College Board AP 2024 (style)-style practice: Section II (FRQ, quantitative). A solenoid has n = 4000 turns per meter, length 0.20 m, and cross-sectional area 5.010^-4 m squared. (a) Derive the inductance. (b) Calculate the energy stored when it carries 2.0 A. (c) State the back-EMF if this current is reduced to zero in 0.010 s. Use _0 = 4π10^-7.

A 5-point FRQ on solenoid inductance and energy. (a) Inductance (2 points): L = _0 n^2 A = (4π10^-7)(4000)^2(5.010^-4)(0.20) = (4π10^-7)(1.610^7)(1.010^-4) = 2.010^-3 H. (b) Energy (2 points): U = 12LI^2 = 12(2.010^-3)(2.0)^2 = 4.010^-3 J. (c) Back-EMF (1 point): || = L|dIdt| = (2.010^-3)2.00.010 = 0.40 V. Markers reward L = _0 n^2 A, the 12LI^2 energy, and the back-EMF.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

What is inductance, and how does an inductor store energy and oppose changes in current?

Topic 13.4 Inductance: define self-inductance, find the inductance and stored energy of a solenoid, and apply the back-EMF of an inductor.

A calculus-based answer to AP Physics C E&M Topic 13.4, covering self-inductance, the back-EMF, the inductance of a solenoid, the energy stored in an inductor, and the magnetic energy density.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Self-inductance $L$ relates the flux linkage of a coil to its own current: $N\Phi_B = LI$ , with $L$ in henries (H). A changing current induces a back-EMF opposing the change: $\varepsilon = -L\dfrac{dI}{dt}$ . For a solenoid, $L = \mu_0 n^2 A \ell$ (turns per meter $n$ , area $A$ , length $\ell$ ). An inductor stores energy in its magnetic field, $U = \tfrac{1}{2}LI^2$ , analogous to a capacitor's $\tfrac{1}{2}CV^2$ . The energy resides in the field with energy density $u = \dfrac{B^2}{2\mu_0}$ . Because of the back-EMF, the current through an inductor cannot change instantaneously: an inductor opposes sudden change in current the way a capacitor opposes sudden change in voltage.

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What this topic is asking
Self-inductance and the back-EMF
Inductance of a solenoid
Energy stored and energy density
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What this topic is asking

The College Board (Topic 13.4) wants you to define self-inductance, derive the inductance of a solenoid, find the energy stored in an inductor, and apply the back-EMF that opposes changes in current. An inductor is to current what a capacitor is to voltage: it resists sudden change and stores energy in the magnetic field.

Self-inductance and the back-EMF

When the current through a coil changes, its own magnetic flux changes, and by Faraday's law this induces an EMF in the coil itself. Lenz's law makes it a back-EMF that fights the change: trying to increase the current meets opposition, and trying to decrease it meets a push to maintain it. This is why the current through an inductor cannot jump instantaneously.

Inductance of a solenoid

For a long solenoid, the interior field is $B = \mu_0 n I$ (from Ampere's law), so the flux through one turn is $\Phi = BA = \mu_0 n I A$ , and the total flux linkage over $N = n\ell$ turns gives

L = \frac{N\Phi}{I} = \mu_0 n^2 A \ell

The inductance depends only on geometry (turns per length squared, area, length), just as capacitance depends only on geometry.

Energy stored and energy density

Building the current in an inductor requires work against the back-EMF. Integrating the power $\varepsilon I = LI\dfrac{dI}{dt}$ over time gives the stored energy:

U = \int_0^I L\,i\,di = \tfrac{1}{2}LI^2

This energy lives in the magnetic field, with energy density

u = \frac{B^2}{2\mu_0} \qquad(\text{J per m cubed})

the magnetic analogue of the electric energy density $\tfrac{1}{2}\varepsilon_0 E^2$ .

Energy and back-EMF of an inductor

A $50$ mH inductor carries a current that is reduced from $4.0$ A to $0$ in $0.020$ s. Find the energy initially stored and the average back-EMF.

step 1 Find the stored energy

$U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(50\times10^{-3})(4.0)^2 = \tfrac{1}{2}(0.050)(16) = 0.40$ J.

step 2 Find the rate of current change

$\dfrac{dI}{dt} = \dfrac{0 - 4.0}{0.020} = -200$ A/s.

step 3 Find the back-EMF

$|\varepsilon| = L\left|\dfrac{dI}{dt}\right| = (0.050)(200) = 10$ V.

step 4 Interpret

The back-EMF opposes the decrease, acting to keep the current flowing. This is why opening an inductive circuit can produce a large voltage spike across the switch.

Try this

Q1. A $0.10$ H inductor's current changes at $5.0$ A/s. Find the back-EMF magnitude. [1 point]

Cue. $|\varepsilon| = L\left|\dfrac{dI}{dt}\right| = (0.10)(5.0) = 0.50$ V.

Q2. A $20$ mH inductor carries $3.0$ A. Find the energy stored. [2 points]

Cue. $U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(20\times10^{-3})(3.0)^2 = 0.090$ J.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). The current through a

0.20

H inductor changes at a rate of

3.0

A/s. The magnitude of the back-EMF across the inductor is (A)

0.60

V (B)

0.067

V (C)

15

V (D)

1.8

V. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the inductor back-EMF. The answer is (A).

The self-induced EMF is $|\varepsilon| = L\left|\dfrac{dI}{dt}\right| = (0.20)(3.0) = 0.60$ V. The trap is (B): dividing instead of multiplying. The back-EMF opposes the change in current, the inductor's defining behavior.

AP 2024 (style)5 marksSection II (FRQ, quantitative). A solenoid has

n = 4000

turns per meter, length

0.20

m, and cross-sectional area

5.0\times10^{-4}

m squared. (a) Derive the inductance. (b) Calculate the energy stored when it carries

2.0

A. (c) State the back-EMF if this current is reduced to zero in

0.010

s. Use

\mu_0 = 4\pi\times10^{-7}

Show worked answer →

A 5-point FRQ on solenoid inductance and energy.

(a) Inductance (2 points): $L = \mu_0 n^2 A \ell = (4\pi\times10^{-7})(4000)^2(5.0\times10^{-4})(0.20) = (4\pi\times10^{-7})(1.6\times10^7)(1.0\times10^{-4}) = 2.0\times10^{-3}$ H.
(b) Energy (2 points): $U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(2.0\times10^{-3})(2.0)^2 = 4.0\times10^{-3}$ J.
(c) Back-EMF (1 point): $|\varepsilon| = L\left|\dfrac{dI}{dt}\right| = (2.0\times10^{-3})\dfrac{2.0}{0.010} = 0.40$ V.

Markers reward $L = \mu_0 n^2 A\ell$ , the $\tfrac{1}{2}LI^2$ energy, and the back-EMF.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)