Inductance — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Self-inductance, mutual inductance, RL circuit charging/discharging, energy stored in inductors, magnetic energy density, inductance of a solenoid, and limit behavior of inductors in steady state.

You should already know: Faraday's law and Lenz's law for electromagnetic induction. Kirchhoff's loop rule for series circuits. Capacitor energy and differential equation solving for RC circuits.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Inductance?

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, the magnetic flux through the conductor (or a nearby conductor) also changes, inducing an emf that acts to oppose the change in current per Lenz’s law. This effect is often described as "electrical inertia": inductance resists changes to current, just like mass resists changes to velocity in mechanics.

Per the AP Physics C: E&M Course and Exam Description (CED), inductance accounts for ~10-15% of the total exam score, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is frequently combined with circuit analysis, energy concepts, and AC circuits. The standard notation is $L$ for self-inductance, $M$ for mutual inductance, and the SI unit is the henry (H), where $1\text{H}=1\text{Wb/A}=1\text{Vcdotps/A}$.

2. Self-Inductance

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: the induced emf is proportional to the rate of change of current, with the proportionality constant equal to the self-inductance $L$: $$\epsilon =-L\frac{dI}{dt}$$ 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 ${\Phi }_B$ 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 $B={\mu }_{0}nI={\mu }_0\frac{NI}{l}$, so flux through one turn is ${\Phi }_B=BA={\mu }_0\frac{NAI}{l}$. 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: $l=0.15\text{m}$, $r=0.02\text{m}$, so cross-sectional area $A=\pi r^2\approx 1.257\times 10^{-3}{\text{m}}^2$.
2. Substitute into the solenoid inductance formula, using ${\mu }_0=4\pi \times 10^{-7}\text{Tcdotpm/A}$: $$L=\frac{(4\pi \times 10^{-7})(500{)}^2(1.257\times 10^{-3})}{0.15}\approx 1.3\times 10^{-3}\text{H}=1.3\text{mH}$$
3. The magnitude of the induced emf is $|\epsilon |=L\left|\frac{dI}{dt}\right|=(1.3\times 10^{-3}\text{H})(120\text{A/s})\approx 0.16\text{V}$.

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 very common distracter in MCQs.

3. Mutual Inductance

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 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 ${\Phi }_{12}$ 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: $${\epsilon }_2=-M\frac{d{I}_1}{dt}$$ $M$ 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 (so both have the same cross-sectional area), $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$.

Worked Example

A 100-turn receiving coil is wrapped tightly around the center of the 500-turn solenoid from the previous worked example, so it shares 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: $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$, where $N_1=500$ (solenoid), $N_2=100$ (receiving coil).
2. Substitute values, using the same $A$ and $l$ from the previous example: $$M=\frac{(4\pi \times 10^{-7})(500)(100)(1.257\times 10^{-3})}{0.15}\approx 2.6\times 10^{-4}\text{H}=260\mu \text{H}$$
3. The magnitude of the induced emf is $|{\epsilon }_2|=M\left|\frac{d{I}_1}{dt}\right|=(2.6\times 10^{-4}\text{H})(80\text{A/s})=0.021\text{V}=21\text{mV}$. The negative sign confirms the emf acts to oppose the decrease in current per Lenz’s law.

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

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 $LdI/dt$.

For a charging RL circuit (battery connected at $t=0$, current is zero at $t=0$), 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: $$I(t)=I_{\text{max}}\left(1-e^{-t/\tau }\right),I_{\text{max}}=\frac{V}{R},\tau =\frac{L}{R}$$ where $\tau =L/R$ 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 $t=0$, when current is $I_0$), the solution is $I(t)=I_0{e}^{-t/\tau }$.

Two key limit cases to memorize: At $t=0$ (just after the switch is closed), current is zero, so the inductor acts like an open circuit. After a long time, current is constant ($dI/dt=0$), so the inductor acts like a short circuit (zero potential drop).

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 $dI/dt$ at $t=0$.

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

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 an Inductor

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 $P=\epsilon I=LIdI/dt=dU/dt$. Integrating from $I=0$ to final current $I$ gives: $$U={\int }_0^{I}L{I}^{\prime }d{I}^{\prime }=\frac{1}{2}L{I}^2$$ 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 electric energy density $u_E=\frac{1}{2}{\epsilon }_0{E}^2$, 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: $U=\frac{1}{2}L{I}^2=0.5(0.0013\text{H})(2.0\text{A}{)}^2=0.0026\text{J}=2.6\text{mJ}$.
2. Calculate the magnetic field inside the solenoid: $B={\mu }_{0}NI/l=(4\pi \times 10^{-7})(500)(2.0)/0.15\approx 0.0084\text{T}$.
3. Use the energy density formula: $u_B=B^2/(2{\mu }_0)=(0.0084{)}^2/(2\cdot 4\pi \times 10^{-7})\approx 28{\text{J/m}}^3$. This matches the value calculated by dividing total energy by the volume of the solenoid, within rounding error.

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 (and how to avoid them)

• Wrong move: Confusing RL circuit time constant $\tau =L/R$ with RC circuit time constant $\tau =RC$, using $\tau =RC$ for RL problems. Why: Students remember time constant is a product of $R$ and another component, so they default to $RC$ regardless of circuit. Correct move: Write $\tau$ explicitly with your circuit components at the start of the problem: $\tau =L/R$ for RL, $\tau =RC$ 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, $dI/dt=0$, so induced emf is zero, and the inductor acts as an ideal wire.
• Wrong move: Using the self-inductance formula $L={\mu }_0{N}^{2}A/l$ 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. Correct move: Always start from the definition $M=N_2{\Phi }_{12}/I_1$ 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 $P=LIdI/dt$ 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 $L={\mu }_{0}NA/l$. Why: Mixes up the flux linkage formula with the flux per turn formula. Correct move: Always confirm that $L$ scales with $N^2$, because $L=N\Phi /I=N(BA)/I=N({\mu }_{0}NIA/l)/I={\mu }_0{N}^{2}A/l$, so the $I$ cancels leaving $N^2$.

7. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

A 5.0 H inductor and 100 Ω resistor are connected in series to a 10 V constant voltage source at $t=0$. What is the current through the inductor at $t=0.10\text{s}$? A) $0\text{A}$ B) $0.063\text{A}$ C) $0.086\text{A}$ D) $0.10\text{A}$

Worked Solution: First calculate the time constant for the RL circuit: $\tau =L/R=5.0\text{H}/100\Omega =0.05\text{s}$. For $t=0.10\text{s}$, $t/\tau =2$. The maximum current is $I_{\text{max}}=V/R=10\text{V}/100\Omega =0.10\text{A}$. Substitute into the charging current formula: $I(t)=I_{\text{max}}(1-e^{-t/\tau })=0.10(1-e^{-2})\approx 0.10(1-0.135)=0.0865\text{A}\approx 0.086\text{A}$. The correct answer is C.

Question 2 (Free Response)

A battery of emf $V$ is connected in series to a resistor $R$ and an inductor $L$ at $t=0$, when the current is zero. (a) Write the differential equation for current $I(t)$ using Kirchhoff's loop rule. (b) Solve the differential equation to get $I(t)$, showing all integration steps. (c) Prove that at any time $t$, the power supplied by the battery equals the sum of the power dissipated in the resistor and the power stored as energy in the inductor.

Worked Solution: (a) Summing potential changes around the loop: the battery contributes $+V$, the resistor contributes $-IR$, the inductor contributes $-LdI/dt$ for the induced emf. The differential equation is: $$V-IR-L\frac{dI}{dt}=0$$ (b) Separate variables: $\frac{dI}{V-IR}=\frac{dt}{L}$. Integrate from $t=0$ ($I=0$) to $t$ ($I$): $${\int }_0^I\frac{d{I}^{\prime }}{V-I^{\prime }R}={\int }_0^t\frac{d{t}^{\prime }}{L}$$ Evaluate the integrals: $$-\frac{1}{R}\ln \left(\frac{V-IR}{V}\right)=\frac{t}{L}⟹\ln \left(\frac{V}{V-IR}\right)=\frac{Rt}{L}$$ Exponentiate both sides and rearrange: $$V=(V-IR)e^{Rt/L}⟹IR=V\left(1-e^{-Rt/L}\right)⟹I(t)=\frac{V}{R}\left(1-e^{-t/\tau }\right),\tau =\frac{L}{R}$$ (c) Power supplied by the battery is $P_{\text{batt}}=VI$. Power dissipated in the resistor is $P_R=I^{2}R$. Power stored in the inductor is $P_L=dU/dt=d/dt\left(\frac{1}{2}L{I}^2\right)=LIdI/dt$. From the differential equation in (a), $V=IR+LdI/dt$. Multiply both sides by $I$: $VI=I^{2}R+LIdI/dt$, so $P_{\text{batt}}=P_R+P_L$, which proves the statement.

Question 3 (Application / Real-World Style)

A wireless phone charging pad uses mutual inductance to transfer power. The charging pad has a coil with 50 turns, and the phone has a receiving coil with 200 turns. Both coils have cross-sectional area $1.0{\text{cm}}^2$, and they are aligned such that all flux from the pad coil passes through the receiving coil. The length of the pad coil is $2.0\text{cm}$. If the current in the pad coil is $I_1(t)=(1.0\text{A})\sin (1000t)$, what is the amplitude of the induced emf in the phone's receiving coil?

Worked Solution: First convert units to SI: $A=1.0\times 10^{-4}{\text{m}}^2$, $l=0.02\text{m}$. Calculate mutual inductance: $$M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}=\frac{(4\pi \times 10^{-7})(50)(200)(1.0\times 10^{-4})}{0.02}\approx 6.28\times 10^{-5}\text{H}$$ Induced emf is ${\epsilon }_2=-Md{I}_1/dt=-M(1.0\cdot 1000)\cos (1000t)$. The amplitude is the magnitude of the coefficient: $$|{\epsilon }_{\text{amp}}|=M\cdot 1000=(6.28\times 10^{-5}\text{H})(1000)\approx 0.063\text{V}=63\text{mV}$$ This 63 mV amplitude is large enough to be rectified and used to trickle-charge a modern smartphone battery, matching real-world performance of small wireless charging coils.

8. Quick Reference Cheatsheet

9. What's Next

Inductance is the foundational concept for the remaining topics in Unit 5 Electromagnetism for AP Physics C: E&M. Without mastering the behavior of inductors, mutual inductance, and energy storage, you cannot understand LC oscillations, AC circuits, or transformers, all of which are heavily tested on the exam. This topic also completes the connection between electric and magnetic energy, setting up the derivation of Maxwell’s equations and electromagnetic waves. Inductance builds directly on Faraday’s law, and in turn is the prerequisite for analyzing time-varying electromagnetic systems. Next you will study oscillations in combined inductor-capacitor circuits, then extend that analysis to driven alternating current circuits.

• LC Circuits and Electromagnetic Oscillations
• Alternating Current (AC) Circuits
• Maxwell's Equations and Electromagnetic Waves