slow make, fast break; the break gives a huge secondary emf; the capacitor speeds the break and reverses the current.