AP Physics C: E&M · Topic 13.4

Topic 13.4: Inductance

Unit 13: Electromagnetic Induction10-20% of the multiple-choice section

Inductance is the tendency of a conductor to oppose a change in electrical current. Change the current in a coil and its own changing flux induces an emf that fights the change. That emf is the inductance times the rate of change of current, and the coil stores energy in its magnetic field.

AP Physics: Unit 13 (topics 13.4 Inductance). AP Physics C: Electricity and Magnetism Unit 13, Topic 13.4. One learning objective, 13.4.A, describe the physical and electrical properties of an inductor. Supported by 13.4.A.1 (inductance is the tendency of a conductor to oppose a change in electrical current), 13.4.A.1.i (inductance of a conductor depends on the physical properties of the conductor, and straight wires are typically modeled as having zero inductance), 13.4.A.1.ii (an inductor, such as a solenoid, is a circuit element that has significant inductance), 13.4.A.1.iii (the inductance of a solenoid depends on the total number of turns, the length, the cross-sectional area, and the magnetic permeability of the core, with the relevant equation L_sol = mu_core N squared A over l), 13.4.A.2 (inductors store energy in the magnetic field generated by current in the inductor, with the relevant equation U_L = one half L I squared), 13.4.A.2.i (that stored energy can be dissipated through a resistor or used to charge a capacitor), 13.4.A.2.ii (the transfer of energy generated in an inductor to other forms obeys conservation laws), and 13.4.A.3 (applying Faraday's law to an inductor and using the definition of inductance relates induced emf to inductance and the rate of change of current, with the relevant equation emf_i = -L dI/dt). Topic 13.4 prints no boundary statement. All three equations are on the equation sheet, which writes the self-induced emf without the subscript. The CED refers to the definition of inductance at 13.4.A.3 but does not print it, so the flux-linkage definition L = N Phi_B / I is not a quotable printed equation. Suggested skills are 1.C, 2.A, 2.C and 3.B. The words inductance and inductor do not appear in the AP Physics 2 course and exam description, so this topic has no algebra-based sibling.

What Topic 13.4 requires

Topic 13.4 has one learning objective and seven essential-knowledge statements. It prints no boundary statement.

13.4.A, describe the physical and electrical properties of an inductor.

13.4.A.1 states that inductance is the tendency of a conductor to oppose a change in electrical current.
13.4.A.1.i states that inductance of a conductor depends on the physical properties of the conductor, and that straight wires are typically modeled as having zero inductance.
13.4.A.1.ii states that an inductor, such as a solenoid, is a circuit element that has significant inductance.
13.4.A.1.iii states that the inductance of a solenoid is dependent on the total number of turns, the length of the solenoid, the cross-sectional area of the solenoid, and magnetic permeability of the solenoid's core, with the relevant equation $L_{\text{sol}} = \dfrac{\mu_{\text{core}} N^2 A}{\ell}$ .
13.4.A.2 states that inductors store energy in the magnetic field that is generated by current in the inductor, with the relevant equation $U_L = \dfrac{1}{2} L I^2$ .
13.4.A.2.i states that the energy stored in the magnetic field generated by an inductor in which current is flowing can be dissipated through a resistor or used to charge a capacitor.
13.4.A.2.ii states that the transfer of energy generated in an inductor to other forms of energy obeys conservation laws.
13.4.A.3 states that by applying Faraday's law to an inductor and using the definition of inductance, induced emf can be related to inductance and the rate of change of current, with the relevant equation $\mathcal{E}_i = -L \dfrac{dI}{dt}$ .

Read 13.4.A.2.i again, because it is a roadmap. "Dissipated through a resistor" is Topic 13.5. "Used to charge a capacitor" is Topic 13.6. The CED is announcing the last two topics of the unit inside the essential knowledge of this one.

The suggested skills are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

All three of this topic's equations are printed on the AP Physics C: E&M equation sheet. The sheet writes the self-induced emf without the subscript, as $\mathcal{E} = -L\,dI/dt$ , while the CED writes $\mathcal{E}_i$ .

Inductance is electrical inertia

Statement 13.4.A.1 is a definition in one line: inductance is the tendency of a conductor to oppose a change in electrical current. Every property of an inductor follows from unpacking why a conductor would do that.

The chain is short and it uses nothing new.

A current in a coil makes a magnetic field. That is Ampere's law from Unit 12, and for a long solenoid it gives $B_{\text{sol}} = \mu_0 n I$ , printed on the sheet.
That field passes through the coil's own turns, so the coil has a flux through itself proportional to its own current.
Change the current and that flux changes.
A changing flux induces an emf, by Faraday's law from Topic 13.2.
By Lenz's law that emf opposes the change, so it fights whatever is trying to alter the current.

That is self-induction: the coil induces an emf in itself. The word "opposes" in 13.4.A.1 is the same word as in Lenz's law, and it is the same physics.

The useful mental model is inertia. An inductor is to current what mass is to velocity.

Mechanics	Circuit
mass $m$	inductance $L$
velocity $v$	current $I$
force $F$	emf $\mathcal{E}$
$F = m\,dv/dt$	$\mathcal{E} = L\,dI/dt$ in magnitude
$K = \frac{1}{2}mv^2$	$U_L = \frac{1}{2}LI^2$

That correspondence is not a coincidence and it is not decoration. It carries the whole of Topic 13.6, where the CED says outright that the charge in an LC circuit can be modeled as simple harmonic motion. Learn the analogy here and that topic costs almost nothing.

The practical consequences worth stating now, because they are what the exam asks about:

The current through an inductor cannot jump. A discontinuous current would need an infinite $dI/dt$ and so an infinite emf. Current through an inductor is continuous through the instant a switch changes, the way velocity is continuous through a collision that lasts a finite time.
The potential difference across an inductor can jump, and routinely does. It is proportional to the rate of change of current, not to the current, so it can go from zero to hundreds of volts the moment a switch opens.
A steady current produces no emf at all. With $dI/dt = 0$ the inductor is electrically invisible, which is statement 13.5.A.4.iii in the next topic: after a long time it behaves as a conducting wire with zero resistance.

Statement 13.4.A.1.i adds the practical note that straight wires are typically modeled as having zero inductance. That is a modelling assumption, not a physical fact, and it is what lets you treat the connecting wires in a circuit diagram as inert and put all the inductance in the labelled element. Statement 13.4.A.1.ii names the element: an inductor, such as a solenoid, is a circuit element that has significant inductance. Winding the wire into a coil is what makes the self-flux large enough to matter, because each turn's field threads all the others.

Where the solenoid formula comes from

L_{\text{sol}} = \frac{\mu_{\text{core}} N^2 A}{\ell}

Statement 13.4.A.1.iii prints this and names its four dependences: total number of turns, length of the solenoid, cross-sectional area, and the magnetic permeability of the core. Skill 2.A is listed for this topic, so being able to produce it matters as much as being able to use it.

The derivation takes four lines and reuses two results you already have.

Field inside. For a long solenoid, $B = \mu_{\text{core}} n I$ with $n = N/\ell$ turns per unit length, so $B = \mu_{\text{core}} N I / \ell$ . The vacuum version, $B_{\text{sol}} = \mu_0 n I$ , is on the sheet as a derived result of Ampere's law from statement 12.4.A.1.iii.
Flux through one turn. The field is uniform across the cross-section, so $\Phi_B = BA = \mu_{\text{core}} N I A / \ell$ .
Total flux linked. Each of the $N$ turns encircles that flux, so the flux linkage is $N\Phi_B = \mu_{\text{core}} N^2 I A / \ell$ .
Divide by the current. Inductance is the flux linkage per unit current, so $L = N\Phi_B / I = \mu_{\text{core}} N^2 A / \ell$ .

One honest note on step 4. The CED refers at 13.4.A.3 to "the definition of inductance" without printing it, so the flux-linkage definition $L = N\Phi_B / I$ is not a printed equation you can quote off the sheet. What is printed is the result of using it, in two places: this solenoid formula and the self-induced emf below. Both are consistent with that definition and nothing else, and the derivation above is the standard route between them. If a question asks you to derive $L_{\text{sol}}$ , say what definition you are using rather than leaving step 4 unexplained.

The $N^2$ is the part worth understanding rather than memorising, and it is the classic functional-dependence question. Doubling the number of turns multiplies the inductance by four, not by two. The turns count twice: once because more turns make a stronger field for the same current, and again because more turns are there to be threaded by that field. Watch for the trap version, though. If you double $N$ by winding twice as many turns onto a solenoid of the same length, the length $\ell$ is fixed and $L$ quadruples. If instead you double $N$ by making the solenoid twice as long at the same turn density, then $N$ and $\ell$ both double and $L$ only doubles.

Change	Effect on $L$
double $N$ at fixed length	$\times 4$
double $N$ and double $\ell$ together, so $n$ is unchanged	$\times 2$
double the radius, so $A$ quadruples	$\times 4$
double the length at fixed $N$	$\times \frac{1}{2}$
replace an air core with one of a thousand times the permeability	$\times 1000$

That last row is the CED's fourth sample activity for Unit 13, which asks students to construct a solenoid, measure its inductance using an LR circuit by measuring the time constant and the resistance, and then repeat the experiment with iron or steel in the core to get the increased inductance. Statement 12.1.C.3 of the previous unit is the caution to carry into that lab: the permeability of matter is not a constant for a material and varies with factors including temperature, orientation and field strength, so an iron core does not have one clean multiplier.

The self-induced emf and its sign

\mathcal{E}_i = -L \frac{dI}{dt}

Statement 13.4.A.3 says how this is reached: by applying Faraday's law to an inductor and using the definition of inductance, induced emf can be related to inductance and the rate of change of current. Written out, that is

\mathcal{E} = -\frac{d(N\Phi_B)}{dt} = -\frac{d(LI)}{dt} = -L\frac{dI}{dt}

with the last step using the fact that $L$ is fixed by geometry and the core, not by the current. That is worth saying explicitly on a derivation: the inductance comes out of the derivative because it depends only on the physical properties of the conductor, which is statement 13.4.A.1.i.

The minus sign is Lenz's law again, and the way to use it is to translate it into a direction rather than to carry it through algebra.

What the current is doing	What the inductor does
increasing	opposes the increase, acting like a source pushing back against the current
decreasing	opposes the decrease, acting like a source pushing to keep the current going
steady	nothing at all, zero emf, behaves as a plain wire

The second row is the surprising one and it produces the most dramatic effect in the unit. Break the circuit of an inductor carrying a steady current and $dI/dt$ becomes enormous in magnitude, so the inductor generates a large emf trying to keep the current flowing. That is why opening a switch on an inductive circuit can produce a spark across the contacts, and why a potential difference far larger than the battery's can appear briefly. The third worked example in Topic 13.5 computes one.

Units: from $\mathcal{E} = L\,dI/dt$ , one henry is one volt-second per ampere. Equivalently, from $L = N\Phi_B/I$ , one henry is one weber per ampere. Both are worth knowing because they let you check an answer's dimensions two ways.

One caution on the sign convention. Whether you write the inductor's contribution to a loop equation as $-L\,dI/dt$ or $+L\,dI/dt$ depends on which way you traverse the loop and which way you defined positive current, and the two conventions give the same physics. What must not happen is a convention that changes halfway through. Declare the positive current direction on the diagram, traverse the loop consistently, and check the result against the physical statement: an inductor always fights the change.

Energy in the magnetic field

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

Statement 13.4.A.2 says inductors store energy in the magnetic field that is generated by current in the inductor. The equation is printed on the sheet, and deriving it is a one-integral exercise that is worth doing once, because it is the same integral that produces the kinetic energy expression in mechanics.

Start from power. To build the current up, an external source has to push against the inductor's back emf, delivering power

P = \mathcal{E}I = \left(L\frac{dI}{dt}\right)I

The energy stored is the time integral of that power, and the $dt$ cancels:

U_L = \int_0^t L \frac{dI}{dt} I \, dt = \int_0^{I} L I' \, dI' = \frac{1}{2}LI^2

The change of variable in the middle step is the whole trick, and it is the same move that turns $\int F\,dx$ with $F = ma$ into $\frac{1}{2}mv^2$ .

Three things to take from the result.

The energy goes as the square of the current. Doubling the current stores four times as much energy. This is a skill 2.C comparison waiting to be asked.

It is stored in the field, not in the wire. Statement 13.4.A.2 puts it in the magnetic field generated by the current. When the current stops, the field collapses and the energy has to go somewhere, which is exactly the point of 13.4.A.2.i: it can be dissipated through a resistor or used to charge a capacitor.

It is recoverable, unlike resistive dissipation. A resistor converts electrical energy irreversibly. An inductor holds it and gives it back. Statement 13.4.A.2.ii says the transfer of energy generated in an inductor to other forms obeys conservation laws, which is the licence to write an energy-balance equation instead of solving a differential equation. In an LC circuit that is the whole method: the energy that leaves the capacitor arrives in the inductor and back again.

The comparison with a capacitor is worth having in one place, because the two elements are mirror images and questions exploit the symmetry.

	Inductor	Capacitor
stores energy in	magnetic field	electric field
energy	$U_L = \frac{1}{2}LI^2$	$U_C = \frac{1}{2}Q\,\Delta V$
cannot change instantaneously	its current	its potential difference
relation between the two variables	$\mathcal{E} = -L\,dI/dt$	$C = Q/\Delta V$ with $I = dq/dt$
behaviour long after a switch closes	a wire with zero resistance	an open branch with no current
behaviour the instant a switch closes, from rest	blocks, carrying no current	conducts freely, no potential difference

Both energy expressions and both defining relations are on the sheet. The last two rows are the ones that get mixed up, and reading them as opposites is the fastest way to keep them straight: an uncharged capacitor starts as a wire and ends as a break, while an inductor with no current starts as a break and ends as a wire.

Graphs and the qualitative sketch

Skill 1.C, creating qualitative sketches of graphs that represent features of a model or the behavior of a physical system, is listed first for this topic. Sketching is assessed on the free-response section, and Topic 13.4's sketches are simpler than the exponential ones in Topic 13.5.

The key relationships to draw:

Current against time	Self-induced emf against time
flat, any value	zero
straight ramp up	constant negative, a horizontal step
steeper ramp up	larger in magnitude, still constant
straight ramp down	constant, opposite in sign to the ramp-up case
curve with a maximum	passes through zero at the maximum
sinusoidal	sinusoidal, a quarter cycle out of phase

Row two is the one to have automatic: a linearly rising current gives a constant emf, so the emf graph is a flat step, not a ramp. That is the same relationship as flux and emf in Topic 13.2, because it is literally the same derivative one level down.

The stored-energy graph is worth a separate look, because it is not proportional to anything you have drawn yet. Since $U_L = \frac{1}{2}LI^2$ , a linearly rising current gives an energy rising as $t^2$ , a parabola through the origin. And because the energy depends on $I^2$ , it is the same for a current of $+2$ A as for a current of $-2$ A: reversing the current direction does not empty the inductor.

One more sketch worth being able to produce: potential difference across an inductor against current. It is not a straight line through the origin, because an inductor is not a resistor. In fact there is no fixed relationship between the two at all. The same current can coexist with any potential difference, depending on how fast the current is changing. If a question offers you a resistance-like graph for an inductor, that is the error being tested.

AP Physics 2 has no inductance at all

There is no AP Physics 2 sibling for this page, and that is a fact about the course rather than about this site. The words inductance and inductor do not appear anywhere in the AP Physics 2 course and exam description, and no AP Physics 2 topic covers self-induction, energy stored in a magnetic field, LR circuits or LC circuits. Half of Unit 13, Topics 13.4 through 13.6, has no algebra-based counterpart.

That makes this one of the clearest places in the two-course sequence where AP Physics C is not a harder version of something but genuinely new material.

The nearest useful algebra-based pages are the ones that supply the prerequisites rather than the content.

AP Physics 2 Topic 12.4, Electromagnetic Induction and Faraday's Law, for Faraday's law and Lenz's law in algebra-based form. That page is for AP Physics 2 students; this one is for AP Physics C students. An inductor is Faraday's law applied to a coil's own current, so if the induction step is shaky, that page is where to shore it up.
AP Physics 2 Topic 10.6, Capacitors, for the mirror-image element. The comparison table above only works if you already know how a capacitor behaves.

If a textbook chapter you are reading covers mutual inductance, transformers as a quantitative topic, or magnetic energy density in terms of $B^2$ , check it against the framework before spending time on it. None of those appears in the required course content for Topic 13.4, which names exactly three equations: the solenoid inductance, the stored energy, and the self-induced emf.

How Topic 13.4 is tested

Inductance rarely stands alone as a full question. It is the setup for Topics 13.5 and 13.6, and it appears in multiple-choice questions on its own terms.

The patterns:

Compute $L$ from the geometry and then a factor of change when a dimension or the core changes (skills 2.C, 3.B). The $N^2$ is the point.
Derive $L_{\text{sol}}$ from the solenoid field and the flux linkage (skill 2.A). Say which definition of inductance you are using.
Find the self-induced emf from a stated current ramp, or the reverse (skill 2.C).
Compute stored energy, or compare the energy at two currents (skill 2.C).
Sketch current, emf or energy against time for a described situation (skill 1.C).
Make and justify a claim about what an inductor does at a particular instant (skill 3.B), for example that the current cannot jump when a switch is thrown.

On that last one, the CED's science-practices page for Unit 13 picks an example from exactly this territory: it says that describing how a solenoid resists a change in electrical current due to the required change in magnetic field within that solenoid is a valuable skill, in addition to deriving a mathematical equation that shows the current in the solenoid as a function of time. The verbal explanation is named as a skill in its own right, not as a warm-up for the algebra.

A complete verbal answer to that has four clauses, and it is worth having them ready. The current in the solenoid produces a magnetic field through its own turns. Changing the current changes that field and so changes the flux through the solenoid. A changing flux induces an emf, by Faraday's law. By Lenz's law that emf opposes the change, so it acts against whatever is trying to alter the current. Four sentences, each one citing a law by name.

The inductance of a real solenoid, and what changes it

A solenoid is wound with 800 turns over a length of 0.25 m, with a cross-sectional radius of 0.015 m and an air core. Find (a) its inductance, (b) the inductance if the number of turns is doubled to 1600 without changing the length, and (c) the inductance if an iron core is inserted whose permeability is 1000 times that of free space, with the original 800 turns.

(a) Use the printed relation from 13.4.A.1.iii, $L_{\text{sol}} = \dfrac{\mu_{\text{core}} N^2 A}{\ell}$ , with $\mu_{\text{core}} = \mu_0 = 4\pi \times 10^{-7} \ (\mathrm{T \cdot m})/\mathrm{A}$ for an air core, taken from the Table of Information.
Cross-sectional area: $A = \pi r^2 = \pi (0.015)^2 = \pi (2.25 \times 10^{-4}) = 7.0686 \times 10^{-4} \ \mathrm{m^2}$ .
Turns squared: $N^2 = (800)^2 = 6.40 \times 10^5$ .
Numerator step by step: $(1.25664 \times 10^{-6})(6.40 \times 10^5) = 0.80425$ , then $\times (7.0686 \times 10^{-4}) = 5.6850 \times 10^{-4}$ .
Divide by the length: $L = \dfrac{5.6850 \times 10^{-4}}{0.25} = 2.2740 \times 10^{-3}$ H, so $L = 2.3$ mH to two significant figures.
Check the units: $(\mathrm{T \cdot m/A})(\mathrm{m^2})/(\mathrm{m}) = \mathrm{T \cdot m^2 / A} = \mathrm{Wb/A} = \mathrm{H}$ .
(b) The length is unchanged, so only $N^2$ changes and $L$ scales by $(1600/800)^2 = 4$ . $L = 4(2.2740 \times 10^{-3}) = 9.096 \times 10^{-3}$ H, so 9.1 mH. Doubling the turns quadruples the inductance, because more turns both make a stronger field and provide more turns for it to thread.
Note the contrast that the question is set up to expose. Had the extra turns been added by making the solenoid twice as long at the same winding density, $N$ and $\ell$ would both double, $N^2/\ell$ would only double, and $L$ would be 4.5 mH rather than 9.1 mH. The four dependences in 13.4.A.1.iii are not independent of how you change the coil.
(c) Only $\mu_{\text{core}}$ changes, and $L$ is directly proportional to it, so $L = 1000(2.2740 \times 10^{-3}) = 2.274$ H, about 2.3 H. That is the CED's fourth sample activity for this unit in numbers: put iron or steel in the core and the inductance rises by orders of magnitude.
Carry one caution from statement 12.1.C.3 of the previous unit: the permeability of matter is not a constant for a material and varies with temperature, orientation and field strength, so a single multiplier like 1000 is a model, not a measured constant.

(a) $L = 2.3$ mH. (b) $L = 9.1$ mH, four times the original, because the inductance goes as the square of the turn count at fixed length. (c) $L = 2.3$ H with the iron core, a thousandfold increase, since $L$ is directly proportional to the core permeability.

Self-induced emf from a current ramp

A 0.050 H inductor carries a current that rises steadily from zero to 4.0 A over 0.020 s. Find (a) the rate of change of current, (b) the magnitude of the self-induced emf during the ramp, (c) the emf after the current settles at a steady 4.0 A, and (d) the magnitude of the emf if instead the current were switched off in 0.0010 s at a steady rate.

(a) The current rises linearly, so the rate is constant: $\dfrac{dI}{dt} = \dfrac{4.0 \ \mathrm{A} - 0}{0.020 \ \mathrm{s}} = 200 \ \mathrm{A/s}$ .
(b) From 13.4.A.3, $\lvert \mathcal{E}_i \rvert = L \left\lvert \dfrac{dI}{dt} \right\rvert = (0.050)(200) = 10$ V, constant for the whole ramp because the rate is constant.
Direction: the current is increasing, so by the minus sign the induced emf opposes the increase. The inductor acts against whatever source is driving the current up.
(c) Once the current is steady, $dI/dt = 0$ and the self-induced emf is exactly zero. The inductor is electrically invisible, behaving as a plain wire. Note that a steady 4.0 A is a large current with a large magnetic field, and still no emf: the emf tracks the rate of change, not the current.
(d) Switching off means $\dfrac{dI}{dt} = \dfrac{0 - 4.0}{0.0010} = -4000 \ \mathrm{A/s}$ , so $\lvert \mathcal{E}_i \rvert = (0.050)(4000) = 200$ V.
That is twenty times the emf during the build-up, from the same inductor and the same current, purely because the change is twenty times faster. Now the sign reverses: the current is decreasing, so the inductor pushes to keep it going, which is why breaking an inductive circuit can arc across the switch contacts.
Sanity check on the units: $\mathrm{H} \cdot \mathrm{A/s} = (\mathrm{V \cdot s/A})(\mathrm{A/s}) = \mathrm{V}$ .

(a) 200 A/s. (b) 10 V, constant during the ramp and directed to oppose the rise. (c) exactly zero, because a steady current has no rate of change. (d) 200 V, twenty times larger, directed to keep the current flowing, which is why opening a switch on an inductive circuit produces a voltage spike.

Energy stored, computed two ways

For the same 0.050 H inductor carrying the current that ramps linearly from zero to 4.0 A in 0.020 s: (a) find the energy stored once the current reaches 4.0 A, (b) confirm that answer by integrating the power delivered during the ramp, and (c) state how much energy would be stored at 8.0 A.

(a) Use the printed relation from 13.4.A.2: $U_L = \dfrac{1}{2}LI^2 = \dfrac{1}{2}(0.050)(4.0)^2 = \dfrac{1}{2}(0.050)(16) = 0.40$ J.
(b) Now the integral, to check it. During the ramp the current is $I(t) = (200 \ \mathrm{A/s})\,t$ and the back emf has magnitude $10$ V, from the previous example.
The source delivers power $P = \mathcal{E}I = (10)(200t) = 2000t$ watts, growing linearly from zero.
Integrate over the ramp: $U = \displaystyle\int_0^{0.020} 2000\,t\,dt = 2000 \left[\frac{t^2}{2}\right]_0^{0.020} = 2000\left(\frac{(0.020)^2}{2}\right) = 2000 (2.0 \times 10^{-4}) = 0.40$ J.
The two agree exactly, which is the numerical version of the derivation: $\displaystyle\int L \frac{dI}{dt} I \, dt = \int_0^I L I' \, dI' = \frac{1}{2}LI^2$ , with the $dt$ cancelling in the change of variable.
(c) $U_L \propto I^2$ , so doubling the current to 8.0 A quadruples the stored energy to 1.6 J. Check directly: $\frac{1}{2}(0.050)(64) = 1.6$ J.
One more reading of part (b) worth keeping. The average power over the ramp is $0.40 \ \mathrm{J} / 0.020 \ \mathrm{s} = 20$ W, which is half the final power of 40 W, exactly as you would expect for a power that grows linearly from zero.
Where does the energy go afterwards? Statement 13.4.A.2.i answers: dissipated through a resistor, which is Topic 13.5, or used to charge a capacitor, which is Topic 13.6. And 13.4.A.2.ii says that transfer obeys conservation laws, so this 0.40 J is the number that must reappear in either destination.

(a) 0.40 J. (b) The same 0.40 J from integrating the delivered power, since the integral of L times dI by dt times I over time equals one half L I squared. (c) 1.6 J at 8.0 A, four times as much, because the stored energy goes as the square of the current.

Frequently asked questions

What is inductance in AP Physics C?

Essential knowledge 13.4.A.1 defines inductance as the tendency of a conductor to oppose a change in electrical current. The mechanism is self-induction: a current in a coil creates a magnetic field that passes through the coil's own turns, so changing the current changes that flux, which induces an emf by Faraday's law, and by Lenz's law that emf fights the change. Inductance depends only on the physical properties of the conductor, so straight wires are typically modeled as having zero inductance and an inductor such as a solenoid is a circuit element with significant inductance. The unit is the henry, one volt-second per ampere.

What does the inductance of a solenoid depend on?

Essential knowledge 13.4.A.1.iii names four things: the total number of turns, the length of the solenoid, the cross-sectional area, and the magnetic permeability of the core. The printed relation is the core permeability times the square of the turn count times the area, all divided by the length. The turn count appears squared, so doubling the number of turns on a solenoid of fixed length multiplies the inductance by four rather than two, because more turns both make a stronger field for a given current and provide more turns for that field to thread. Inserting an iron core raises the permeability and can increase the inductance by orders of magnitude.

Why does the current through an inductor not change instantly?

Because a sudden jump would require an infinite rate of change of current, and the self-induced emf equals the inductance times that rate. An infinite rate would demand an infinite emf, which no real source can supply, so the current is continuous through the instant a switch is thrown. The potential difference across an inductor is a different matter and can jump freely, because it depends on how fast the current is changing rather than on the current itself. This is the mirror image of a capacitor, whose potential difference cannot jump but whose current can.

How much energy is stored in an inductor?

One half the inductance times the square of the current, and essential knowledge 13.4.A.2 says that energy is stored in the magnetic field generated by the current in the inductor. The expression comes from integrating the power delivered against the back emf: the power is the inductance times the rate of change of current times the current, and integrating that over time turns into the integral of L times I with respect to I, giving one half L I squared. Because the dependence is quadratic, doubling the current stores four times as much energy. Essential knowledge 13.4.A.2.i adds that this stored energy can later be dissipated through a resistor or used to charge a capacitor.

What is the difference between an inductor and a resistor?

A resistor's potential difference depends on the current through it, and it converts electrical energy irreversibly into thermal energy. An inductor's potential difference depends on the rate of change of the current, not on the current, and it stores energy in a magnetic field rather than dissipating it, so the energy can be recovered later. A steady current through an inductor produces no potential difference at all, which is why an inductor behaves as a conducting wire with zero resistance long after a switch has closed. There is no fixed relationship between an inductor's current and its potential difference, so a straight-line graph of one against the other would be wrong.

Is inductance covered in AP Physics 2?

No. The words inductance and inductor do not appear anywhere in the AP Physics 2 course and exam description, and no AP Physics 2 topic covers self-induction, energy stored in a magnetic field, LR circuits or LC circuits. Topics 13.4 through 13.6 of AP Physics C: Electricity and Magnetism have no algebra-based counterpart at all. The nearest AP Physics 2 material is Topic 12.4 on electromagnetic induction and Faraday's law, which supplies the prerequisite, and Topic 10.6 on capacitors, which is the mirror-image circuit element.

How is an inductor like a mass in mechanics?

The correspondence is exact in the equations. Inductance plays the role of mass, current plays the role of velocity, and emf plays the role of force, so that emf equals inductance times the rate of change of current in the same way that force equals mass times the rate of change of velocity. The stored energy follows the same pattern: one half the inductance times the current squared, matching one half the mass times the velocity squared. This analogy is the reason essential knowledge 13.6.A.2 can say that the charge in an LC circuit can be modeled as simple harmonic motion, since an inductor and a capacitor together behave like a mass on a spring.