Inductance

AP Physics C: E&M· AP Physics C: E&M CED — Electromagnetism· 14 min read

1. Core Definition of Inductance★★☆☆☆⏱ 2 min

Inductance is the inherent property of any current-carrying circuit that opposes changes in current, arising directly from Faraday’s law of induction. When current through a conductor changes, magnetic flux through the conductor (or a nearby conductor) also changes, inducing an emf that opposes the change in current per Lenz’s law. This effect is often called "electrical inertia": inductance resists changes to current, just like mass resists changes to velocity in mechanics.

📘 Definition

Inductance

$L$ (self-inductance), $M$ (mutual inductance)

Property of a current-carrying system that opposes changes in current, measured in henries (H), where $1 H = 1 Wb/A = 1 V \cdotp s/A$ .

Example:

A coiled wire has a larger inductance than a straight wire of the same length.

2. Self-Inductance★★★☆☆⏱ 3 min

Self-inductance occurs when a changing current in a coil or conductor induces an emf in the same conductor. The defining relation comes directly from Faraday’s law, with the proportionality constant equal to the self-inductance $L$ :

\varepsilon = -L \frac{dI}{dt}

By definition, $L$ is also given by the ratio of total flux linkage to current:

L = \frac{N\Phi_B}{I}

where $N$ is the number of turns in the coil and $Φ_{B}$ is the magnetic flux through one turn. For a long solenoid, the most common geometry on the exam, we can derive an explicit formula for $L$ : for a solenoid of length $l$ , $N$ total turns, cross-sectional area $A$ , the magnetic field inside is $B = μ_{0} n I = μ_{0} \frac{N I}{l}$ , so flux through one turn is $Φ_{B} = B A = μ_{0} \frac{N A I}{l}$ . Substituting into the definition of $L$ gives:

L = \frac{\mu_0 N^2 A}{l}

Intuition: Inductance increases with the square of the number of turns, because more turns give more flux linkage, and each turn contributes flux to every other turn.

📐 Worked Example

A 15 cm long solenoid has 500 turns and radius 2.0 cm. Find (1) its self-inductance, and (2) the magnitude of the induced emf when the current through the solenoid increases at 120 A/s.

1
Convert all units to SI:
2
$l = 0.15\ \text{m}, \ r = 0.02\ \text{m}, \ A = \pi r^2 \approx 1.257 \times 10^{-3}\ \text{m}^2$
3
Substitute into the solenoid inductance formula, using $μ_{0} = 4 π \times 1 0^{- 7} T \cdotp m/A$ :
4
$L = \frac{(4\pi \times 10^{-7})(500)^2(1.257 \times 10^{-3})}{0.15} \approx 1.3 \times 10^{-3}\ \text{H} = 1.3\ \text{mH}$
5
Calculate the magnitude of the induced emf:
6
$|\varepsilon| = L \left|\frac{dI}{dt}\right| = (1.3 \times 10^{-3}\ \text{H})(120\ \text{A/s}) \approx 0.16\ \text{V}$

Exam tip:

Always convert length units to meters before calculating inductance; small cm lengths will give a 100x incorrect result if you forget, which is a common distracter in MCQs.

3. Mutual Inductance★★★☆☆⏱ 3 min

Mutual inductance describes the effect where a changing current in one coil induces an emf in a second, nearby coil. This is the operating principle for transformers and wireless power transfer, both common AP exam topics. By definition, mutual inductance $M$ between two coils is:

M = \frac{N_2 \Phi_{12}}{I_1} = \frac{N_1 \Phi_{21}}{I_2}

where $Φ_{12}$ is the flux through one turn of coil 2 caused by current $I_{1}$ in coil 1. A key property is that $M_{12} = M_{21} = M$ , it is symmetric regardless of which coil carries the current. The induced emf in coil 2 is:

\varepsilon_2 = -M \frac{dI_1}{dt}

$M$ depends strongly on geometry: if coils are aligned and close together, all flux from the first coil passes through the second, so $M$ is large; if they are perpendicular or far apart, $M$ is near zero. For two coaxial coils where one fits tightly inside the other (sharing the same cross-sectional area), $M = \frac{μ _{0} N _{1} N _{2} A}{l}$ .

📐 Worked Example

A 100-turn receiving coil is wrapped tightly around the center of the 500-turn solenoid from the previous worked example, sharing the same cross-sectional area. Find (1) the mutual inductance between the two coils, and (2) the magnitude of the emf induced in the receiving coil when the current in the solenoid decreases at 80 A/s.

1
Use the formula for mutual inductance of two aligned coaxial coils: $M = \frac{μ _{0} N _{1} N _{2} A}{l}$ , where $N_{1} = 500$ (solenoid), $N_{2} = 100$ (receiving coil).
2
Substitute values, using the same $A = 1.257 \times 1 0^{- 3} m^{2}$ and $l = 0.15 m$ from the previous example:
3
$M = \frac{(4\pi \times 10^{-7})(500)(100)(1.257 \times 10^{-3})}{0.15} \approx 2.6 \times 10^{-4}\ \text{H} = 260\ \mu\text{H}$
4
Calculate the magnitude of the induced emf. The negative sign confirms the emf opposes the decrease in current per Lenz's law:
5
$|\varepsilon_2| = M \left|\frac{dI_1}{dt}\right| = (2.6 \times 10^{-4}\ \text{H})(80\ \text{A/s}) = 0.021\ \text{V} = 21\ \text{mV}$

Exam tip:

$M$ is always symmetric, so you can calculate it by computing flux from either coil. Always choose the easier calculation (usually flux from the larger coil through the smaller coil, which avoids complicated geometry).

4. RL Circuits★★★☆☆⏱ 3 min

An RL circuit is a series circuit containing a resistor $R$ , inductor $L$ , and usually a voltage source. We analyze RL circuits using Kirchhoff’s loop rule, just like RC circuits, with the inductor contributing a potential drop of $L d I / d t$ .

For a charging RL circuit (battery connected at $t = 0$ , initial current $I (0) = 0$ ), the loop rule gives:

V - IR - L \frac{dI}{dt} = 0

Solving this first-order differential equation gives the current as a function of time:

I(t) = I_{\text{max}} \left(1 - e^{-t/\tau}\right), \quad I_{\text{max}} = \frac{V}{R}, \quad \tau = \frac{L}{R}

where $τ = L / R$ is the time constant for the RL circuit, the time for the current to reach ~63% of its maximum value. For a discharging RL circuit (the battery is removed and the RL combination is shorted at $t = 0$ , initial current $I (0) = I_{0}$ ), the solution is $I (t) = I_{0} e^{- t / τ}$ .

📐 Worked Example

A 12 V battery is connected to a 200 Ω resistor and 4.0 H inductor in series. Find (a) the maximum current after a long time, (b) the current after one time constant, (c) the rate of change of current $d I / d t$ at $t = 0$ .

1
After a long time, $d I / d t = 0$ , so the inductor acts as a short circuit. Maximum current is:
2
$I_{\text{max}} = \frac{V}{R} = \frac{12\ \text{V}}{200\ \Omega} = 0.060\ \text{A} = 60\ \text{mA}$
3
At $t = τ$ , substitute into the charging formula:
4
$I(\tau) = I_{\text{max}} (1 - e^{-1}) \approx 0.060 (1 - 0.368) \approx 0.038\ \text{A} = 38\ \text{mA}$
5
At $t = 0$ , $I = 0$ , so substitute into the loop rule to solve for $d I / d t$ :
6
$V = L \frac{dI}{dt} \implies \frac{dI}{dt} = \frac{V}{L} = \frac{12\ \text{V}}{4.0\ \text{H}} = 3.0\ \text{A/s}$

Exam tip:

Always use the two limit cases to check your answer: if you get a non-zero current at $t = 0$ or non-maximum current after infinite time, you have mixed up charging and discharging formulas.

5. Energy Stored in Inductors★★★★☆⏱ 3 min

To build up current in an inductor, work must be done against the induced emf. This work is stored as magnetic energy in the inductor’s magnetic field. Starting from power: power supplied to the inductor is $P = ε I = L I d I / d t = d U / d t$ . Integrating from $I = 0$ to final current $I$ gives the total stored energy:

U = \frac{1}{2} L I^2

This is the total magnetic energy stored in the inductor at current $I$ , analogous to the $U = \frac{1}{2} C V^{2}$ energy stored in a capacitor. We can also derive the magnetic energy density (energy per unit volume) for any magnetic field, which for uniform $B$ is:

u_B = \frac{1}{2} \frac{B^2}{\mu_0}

This matches the form of electric energy density $u_{E} = \frac{1}{2} ε_{0} E^{2}$ , and is a core relation for understanding electromagnetic energy.

📐 Worked Example

Find the total energy stored in the 1.3 mH solenoid from the self-inductance example when it carries a steady current of 2.0 A, then calculate the magnetic energy density inside the solenoid.

1
Use the total energy formula:
2
$U = \frac{1}{2} L I^2 = 0.5 (0.0013\ \text{H}) (2.0\ \text{A})^2 = 0.0026\ \text{J} = 2.6\ \text{mJ}$
3
Calculate the magnetic field inside the solenoid:
4
$B = \frac{\mu_0 N I}{l} = \frac{(4\pi \times 10^{-7})(500)(2.0)}{0.15} \approx 0.0084\ \text{T}$
5
Use the energy density formula. This matches the value calculated by dividing total energy by the solenoid volume, within rounding error:
6
$u_B = \frac{B^2}{2\mu_0} = \frac{(0.0084)^2}{2 \cdot 4\pi \times 10^{-7}} \approx 28\ \text{J/m}^3$

Exam tip:

If you are not given $L$ for a problem asking for stored energy, calculate $B$ first, then use energy density to find total energy instead of solving for $L$ first; it is often faster.

6. Common Pitfalls

Wrong move:

Confusing RL circuit time constant $τ = L / R$ with RC circuit time constant $τ = R C$ , using $τ = R C$ for RL problems.

Why:

Students remember time constant is a product of $R$ and another component, so they default to $R C$ regardless of circuit type.

Correct move:

Write $τ$ explicitly with your circuit components at the start of the problem: $τ = L / R$ for RL, $τ = R C$ for RC.

Wrong move:

Assuming inductors always block current, so current through an inductor is always zero.

Why:

Confuses the $t = 0$ open-circuit limit with steady-state behavior.

Correct move:

Always check if current is changing: if current is constant, $d I / d t = 0$ , so induced emf is zero, and the inductor acts as an ideal wire.

Wrong move:

Using the self-inductance formula $L = μ_{0} N^{2} A / l$ for mutual inductance, squaring only one coil’s turn count instead of multiplying the two turn counts.

Why:

Memorized formulas for solenoid self-inductance are confused with mutual inductance formulas.

Correct move:

Always start from the definition $M = N_{2} Φ_{12} / I_{1}$ to derive the formula for your specific case, instead of relying on memorized results.

Wrong move:

Missing the factor of $1/2$ when calculating stored energy, writing $U = L I^{2}$ instead of $U = \frac{1}{2} L I^{2}$ .

Why:

Confuses the instantaneous power relation $P = L I d I / d t$ with the integrated total energy.

Correct move:

Always remember energy for inductors and capacitors both have a factor of 1/2, from integrating from zero to final current/voltage.

Wrong move:

Forgetting to square the number of turns when calculating solenoid self-inductance, writing $L = μ_{0} N A / l$ .

Why:

Mixes up the flux linkage formula with the flux per turn formula.

Correct move:

Always confirm that $L$ scales with $N^{2}$ , from $L = N Φ/ I = N (μ_{0} N I A / l) / I = μ_{0} N^{2} A / l$ , so the $I$ cancels leaving $N^{2}$ .

7. Quick Reference Cheatsheet

Category	Formula	Notes
Defining Self-Inductance	$L = \frac{N Φ _{B}}{I}, ε = - L \frac{d I}{d t}$	Unit: henry (H). Negative sign follows Lenz's law, opposes change in current.
Inductance of Long Solenoid	$L = \frac{μ _{0} N ^{2} A}{l} = μ_{0} n^{2} A l$	$n = N / l$ = turns per unit length. Valid for solenoids much longer than diameter.
Defining Mutual Inductance	$M = \frac{N _{2} Φ _{12}}{I _{1}}, ε_{2} = - M \frac{d I _{1}}{d t}$	$M_{12} = M_{21}$ , always symmetric for two coils.
RL Charging Current	$I (t) = I_{max} (1 - e^{- t / τ}), τ = \frac{L}{R}$	Connected to constant voltage source, $I (0) = 0$ , $I_{max} = V / R$ .
RL Discharging Current	$I (t) = I_{0} e^{- t / τ}$	Source removed, RL shorted, $I (0) = I_{0}$ .
Total Energy in Inductor	$U = \frac{1}{2} L I^{2}$	Total magnetic energy stored at steady current $I$ .
Magnetic Energy Density	$u_{B} = \frac{1}{2} \frac{B ^{2}}{μ _{0}}$	Energy per unit volume, valid for any uniform magnetic field.
Inductor Limit Behavior	$t = 0$ : open circuit ( $I = 0$ ), $t \to \infty$ : short circuit ( $d I / d t = 0$ )	Use for quick limit checks in MCQ and FRQ.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
RL circuit time constant calculation
2022 · FRQ
Mutual inductance and induced emf
2021 · MCQ
Energy stored in inductor

What's Next

Inductance is the foundational concept for the remaining topics in Unit 5 Electromagnetism for AP Physics C: E&M. Mastering the behavior of inductors in DC circuits and the energy relations of magnetic fields prepares you to analyze more advanced topics like LC oscillations and AC circuits, which are also heavily tested on the AP exam. Inductance also connects to broader electromagnetic concepts, including Maxwell's equations, which unify electricity and magnetism. Practicing inductance problems, especially differential equation setup for RL circuits, will build the problem-solving skills you need for all circuit-related FRQ questions on the exam.