The energy stored in an inductor is the energy held in the magnetic field created by current through it, given by U_L = ½LI². In AP Physics C: E&M, it explains where energy goes as current builds in an RL circuit and parallels the capacitor formula U_C = ½CV².
When current flows through an inductor, it creates a magnetic field, and that field stores energy. The amount is U_L = ½LI², where L is the inductance and I is the current. The key physical idea is that the energy doesn't live in the wire or the charges. It lives in the magnetic field itself, mostly concentrated inside the coil (for a solenoid, almost entirely in the interior).
Where does that energy come from? Work. As current ramps up, the inductor's back-emf opposes the change, so the source has to push charge against that emf. The power delivered to the inductor is P = LI(dI/dt), and if you integrate that from zero current up to I, you get exactly ½LI². That's the same logic used to derive ½CV² for a capacitor, just with current playing the role that charge played there. The formula also tells you something useful immediately. Energy depends on I squared, so tripling the current means nine times the stored energy.
This term lives in Topic 13.4 (Inductance) and Topic 13.5 (Circuits with Resistors and Inductors) in Unit 13, Electromagnetic Induction. It's the energy bookkeeping piece of inductance. When you analyze an RL circuit, Kirchhoff's loop rule tells you the math, but ½LI² tells you the story. Energy from the battery splits between resistor dissipation (I²R) and energy banked in the inductor's magnetic field. At steady state, dI/dt = 0, the inductor stops absorbing energy, and U_L locks in at ½LI². When the circuit is disconnected, that stored energy is what drives the decaying current and gets burned off in the resistor. It's also one half of the big symmetry the exam loves, magnetic field energy in inductors mirroring electric field energy in capacitors.
Keep studying AP® Physics C: E&M Unit 13
RL Circuit (Unit 13)
An RL circuit is where stored inductor energy actually does something. While current grows toward its steady-state value, the battery's energy splits two ways, some dissipated in the resistor and some stored as ½LI² in the inductor. During current decay, that stored energy flows back out and gets dissipated.
Energy Stored in a Capacitor (Unit 10)
U_C = ½CV² and U_L = ½LI² are mirror images. A capacitor stores energy in an electric field and resists changes in voltage; an inductor stores energy in a magnetic field and resists changes in current. If you know one derivation, you basically know both.
dI/dt and Induced EMF (Unit 13)
Power delivered to an inductor is P = LI(dI/dt), which is just the time derivative of ½LI². That means the inductor only gains or loses energy while the current is changing. Constant current means constant stored energy, even though current is still flowing.
Steady State (Unit 13)
At steady state an ideal inductor acts like a plain wire with no voltage across it, but it's not 'empty.' It's holding ½LI² in its magnetic field the whole time. That distinction, no voltage but plenty of stored energy, is a classic conceptual MCQ trap.
Multiple-choice questions hit this from a few angles. Ratio reasoning is common, like a problem where 0.5 J is stored at 2 A and you're asked what current gives 4.5 J (energy scales with I², so the answer is 6 A). Conceptual stems ask where the energy is physically stored, and the answer is the magnetic field, not the wire or the charges. Calculation problems combine formulas, like finding the inductance of a solenoid from its turns and dimensions, then plugging the peak of a sinusoidal current into ½LI² for maximum stored energy. On free-response, energy storage typically shows up inside RL circuit analysis. You might derive I(t), compute the steady-state energy ½LI², or write an energy-conservation statement showing battery energy equals resistor dissipation plus stored field energy. Be ready to connect U_L to emf too, since ε = L(dI/dt) and ½LI² come as a package.
Both are ½ times something times something squared, so they blur together under exam pressure. A capacitor stores energy in an electric field and the formula uses voltage, U_C = ½CV². An inductor stores energy in a magnetic field and the formula uses current, U_L = ½LI². Quick memory check, capacitors care about voltage (they fight voltage changes), inductors care about current (they fight current changes). Each one's energy formula squares the thing it cares about.
The energy stored in an inductor is U_L = ½LI², and it's physically stored in the magnetic field created by the current, not in the wire itself.
Because energy scales with the square of the current, doubling the current quadruples the stored energy, which makes ratio problems fast if you spot the pattern.
The formula comes from integrating the power P = LI(dI/dt) delivered while current builds from zero, the same derivation style as ½CV² for a capacitor.
In an RL circuit, battery energy splits between resistor dissipation and stored magnetic field energy, and at steady state the inductor holds a constant ½LI².
An inductor only exchanges energy with the circuit while current is changing; with constant current it stores energy but absorbs no power.
When current decays in an RL circuit, the inductor's stored energy is what keeps the current flowing, and it ends up dissipated in the resistor.
It's the energy held in the magnetic field produced by current through the inductor, equal to U_L = ½LI². It builds up as current increases and is released when current decreases.
In the magnetic field, not in the wire or the moving charges. For a solenoid, the energy is concentrated almost entirely in the field inside the coil. This exact question shows up as a conceptual MCQ.
No. At steady state the inductor acts like a plain wire (zero voltage across an ideal inductor), but it still holds ½LI² in its magnetic field. It just stops absorbing new energy because dI/dt = 0.
An inductor stores energy in a magnetic field and the formula uses current, U_L = ½LI². A capacitor stores energy in an electric field and uses voltage, U_C = ½CV². Same ½(thing)(variable)² structure, different field and different variable.
Use the fact that U is proportional to I². For example, if 0.5 J is stored at 2 A, then 4.5 J is 9 times the energy, so the current is 3 times bigger, which gives 6 A. No need to solve for L.
Connect this key term to the AP exam workflow: review the course, practice questions, and check related study tools.
Review units, study guides, and course resources.
Check this vocabulary in multiple-choice context.
Apply key concepts in written AP responses.
Estimate the exam score you are working toward.
Review the highest-yield facts before practice.
Put the full course together before test day.