Inductance - AP Physics C Electricity and Magnetism

SECTION 1

Historical Context & Motivation

The story of inductance begins with a deceptively simple observation: a coil of wire resists sudden changes in the current flowing through it, much like a massive flywheel resists changes in its rotational speed. This property, rooted in the deep connection between electricity and magnetism, was not immediately apparent to early experimenters. It took decades of discovery—from Faraday's first experiments on electromagnetic induction to Neumann's formal mathematical treatment—before physicists understood that a changing current in a conductor induces an electromotive force (EMF) that opposes the change, and that this opposition can be quantified by a single circuit parameter we now call inductance.

1831 • Faraday's Discovery of Electromagnetic Induction

Michael Faraday demonstrated that a changing magnetic flux through a loop of wire produces an EMF. His experiments with iron ring transformers and moving magnets laid the experimental foundation for the concept of inductance.

1834 • Lenz's Law

Heinrich Lenz formulated the rule that the direction of an induced current is always such as to oppose the change in flux that produced it. This principle explains why inductors resist rapid current changes.

1845 • Neumann's Mathematical Formulation

Franz Neumann introduced the concept of mutual inductance and derived formal expressions for the coefficients of induction, providing the mathematical framework still used today.

1886 • Heaviside and the Henry

Oliver Heaviside reformulated Maxwell's equations into modern vector form and helped standardize circuit analysis. The SI unit of inductance, the henry (H), was later named after Joseph Henry, who independently discovered self-induction in 1832.

The central question that inductance answers is this: when the current through a coil changes, how large is the back-EMF that the coil generates in response? Understanding inductance is essential not only for analyzing RL circuits on the AP exam but also for grasping how transformers, motors, and modern power electronics operate. The concept naturally extends to mutual inductance, which describes how a changing current in one coil can induce an EMF in a neighboring coil—the principle behind every transformer and wireless charger.

SECTION 2

Core Principles & Definitions

Inductance is fundamentally a geometric and material property of a conductor arrangement. It quantifies how effectively a given configuration of conductors converts current into magnetic flux linkage. When that current changes in time, the resulting change in flux linkage produces an EMF according to Faraday's law. Two distinct but related quantities arise: self-inductance (often simply called inductance), which characterizes a single coil's response to changes in its own current, and mutual inductance, which characterizes the coupling between two distinct coils.

Self-Inductance (L)

The ratio of total magnetic flux linkage through a coil to the current producing that flux: L = NΦ/I. A larger L means more flux per ampere, and hence a stronger back-EMF for a given rate of current change.

Mutual Inductance (M)

The ratio of flux linkage in coil 2 due to current in coil 1: M = N₂Φ₂₁/I₁. By Neumann's reciprocity theorem, M₁₂ = M₂₁, so the coupling is symmetric regardless of which coil carries the current.

Back-EMF (Faraday–Lenz)

A changing current through an inductor produces an EMF: ε = −L(dI/dt). The negative sign (Lenz's law) indicates the EMF opposes the change in current, acting as electromagnetic inertia.

Energy Storage

An inductor carrying current I stores energy U = ½LI² in its magnetic field. This is analogous to a capacitor storing energy ½CV² in its electric field, and it is recoverable when the current decreases.

RL Time Constant (τ = L/R)

In a series RL circuit, the inductance and resistance together set the exponential time scale for current growth or decay: τ = L/R. After one time constant, the current reaches about 63% of its final value.

✦

KEY TAKEAWAY

Think of an inductor as the electrical analog of a massive spinning flywheel. Just as a flywheel resists sudden changes in rotational speed due to its moment of inertia, an inductor resists sudden changes in current due to its inductance. The larger the inductance (or the larger the flywheel), the more energy is stored and the more vigorously the system opposes change. This electromagnetic inertia is the defining characteristic of every inductor.

SECTION 3

Visual Explanation — The Solenoid Inductor

The ideal solenoid is the canonical example of a self-inducting device. When a steady current flows through a tightly wound solenoid of N turns, length ℓ, and cross-sectional area A, the interior magnetic field is nearly uniform with magnitude B = μ₀nI, where n = N/ℓ is the turn density. Because the total flux linkage through all N turns is NΦ = N(BA) = μ₀n²ℓAI, the self-inductance is L = μ₀n²ℓA = μ₀N²A/ℓ. The diagram below illustrates this geometry and the resulting magnetic field.

A solenoid of N turns, length ℓ, and cross-sectional area A carrying current I. The cyan arrows represent the nearly uniform interior magnetic field B = μ₀nI. The gold path indicates the current direction through the external circuit. The resulting self-inductance depends on geometry and the permeability of the core material.

Notice that the inductance L = μ₀N²A/ℓ scales as the square of the number of turns. Doubling the number of turns quadruples the inductance, because both the field strength (proportional to N) and the number of turns through which that field links (also proportional to N) increase simultaneously. This N² dependence is a hallmark of inductive devices and explains why practical inductors often use many tightly wound turns. Inserting a ferromagnetic core with relative permeability κ_m replaces μ₀ with κ_mμ₀, dramatically boosting the inductance without changing the geometry.

SECTION 4

Mathematical Framework

The mathematical description of inductance begins with Faraday's law and unfolds into a set of equations that govern transient behavior in circuits containing inductors. We present the key results here with their derivations, as the AP Physics C exam expects fluency with both the formulas and the calculus behind them.

SELF-INDUCTANCE DEFINITION

L = NΦ_B / I

L = self-inductance (henrys, H), N = number of turns, Φ_B = magnetic flux through one turn (Wb), I = current (A). This is a definition valid when the medium is linear so that Φ_B is proportional to I.

BACK-EMF OF AN INDUCTOR

ε = −L (dI/dt)

This follows from Faraday's law: ε = −d(NΦ_B)/dt. Since NΦ_B = LI for constant L, we obtain ε = −L(dI/dt). The negative sign embodies Lenz's law: the induced EMF opposes the current change.

ENERGY STORED IN AN INDUCTOR

U = ½LI²

Derived by integrating the power delivered to the inductor: P = εI = LI(dI/dt). Integrating from 0 to I gives U = ∫₀ᴵ LI dI = ½LI². Equivalently, this energy resides in the magnetic field; for a solenoid U = (B²/2μ₀)(volume).

RL CIRCUIT — CURRENT GROWTH

I(t) = (ε/R)(1 − e^(−t/τ)), τ = L/R

When a battery of EMF ε is connected to a series RL circuit at t = 0, Kirchhoff's loop rule gives ε − IR − L(dI/dt) = 0. Solving this first-order linear ODE with initial condition I(0) = 0 yields the exponential growth shown. The time constant τ = L/R has units of seconds.

💡

AP Exam Tip

The AP Physics C exam frequently asks you to apply Kirchhoff's voltage law to RL circuits and solve the resulting differential equation. Remember: the voltage across an inductor is L(dI/dt) with a sign determined by your chosen current direction. Practice writing the loop equation, separating variables, and integrating to obtain the exponential solution.

SECTION 5

RL Circuit Behavior — Growth and Decay

Understanding how current evolves in an RL circuit is one of the most heavily tested inductance topics on the AP Physics C exam. Two canonical scenarios appear repeatedly: current growth (when a battery is first connected) and current decay (when the battery is disconnected and the inductor drives current through the resistor). In both cases, the time constant τ = L/R governs how quickly the system approaches its steady state. The graph below shows both curves, with key time-constant markers annotated.

Current vs. time for a series RL circuit. The green curve shows current growth when a battery is connected: I rises from 0 toward ε/R exponentially, reaching 63% of the final value at t = τ. The red curve shows current decay after the battery is removed: I falls from ε/R toward 0, retaining 37% at t = τ. Both curves share the same time constant τ = L/R.

For the growth phase, at t = τ = L/R, the current has reached (1 − e⁻¹) ≈ 63.2% of its steady-state value ε/R. After about 5τ, the current is within 1% of ε/R and is considered to be at steady state. During decay, the current drops to e⁻¹ ≈ 36.8% after one time constant. The symmetry between growth and decay is a direct consequence of the linearity of the differential equation. It is also important to recognize that the voltage across the inductor is maximal at t = 0 (during growth, V_L = ε; during decay, V_L = −ε/R × R = −ε) and decays exponentially as well.

Time (multiples of τ)	Growth: I / (ε/R)	Decay: I / (ε/R)
0	0	1.000
1τ	0.632	0.368
2τ	0.865	0.135
3τ	0.950	0.050
5τ	0.993	0.007

Fraction of steady-state current at key multiples of the RL time constant

SECTION 6

Worked Example — RL Circuit Analysis

A 12 V battery is connected in series with a 4.0 Ω resistor and a 20 mH inductor at t = 0. Find (a) the time constant, (b) the current at t = 5.0 ms, (c) the energy stored in the inductor at t = 5.0 ms, and (d) the voltage across the inductor at t = 5.0 ms.

RL Circuit Growth Analysis

1
Step 1 — Identify Given Values

We have ε = 12 V, R = 4.0 Ω, L = 20 mH = 0.020 H, and we need to evaluate quantities at t = 5.0 ms = 0.0050 s.
2
Step 2 — Calculate the Time Constant

The RL time constant is τ = L/R = 0.020 H / 4.0 Ω = 0.0050 s = 5.0 ms. Notice that our evaluation time t = 5.0 ms corresponds to exactly one time constant.

τ = 5.0 ms
3
Step 3 — Find the Current at t = τ

Using the growth equation: I(t) = (ε/R)(1 − e^−t/τ). At t = τ: I = (12/4.0)(1 − e⁻¹) = 3.0 × (1 − 0.368) = 3.0 × 0.632 = 1.90 A.

I = 1.90 A
4
Step 4 — Calculate Energy Stored

The energy stored in the magnetic field of the inductor is U = ½LI² = ½(0.020)(1.90)² = ½(0.020)(3.61) = 0.036 J = 36 mJ.

U = 36 mJ
5
Step 5 — Find Voltage Across the Inductor

The voltage across the inductor is V_L = L(dI/dt). Using the derivative of the growth equation: dI/dt = (ε/L)e^−t/τ. Therefore V_L = εe^−t/τ = 12 × e⁻¹ = 12 × 0.368 = 4.4 V. We can verify: V_R = IR = 1.90 × 4.0 = 7.6 V, and V_R + V_L = 7.6 + 4.4 = 12 V = ε. ✓

$V_L$ = 4.4 V

SECTION 7

Inductors vs. Capacitors — A Duality

Inductors and capacitors are dual circuit elements: both store energy, both produce transient exponential behavior in combination with resistors, but they do so through fundamentally different mechanisms. Recognizing the structural parallels between L and C—and where those parallels break—is an efficient study strategy for the AP exam, where both RC and RL circuits appear. The table below highlights the key correspondences.

Property	Inductor (L)	Capacitor (C)
Energy stored	U = ½LI²	U = ½CV²
Field type	Magnetic (B)	Electric (E)
Opposes changes in	Current (dI/dt)	Voltage (dV/dt)
Voltage–current relation	V = L(dI/dt)	I = C(dV/dt)
Time constant with R	τ = L/R	τ = RC
DC steady state	Short circuit (wire)	Open circuit (gap)
Instantaneous constraint	Current cannot jump	Voltage cannot jump

Inductor–capacitor duality comparison

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KEY TAKEAWAY

The inductor–capacitor duality is not just a mnemonic convenience; it reflects a deep mathematical symmetry in Maxwell's equations. In every RC equation, swapping V ↔ I, C ↔ L, and R ↔ 1/R produces the corresponding RL equation. On the exam, if you can solve one type of circuit, you can solve the other by applying these substitutions. The key physical constraint to remember: current through an inductor cannot change instantaneously, just as voltage across a capacitor cannot change instantaneously.

SECTION 8

Connection to Advanced Theory — Mutual Inductance & LC Oscillations

Self-inductance is only the beginning of the story. When two inductors are placed near each other, the changing current in one coil generates a changing flux through the other, inducing an EMF governed by the mutual inductance M. This principle underlies transformers, which step voltages up or down by adjusting the turns ratio N₂/N₁. Furthermore, combining an inductor and a capacitor in a circuit creates an LC oscillator, in which energy sloshes back and forth between the magnetic field of the inductor and the electric field of the capacitor at angular frequency ω = 1/√(LC). This oscillatory behavior is the electromagnetic analog of a frictionless spring-mass system and is foundational to radio tuning, signal processing, and quantum electrodynamics.

Concept	AP Physics C Level	Advanced / Engineering Level
Self-inductance	L = μ₀N²A/ℓ for solenoid; ε = −L(dI/dt)	Neumann formula: L = (μ₀/4π) ∮∮ (dl₁ · dl₂)/\|r₁ − r₂\|
Mutual inductance	M = N₂Φ₂₁/I₁; ε₂ = −M(dI₁/dt)	Coupling coefficient k = M/√(L₁L₂); transformer theory
Energy	U = ½LI²	u = B²/(2μ₀) integrated over all space; includes mutual energy terms
Oscillations	LC circuit: ω = 1/√(LC)	RLC damped oscillations; quality factor Q = ωL/R

Inductance concepts at AP vs. advanced levels

While the AP exam does not require the Neumann integral formula or full RLC analysis, it does test mutual inductance qualitatively and expects familiarity with LC oscillation frequency. Understanding how self-inductance extends to these richer scenarios provides the conceptual scaffolding for upper-division electrodynamics and circuit theory courses.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A steady current of 3 A flows through a 50 mH inductor. What is the voltage across the inductor?

A. 150 V
B. 0.15 V
C. 0 V
D. The answer depends on the resistance of the inductor wire

Show answer (C)

The correct answer is C. The voltage across an ideal inductor is V = L(dI/dt). For a steady (constant) current, dI/dt = 0, so V = 0. Choice A (150 V) incorrectly multiplies L × I, treating inductance like resistance—but V = LI is not a valid equation. Choice B (0.15 V) similarly confuses the relationship. Choice D is incorrect because for an ideal inductor (zero wire resistance), the voltage is exactly zero regardless of wire properties; the question specifies an inductor, not a real component with parasitic resistance.

Question 2: BASIC CALCULATION

A solenoid has 500 turns, a length of 0.25 m, and a cross-sectional area of 4.0 × 10⁻⁴ m². What is its self-inductance? (μ₀ = 4π × 10⁻⁷ T·m/A)

A. 5.0 × 10⁻⁴ H
B. 2.5 × 10⁻⁴ H
C. 1.3 × 10⁻³ H
D. 5.0 × 10⁻⁷ H

Show answer (A)

The correct answer is A. Using L = μ₀N²A/ℓ: L = (4π × 10⁻⁷)(500)²(4.0 × 10⁻⁴)/(0.25) = (4π × 10⁻⁷)(2.5 × 10⁵)(4.0 × 10⁻⁴)/(0.25) = (4π × 10⁻⁷)(4.0 × 10²) = 1.6π × 10⁻⁴ ≈ 5.0 × 10⁻⁴ H. Choice B (2.5 × 10⁻⁴ H) likely results from using N instead of N² in the formula. Choice C (1.3 × 10⁻³ H) could arise from an arithmetic error with the powers of ten. Choice D (5.0 × 10⁻⁷ H) reflects omitting the N² factor altogether.

Question 3: INTERMEDIATE

In a series RL circuit with L = 0.40 H and R = 8.0 Ω, connected to a 24 V battery at t = 0, how long does it take for the current to reach 2.0 A?

A. 0.035 s
B. 0.050 s
C. 0.055 s
D. 0.10 s

Show answer (C)

The correct answer is C. The steady-state current is I₀ = ε/R = 24/8.0 = 3.0 A. The time constant is τ = L/R = 0.40/8.0 = 0.050 s. Using I(t) = I₀(1 −

e^{−t/τ}

), set 2.0 = 3.0(1 −

e^{−t/0.050}

). Solving: 2/3 = 1 −

e^{−t/0.050}

→

e^{−t/0.050}

= 1/3 → −t/0.050 = ln(1/3) = −1.099 → t = 0.050 × 1.099 ≈ 0.055 s. Choice A (0.035 s) results from using the wrong fraction. Choice B (0.050 s) is simply one time constant, which gives 63.2% of 3.0 A ≈ 1.9 A, not 2.0 A. Choice D (0.10 s = 2τ) gives about 2.6 A, overshooting the target.

Question 4: APPLIED

5 pts

A series RL circuit consists of a battery with EMF ε, a resistor R, an inductor L, and a switch. The switch is closed at t = 0. (a) Starting from Kirchhoff's voltage law, derive the expression for the current I(t) as a function of time. Show all steps of solving the differential equation. (b) Determine the rate at which energy is being stored in the inductor at t = L/R. (c) At t = L/R, what fraction of the power delivered by the battery is dissipated in the resistor?

Show answer

Rubric (5 points): Point 1 — (a) Correctly writing Kirchhoff's loop equation: ε − IR − L(dI/dt) = 0. Point 2 — (a) Separating variables and integrating: dI/(ε/L − RI/L) = dt → −(L/R)ln(ε − IR) evaluated with limits gives I(t) = (ε/R)(1 −

e^{−Rt/L}

). Point 3 — (b) The rate of energy storage is dU/dt = d(½LI²)/dt = LI(dI/dt). At t = τ = L/R: I = (ε/R)(1 − e⁻¹) and dI/dt = (ε/L)e⁻¹. So dU/dt = L × (ε/R)(1 − e⁻¹) × (ε/L)e⁻¹ = (ε²/R)e⁻¹(1 − e⁻¹) = (ε²/R)(0.368)(0.632) ≈ 0.232ε²/R. Point 4 — (c) At t = τ, P_battery = εI = ε(ε/R)(1 − e⁻¹) = (ε²/R)(0.632). P_resistor = I²R = (ε/R)²(1 − e⁻¹)²R = (ε²/R)(1 − e⁻¹)² = (ε²/R)(0.400). Point 5 — (c) Fraction dissipated =

P_R

/P_battery = (1 − e⁻¹)²/(1 − e⁻¹) = (1 − e⁻¹) ≈ 0.632 or 63.2%. This result makes physical sense: the fraction of battery power going to the resistor equals the fraction (1 −

e^{−t/τ}

), which increases toward 1 as t → ∞.

Question 5: CRITICAL THINKING

4 pts

Two coaxial solenoids share the same axis. Solenoid 1 (inner) has N₁ = 200 turns, radius r₁ = 2.0 cm, and length ℓ = 0.30 m. Solenoid 2 (outer) has N₂ = 800 turns, radius r₂ = 5.0 cm, and the same length ℓ = 0.30 m. (a) Derive an expression for the mutual inductance M between the two solenoids. (b) If the current in solenoid 1 changes at a rate of 50 A/s, find the magnitude of the EMF induced in solenoid 2. (c) Explain why M depends on the cross-sectional area of the inner solenoid rather than the outer one.

Show answer

Rubric (4 points): Point 1 — (a) The field inside solenoid 1 is B₁ = μ₀(N₁/ℓ)I₁. This field passes through each turn of solenoid 2 only over the area of solenoid 1 (A₁ = πr₁²), because B₁ ≈ 0 outside the inner solenoid. Point 2 — (a) The flux through solenoid 2 is Φ₂₁ = B₁ × A₁ = μ₀(N₁/ℓ)I₁ × πr₁². The total flux linkage is N₂Φ₂₁. Thus M = N₂Φ₂₁/I₁ = μ₀N₁N₂πr₁²/ℓ. Substituting: M = (4π × 10⁻⁷)(200)(800)(π)(0.020)²/(0.30) = (4π × 10⁻⁷)(1.6 × 10⁵)(1.257 × 10⁻³)/(0.30) ≈ 8.4 × 10⁻⁴ H ≈ 0.84 mH. Point 3 — (b) |ε₂| = M|dI₁/dt| = (8.4 × 10⁻⁴ H)(50 A/s) = 0.042 V = 42 mV. Point 4 — (c) The magnetic field produced by the inner solenoid is confined to its interior (radius r₁). Outside r₁ but inside r₂, the field is essentially zero for an ideal solenoid. Therefore, only the cross-sectional area of the inner solenoid (A₁ = πr₁²) determines how much flux links with each turn of the outer solenoid. The area of the outer solenoid is irrelevant because the additional area between r₁ and r₂ intercepts no flux from solenoid 1.

SUMMARY

Inductance — Key Concepts at a Glance

Summary

Self-inductance (L) quantifies a coil's ability to oppose changes in its own current through a back-EMF ε = −L(dI/dt). For a solenoid with N turns, length ℓ, and area A, L = μ₀N²A/ℓ. The inductor stores energy U = ½LI² in its magnetic field. In a series RL circuit, the time constant τ = L/R governs exponential current growth and decay, with the current reaching 63% of its final value after one time constant.

Mutual inductance (M) extends the concept to coupled coils, with ε₂ = −M(dI₁/dt) and the reciprocity relation M₁₂ = M₂₁. The inductor–capacitor duality (L ↔ C, I ↔ V) provides a powerful framework for translating between RL and RC circuit analysis. Remember: current through an inductor cannot change instantaneously, energy is stored in the magnetic field, and the fundamental cause of inductance is Faraday's law of electromagnetic induction.