Inductance — AP Physics C: E&M
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.
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 $\Phi_B$ 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 $B = \mu_0 n I = \mu_0 \frac{N I}{l}$, so flux through one turn is $\Phi_B = BA = \mu_0 \frac{N A I}{l}$. 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.
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 $\Phi_{12}$ 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:
\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), $M = \frac{\mu_0 N_1 N_2 A}{l}$.
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 $L dI/dt$.
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 $\tau = L/R$ 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 $t=0$, initial current $I(0) = I_0$), the solution is $I(t) = I_0 e^{-t/\tau}$.
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 $P = \varepsilon I = L I dI/dt = dU/dt$. 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 $u_E = \frac{1}{2} \varepsilon_0 E^2$, and is a core relation for understanding electromagnetic energy.
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.
Common Pitfalls
Why: Students remember time constant is a product of $R$ and another component, so they default to $RC$ regardless of circuit type.
Why: Confuses the $t=0$ open-circuit limit with steady-state behavior.
Why: Memorized formulas for solenoid self-inductance are confused with mutual inductance formulas.
Why: Confuses the instantaneous power relation $P = L I dI/dt$ with the integrated total energy.
Why: Mixes up the flux linkage formula with the flux per turn formula.