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UNIT 6About 12 min + practice

Electromagnetic Induction

Electricity & Magnetism track · College Board Unit 13. Build the model, derive its consequences, and test the assumptions.

What you’ll learn

  • Relate induced emf to changing magnetic flux.
  • Use Lenz’s law and energy accounting consistently.
  • Model RL transients and ideal LC oscillations.
01

Flux can change in more than one way

Magnetic flux through a loop is Φ_B=∫B·dA. For a flat loop in uniform B, Φ_B=BA cosθ, with θ measured from the area normal. Flux changes if field strength, enclosed area, or orientation changes. A moving loop entirely inside a uniform field can have constant flux and zero net induced emf.

Faraday’s law gives ε=−N dΦ_B/dt for N identical linked turns. The minus sign expresses opposition to change, not an automatically negative measured voltage. Choose a loop direction and the associated normal before interpreting signed flux.

ε=−N dΦ_B/dt
The angle used in magnetic flux

The area vector is normal to the loop. Changing angle, area, or field can change flux.

The angle used in magnetic fluxArea normalBθΦ = BA cos θθ is measured from the normal, not the plane
Original ScienceHub diagram · Schematic, not to scale.

PAUSE & TRY IT

A loop’s flux is large but constant. Is there an induced emf?

Reveal answer

No. Faraday’s law depends on the rate of change of flux.

02

Lenz’s law opposes the change, not necessarily the field

If outward flux increases, an induced current produces an inward field. If outward flux decreases, the induced field is outward to resist the decrease. Use the right-hand grip rule to turn that field direction into a current direction. The induced field need not cancel the original field completely.

A loop crossing a finite field region has different phases: entering changes overlap area, being fully inside can keep flux constant, and leaving reverses the flux change. Track the signed flux as a function of position and then time; this is more reliable than memorizing one current direction.

PAUSE & TRY IT

As an outward field decreases, what field does the induced current produce?

Reveal answer

Outward, opposing the decrease.

03

Motional emf transfers mechanical energy

For a conducting rod of length ℓ moving perpendicular to uniform B and itself at speed v, charge separation produces magnitude ε=Bℓv. With a completed circuit of resistance R, I=Bℓ. The current-carrying rod experiences magnetic drag F=B2ℓ.

To maintain constant speed, an external force supplies power Fv=I2R. The electrical energy is supplied by mechanical work. If the rod slows freely, kinetic energy decreases while thermal energy increases. Omitting the induced force would allow energy to appear without a source.

04

Inductors resist changes in current

A coil’s flux linkage is proportional to current in a linear model: NΦ=LI. Its induced emf is −L dI/dt. A current cannot jump instantaneously through an ideal inductor with finite voltage. Unlike a resistor, an ideal inductor does not continually dissipate power; it stores magnetic energy U=.

For a series RL circuit connected to a source, ε−IR−L dI/dt=0. Starting at zero, I=()(1−e−tR/L). The time constant is . Immediately after closing the switch, current is zero; long after, the ideal inductor has zero voltage drop while the resistor carries .

PAUSE & TRY IT

At fixed L, doubling R changes an RL time constant how?

Reveal answer

τ= halves.

05

A switch changes the differential equation

When the battery is removed and the inductor discharges through a resistor, I=I0e−tR/L. The inductor’s polarity reverses so that current continues in its original direction while decreasing. The resistor converts stored magnetic energy into thermal energy.

The relevant resistance is the entire resistance in the actual discharge path. A switch diagram must show that path; simply removing the source without a closed path can produce a large voltage rather than the assumed gentle exponential. Check both continuity at t=0 and the long-time state.

RL current growth, ε/R=2 A and L/R=0.5 s
RL current growth, ε/R=2 A and L/R=0.5 s00.511.5200.6251.251.8752.5 t (s)I (A)RL current growth, ε/R=…
Read figure values as text

RL current growth, ε/R=2 A and L/R=0.5 s: 0: 0; 0.041666666666666664: 0.15991117074135341; 0.08333333333333333: 0.30703655021877174; 0.125: 0.44239843385719024; 0.16666666666666666: 0.5669373788524215; 0.20833333333333334: 0.6815187395991125; 0.25: 0.7869386805747332; 0.2916666666666667: 0.8839297084599058; 0.3333333333333333: 0.973165761934816; 0.375: 1.0552668945179706; 0.4166666666666667: 1.1308035829858436; 0.4583333333333333: 1.2003006913103054; 0.5: 1.2642411176571153; 0.5416666666666666: 1.3230691497865155; 0.5833333333333334: 1.3771935521708047; 0.625: 1.4269904062796197; 0.6666666666666666: 1.4728057237685466; 0.7083333333333334: 1.5149578507287027; 0.75: 1.5537396797031404; 0.7916666666666666: 1.5894206848401815; 0.8333333333333334: 1.6222487943248765; 0.875: 1.6524521130991097; 0.9166666666666666: 1.6802405078406122; 0.9583333333333334: 1.7058070652140465; 1: 1.7293294335267746; 1.0416666666666667: 1.750971057111754; 1.0833333333333333: 1.7708823120146246; 1.125: 1.7892015508762713; 1.1666666666666667: 1.8060560642711898; 1.2083333333333333: 1.8215629651814798; 1.25: 1.8358300027522023; 1.2916666666666667: 1.8489563109824525; 1.3333333333333333: 1.861033097554397; 1.375: 1.8721442775865849; 1.4166666666666667: 1.8823670567151403; 1.4583333333333333: 1.8917724675543568; 1.5: 1.900425863264272; 1.5416666666666667: 1.908387371654757; 1.5833333333333333: 1.9157123129814473; 1.625: 1.922451584336556; 1.6666666666666667: 1.9286520133054952; 1.7083333333333333: 1.9343566833466737; 1.75: 1.939605233155363; 1.7916666666666667: 1.9444341320917515; 1.8333333333333333: 1.948876933586985; 1.875: 1.9529645082879818; 1.9166666666666667: 1.9567252585610138; 1.9583333333333333: 1.9601853158445326; 2: 1.9633687222225316; 2.0416666666666665: 1.966297597480105; 2.0833333333333335: 1.9689922928019814; 2.125: 1.9714715321820016; 2.1666666666666665: 1.973752542526118; 2.2083333333333335: 1.9758511733529343; 2.25: 1.9777820069235155; 2.2916666666666665: 1.9795584595657074; 2.3333333333333335: 1.9811928748970096; 2.375: 1.9826966095937588; 2.4166666666666665: 1.984080112302587; 2.4583333333333335: 1.9853529962424692; 2.5: 1.986524106001829

06

An ideal LC circuit exchanges two energy stores

For an ideal isolated LC circuit, L +=0. The charge oscillates with angular frequency ω=(LC); current is dq/dt and is one quarter-cycle out of phase. Total energy + remains constant when resistance is neglected.

At maximum capacitor charge, current and magnetic energy are zero. When capacitor charge is zero, current magnitude and magnetic energy are largest. A real circuit’s resistance damps the oscillation, so the ideal solution is a model with a specified approximation rather than a claim that energy losses never occur.

ω=(LC)
U=+

PAUSE & TRY IT

In an ideal LC circuit, when is current magnitude largest?

Reveal answer

When capacitor charge passes through zero and all stored energy is magnetic.

07

Compute magnetic flux with the correct area normal

Magnetic flux through a surface is the integral of B·dA. For a uniform field and flat loop it becomes BA cos θ, where θ is between B and the area normal. If the angle is given relative to the plane, convert it. A field parallel to the plane gives zero flux through that plane.

Flux can change because field magnitude, loop area or orientation changes. A conducting loop moving through a nonuniform region can experience changing flux even if the source field is steady in time. Conversely, a loop moving entirely within a uniform field at fixed area and orientation can have constant flux.

Flux is a signed quantity defined by the chosen normal. A consistent normal and loop orientation are essential when assigning an emf sign. The observable current direction must ultimately agree with the physical opposition described by Lenz’s law.

  1. Choose the loopDefine its area vector and positive circulation consistently.
  2. Find flux changeInclude changes in field, area, and orientation.
  3. Use Faraday–LenzInduced emf opposes the flux change; current also depends on the circuit.
08

Apply Faraday’s and Lenz’s laws in separate steps

Faraday’s law relates induced emf to the rate of change of linked flux, with the number of turns included for a coil. A large flux does not necessarily produce an emf if it is constant. A small flux changing rapidly can produce a substantial emf.

First determine whether the original flux in a chosen direction is increasing or decreasing. Then choose an induced field that opposes that change, and finally use the right-hand rule to find the current direction. The induced field does not always oppose the original field: it reinforces an original field that is decreasing.

Emf is not automatically current. Current additionally depends on a closed conducting path and its electrical properties. An open loop can have an induced emf without the same circulating current that would flow through a closed low-resistance loop.

PAUSE & TRY IT

A loop has a large, constant magnetic flux. Must it have an induced emf?

Reveal answer

No. Faraday’s law depends on the time rate of change of flux, not flux magnitude alone.

09

Connect motional emf to forces and energy

For a conducting rod moving appropriately through a uniform field, magnetic forces separate charges until an electric field develops. In the standard perpendicular geometry, emf magnitude is BLv. This expression requires the relevant perpendicular components and should not be applied unchanged to every orientation.

If the rod completes a circuit, the resulting current can experience a magnetic force opposing the driven motion. An external agent must supply work to maintain constant speed. In an ideal resistive circuit, mechanical power can equal the electrical heating rate, demonstrating energy conservation.

Lenz’s law prevents a self-amplifying process that would generate useful electrical energy without an input. When a magnet slows near a conducting region, kinetic energy can become electrical and thermal energy. Explain the transfer rather than treating the magnetic field as an unlimited fuel source.

10

Understand inductors and time-dependent current

A changing current produces a changing magnetic field and a self-induced emf. An ideal inductor opposes changes in current through its L di/dt relation; it does not simply oppose any current forever. In an ordinary ideal finite-voltage circuit, inductor current cannot jump instantaneously.

For a simple RL step response, sets the time constant. Initially, a previously unenergized inductor prevents an instantaneous current jump; at long time under DC, an ideal inductor’s voltage approaches zero while current can remain nonzero. This is complementary to the long-time behavior of an ideal capacitor.

Stored magnetic energy is ½LI2. When a source is removed, that energy can sustain current temporarily and be dissipated in a resistor. A switching analysis must use the initial inductor current, the new circuit topology and the final state. Confusing current continuity with voltage continuity swaps the roles of inductors and capacitors.

An inductor opposes a change in currentIllustrative model, not collected experimental data. For an ideal series RL circuit connected to a DC source, i/(V/R)=1−e^(−tR/L). The final current is limited by resistance.
An inductor opposes a change in current00.250.50.75101.252.53.755 Time / (L/R)Current / final currentRL current
Read figure values as text

RL current: 0: 0; 0.10416666666666667: 0.09892489427870943; 0.20833333333333334: 0.18806365384936508; 0.3125: 0.2683843710533582; 0.4166666666666667: 0.34075936979955623; 0.5208333333333334: 0.405974679446365; 0.625: 0.4647385714810097; 0.7291666666666666: 0.5176892517087218; 0.8333333333333334: 0.5654017914929218; 0.9375: 0.608394373323201; 1.0416666666666667: 0.6471339185411511; 1.1458333333333333: 0.6820411583440102; 1.25: 0.7134952031398099; 1.3541666666666667: 0.7418376598795473; 1.4583333333333333: 0.7673763420827073; 1.5625: 0.7903886128489022; 1.6666666666666667: 0.8111243971624382; 1.7708333333333333: 0.8298088962049714; 1.875: 0.8466450331550716; 1.9791666666666667: 0.8618156570373211; 2.0833333333333335: 0.875485528555877; 2.1875: 0.8878031094796562; 2.2916666666666665: 0.8989021750127812; 2.3958333333333335: 0.9089032666614493; 2.5: 0.9179150013761012; 2.6041666666666665: 0.9260352511868384; 2.7083333333333335: 0.9333522061435318; 2.8125: 0.939945332104692; 2.9166666666666665: 0.9458862337771784; 3.0208333333333335: 0.9512394323797938; 3.125: 0.9560630663765926; 3.2291666666666665: 0.9604095228902189; 3.3333333333333335: 0.9643260066527476; 3.4375: 0.9678550526731239; 3.5416666666666665: 0.9710349881890298; 3.6458333333333335: 0.9739003489202116; 3.75: 0.9764822541439909; 3.8541666666666665: 0.9788087446664702; 3.9583333333333335: 0.9809050873599727; 4.0625: 0.9827940495741486; 4.166666666666667: 0.9844961464009907; 4.270833333333333: 0.9860298634791853; 4.375: 0.987411857757566; 4.479166666666667: 0.9886571383980641; 4.583333333333333: 0.9897792297828537; 4.6875: 0.9907903183960318; 4.791666666666667: 0.991701385175045; 4.895833333333333: 0.9925223247692634; 5: 0.9932620530009145

11

Derive motional emf and check the energy source

A conducting rod of length ℓ moving at speed v perpendicular to a uniform B experiences charge separation through magnetic force. In the standard perpendicular geometry, the motional emf magnitude is Bℓv. If it forms a closed circuit of resistance R, current magnitude is Bℓ under the ideal assumptions.

The induced current produces a magnetic force opposing the motion that changes flux. An external agent maintaining constant speed must supply mechanical power. Using F=IℓB gives Fv=I(Bℓv)=I2R, matching resistive heating. This energy check connects Lenz’s law with conservation rather than treating its direction rule as arbitrary.

An open rod can have an emf without a sustained closed-loop current. A changing magnetic flux can also induce emf in a stationary loop. Do not assume that all induction requires a moving conductor, or that the presence of emf alone guarantees a particular current without knowing the circuit.

PAUSE & TRY IT

Does Lenz’s law oppose the magnetic field or its flux change?

Reveal answer

It opposes the change in flux. If an existing flux is decreasing, the induced effect can reinforce its original direction.

12

Connect inductance, stored energy, and switch behavior

An inductor’s self-induced emf opposes change in current: the sign follows −L di/dt under the usual convention. Its stored magnetic energy is . A large current is not necessarily a large induced emf; an unchanging current has di/dt=0 in the ideal inductor.

For a series RL circuit connected to a constant source, Kirchhoff’s rule gives V−IR−L di/dt=0. Starting from zero current, I=()(1−e−tR/L); the time constant is . At long times, an ideal inductor has zero voltage drop while the resistor limits current.

When the source is removed but a resistive discharge path remains, current decays exponentially and stored magnetic energy becomes thermal energy. Current cannot jump instantaneously in the ordinary finite-voltage ideal model. Opening a path abruptly can produce a large voltage, showing why a complete switching circuit matters to the physical prediction.

FROM IDEA TO APPLICATION

Worked examples

EXAMPLE 1

A changing field through a fixed loop

A 20-turn loop has area 0.030 m2 with its normal parallel to B(t)=0.40t2 T for t in seconds. Find induced emf magnitude at t=2 s.

Reveal worked solution
  1. dB/dt=0.80t, so it is 1.60 at 2 s.
  2. |ε|=NA|dB/dt|.
Result & interpretation

|ε|=0.960 V. The induced field opposes the increasing applied flux.

EXAMPLE 2

An LC energy transfer

A 5 μF capacitor charged to 20 V is connected to a 0.20 H ideal inductor. Find maximum current.

Reveal worked solution
  1. Initial energy is .
  2. At maximum current, LI_ equals that energy.
  3. I_max=V√().
Result & interpretation

I_max=0.100 A.

EXAMPLE 3

Use a flux slope to find emf

A 50-turn coil has flux per turn decreasing uniformly from 0.012 Wb to 0.004 Wb in 0.20 s. Find emf magnitude.

Reveal worked solution
  1. Flux change per turn is −0.008 Wb.
  2. Rate is =−0.040 Wb/s.
  3. Magnitude is N||=50(0.040)=2.0 V.
Result & interpretation

The induced emf magnitude is 2.0 V. Current direction requires the flux orientation and opposes its decrease.

EXAMPLE 4

Energy check for a moving rod

A 0.5 m rod moves at 4 through 0.2 T in a circuit of resistance 2 Ω, all perpendicular as required.

Reveal worked solution
  1. Emf=Bℓv=0.4 V and I==0.2 A.
  2. Magnetic drag magnitude IℓB=0.02 N.
  3. Mechanical power Fv=0.08 W; I2R=0.08 W as well.
Result & interpretation

The external work supplies the resistor’s heating in the ideal steady-speed model.

EXAMPLE 5

An inductor at steady current

An ideal 0.3 H inductor carries a constant 2 A.

Reveal worked solution
  1. di/dt=0, so induced voltage is zero.
  2. Stored energy is =0.3=0.6 J.
Result & interpretation

Zero induced voltage does not mean zero stored magnetic energy.

MAKE THE DISTINCTION

Common mistakes, clearer reasoning

The trapInduced current always opposes the applied magnetic field.

The better explanationIt opposes the change in flux, so its field can reinforce a decreasing applied field.

The trapAn inductor prevents current from ever flowing.

The better explanationIt resists changes; in steady DC an ideal inductor can carry constant current with zero voltage drop.

RETRIEVE BEFORE YOU REVEAL

Practice checkpoints

Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.

1. A loop’s flux is large but constant. Is there an induced emf?

Reveal answer

No. Faraday’s law depends on the rate of change of flux.

2. As an outward field decreases, what field does the induced current produce?

Reveal answer

Outward, opposing the decrease.

3. At fixed L, doubling R changes an RL time constant how?

Reveal answer

τ= halves.

4. In an ideal LC circuit, when is current magnitude largest?

Reveal answer

When capacitor charge passes through zero and all stored energy is magnetic.

5. A loop has a large, constant magnetic flux. Must it have an induced emf?

Reveal answer

No. Faraday’s law depends on the time rate of change of flux, not flux magnitude alone.

6. Does Lenz’s law oppose the magnetic field or its flux change?

Reveal answer

It opposes the change in flux. If an existing flux is decreasing, the induced effect can reinforce its original direction.

Key language

Magnetic flux
Surface integral ∫B·dA.
Faraday’s law
Induced emf is the negative time derivative of flux linkage.
Lenz’s law
Induction opposes the change in magnetic flux.
Inductance
Flux linkage per current in a linear system.
RL time constant
Characteristic transient time L/R.
LC oscillation
Periodic exchange of electric and magnetic stored energy.
Connect it to the course

Connect the equations to the physical assumptions. Explain directions, signs, limiting cases, and units before accepting a numerical result.

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Written for ScienceHub · Original instructional material. Course framework reference ↗. These notes are independently authored and are not College Board materials. External photographs retain their credited licenses.

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