3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Which statement best describes the effect of an inductor on the current in the circuit that contains it?
Answer and reasoning
AIt opposes the current itself, reducing even a steady current. A student who thinks an inductor acts like a resistor picks this. With a steady current, dI/dt = 0 and an ideal inductor has no effect; it opposes only changes.
BIt opposes increases in its current but lets a decrease happen freely. A student who links an inductor only to delaying a rising current picks this. When the current decreases, the induced emf acts to keep it going, opposing the decrease.
CIt passes current only until charge builds up on its turns, then stops it. A student who thinks an inductor stores charge as a capacitor does expects the current to stop once the inductor is charged. An inductor stores no net charge; once its current is steady, an ideal inductor behaves as a wire.
DIt opposes any change in its current, increase or decrease.Correct Inductance is the tendency to oppose a change in current. The induced emf, −L dI/dt, opposes increases and decreases alike, and it is zero when the current is steady.
An ideal inductor that carries a current is connected across a resistor, with no battery in the circuit, and its current decreases to zero. Which statement about energy during this process is correct?
Answer and reasoning
AThe fall in the inductor's stored energy equals the energy the resistor dissipates.Correct Energy is conserved. The ideal inductor dissipates nothing; as its current falls, the energy stored in its magnetic field is transferred to the resistor, which dissipates exactly that amount as thermal energy.
BThe inductor itself turns its stored energy into thermal energy as it opposes the change. A student who thinks opposing a change means dissipating energy, as a resistor does, picks this. An ideal inductor has no resistance and dissipates nothing; the resistor dissipates the energy.
CThe inductor's stored energy disappears once the current and its magnetic field reach zero. A student who thinks the energy vanishes with the field picks this. Energy is conserved: it is transferred to the resistor as the current decreases.
DThe resistor dissipates more energy than was stored, as the inductor's emf adds more. A student who treats every emf as an energy source picks this. With no battery, the only energy available is the (1/2)LI² stored at the start; the inductor's emf only transfers it.
The graph shows the current I in an ideal inductor of inductance 0.040 H as a function of time t. What is the magnitude of the emf induced across the inductor at t = 9.0 ms?
Answer and reasoning
A1.3 V A student who divides the current at 9.0 ms by the elapsed time picks this: (0.040 H)(0.30 A ÷ 0.0090 s) = 1.3 V. The current did not change at a constant rate since t = 0; the emf uses the slope at 9.0 ms.
B0.024 V A student who multiplies L by the change in current, without dividing by the time taken, picks this: (0.040 H)(0.60 A) = 0.024 V·s. The emf depends on the rate of change, 0.60 A in 2.0 ms.
C0 V A student who thinks an inductor has an emf only while its current increases picks this. The current is decreasing at 9.0 ms, and the inductor opposes that decrease with an emf of magnitude L|dI/dt|.
D12 VCorrect At t = 9.0 ms the current is on the segment falling from 0.60 A at 8 ms to 0 at 10 ms, so |dI/dt| = 0.60 A ÷ 0.0020 s = 300 A/s. |ε| = L|dI/dt| = (0.040 H)(300 A/s) = 12 V.
Working Between 8 and 10 ms, I falls 0.60 A → 0: |dI/dt| = 0.60/0.0020 = 300 A/s. |ε| = 0.040 × 300 = 12 V.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
13.4.A.1 Inductance, L Fix
Inductance, L
The tendency of a conductor to oppose a change in the current in it. It is the ratio of the total magnetic flux through the conductor's turns (the flux linkage NΦB) to the current that produces it, L = NΦB/I. SI unit: henry (H = Wb/A = V·s/A).
Opposition to a change, not to the current
An inductor's induced emf depends on dI/dt, so it opposes increases and decreases in the current alike; it does not oppose a steady current. An ideal inductor (zero resistance) carrying a steady current has zero potential difference across it and behaves as a wire.
Dependence of inductance on physical properties
A conductor's inductance is set by its shape, size and number of turns and by the magnetic permeability of the material inside it, not by the current, the potential difference or the rest of the circuit. A straight wire is usually modeled as having no inductance.
Inductor
A circuit element with significant inductance, such as a solenoid (a wire wound into a helical coil). Its circuit symbol is a row of loops.
Inductance of a solenoid, Lsol
For a long solenoid of N turns, length ℓ and cross-sectional area A with a core of permeability μcore, Lsol = μcore N²A/ℓ. It grows as the square of the number of turns because both the field inside and the number of turns the flux passes through are proportional to N. SI unit: H.
Magnetic permeability of the core, μcore
A property of the material inside a solenoid that sets how large a field a given current produces there. For an air (or vacuum) core, μcore = μ0 = 4π × 10⁻⁷ T·m/A; a ferromagnetic core has a much larger permeability. SI unit: T·m/A.
Students often think An inductor opposes the current itself, so it acts like an extra resistance and reduces even a steady current. In fact No. An inductor opposes changes in the current. Once the current is steady, an ideal inductor has no potential difference across it and behaves as a wire; it does not reduce the steady current.
Students often think An inductor opposes only increases in current; when the current decreases, the inductor has no effect and no emf. In fact No. It opposes decreases just as much: when the current falls, the induced emf acts to keep it going. Its emf, −L dI/dt, is nonzero whenever the current changes in either direction.
13.4.A.2 Energy stored in an inductor, ULFix
Energy stored in an inductor, UL
UL = (1/2)LI², the energy stored in the magnetic field produced by current I in an inductor of inductance L. It is the work done against the induced emf while the current is built up from zero. SI unit: joule (J).
Transfer of an inductor's energy
When an inductor's current decreases, the energy released from its magnetic field goes to the rest of the circuit: a resistor dissipates it as thermal energy, or a capacitor stores it as electric potential energy.
Energy conservation with an inductor
The decrease in UL equals the energy delivered to the rest of the circuit; an ideal inductor itself dissipates nothing.
Students often think An inductor has a magnetic field and stores energy only while its current is changing; with a steady current the field and the stored energy are zero. In fact No. Any current in an inductor produces a magnetic field, and UL = (1/2)LI² depends on the current, not on its rate of change. A steady current stores a steady energy.
Students often think The energy stored in an inductor is proportional to its current, so doubling the current doubles the energy and halving the current halves it. In fact No. UL = (1/2)LI² is proportional to the square of the current: doubling the current makes the stored energy four times as large, and halving it leaves one-quarter.
13.4.A.3 Self-induced emf, εi = −L dI/dt Fix
Self-induced emf, εi = −L dI/dt
Faraday's law applied to an inductor, with the flux linkage equal to LI, gives the emf induced by the inductor's own changing current: εi = −L dI/dt. Its magnitude depends on the rate of change of the current, not on the current; its direction opposes the change. Going through an ideal inductor in the direction of the current, the potential drops by L dI/dt.
Students often think While the current increases, the inductor's induced emf acts in the direction of the current, so the potential rises across the inductor in the direction of the current. In fact No. The induced emf opposes the increase, so going through the inductor in the direction of the increasing current the potential drops by L dI/dt, as it does across a resistor.
Students often think The potential difference across an inductor is proportional to the current in it, like IR for a resistor, with L in the role of R. In fact No. It is proportional to the rate of change of the current: |ΔVL| = L|dI/dt|. A large steady current gives zero potential difference across an ideal inductor, and a zero current that is changing gives a nonzero one.
13 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 13
In the circuit shown, the bulbs B1 and B2 are identical and the inductor is ideal. When switch S is closed, B2 lights at once while B1 brightens gradually; a long time later, B1 and B2 are equally bright. Which claim about the inductor do these observations support?
Answer and reasoning
AIt opposes the current in its branch, so it acts as an extra resistance. A student who thinks an inductor opposes the current itself picks this. An extra resistance would leave B1 dimmer than B2 for as long as S is closed; the observation that they end equally bright contradicts it.
BIt stores charge while S is closed, just as a capacitor in its place would. A student who treats an inductor like a capacitor picks this. A capacitor in B1's branch would let B1 light at once and then go out as it charged; the observations show the opposite.
CIt opposes changes in the current in its branch, not a steady current.Correct B1's gradual brightening shows the inductor opposing the rise of the current in its branch. Once the current is steady, B1 is as bright as B2, which has no inductor, so the inductor does not reduce a steady current. Both observations fit a device that opposes only changes.
DIt has a magnetic field only while the current in its branch is changing. A student who thinks an inductor's field exists only while its current changes picks this. The observations say nothing about the field in the steady state, and the claim is false: a steady current in the inductor produces a steady field.
A coil of copper wire is connected in a circuit with a battery. Which of the following changes would change the inductance of the coil?
Answer and reasoning
ADoubling the current in the coil by using a battery of greater emf A student who thinks inductance depends on the current picks this. Doubling the current doubles the flux linkage, so their ratio, the inductance, is unchanged.
BStretching the coil to twice its length, keeping the same number of turnsCorrect Inductance depends on the coil's physical properties. Stretching the coil spreads the same turns over twice the length, which halves the field inside for a given current and so halves the inductance (Lsol = μcore N²A/ℓ).
CAdding a resistor in series with the coil in the same circuit A student who treats inductance as a kind of resistance, which adds in series, picks this. A resistor changes the circuit's resistance, not the coil's inductance, which depends only on the coil.
DRewinding the coil identically with aluminum wire in place of copper A student who thinks inductance depends on the wire's material, as resistance does, picks this. Aluminum is not magnetic, and the coil's turns, length and area are the same, so the inductance is unchanged; only its resistance changes.
In analyzing a circuit, which of the following elements is modeled as having significant inductance?
Answer and reasoning
AA tightly wound solenoidCorrect A solenoid links the field of its own current through many turns, so it has significant inductance; that is what makes it an inductor. Straight wires are modeled as having zero inductance, and capacitors and resistors are modeled without inductance.
BA long straight wire A student who thinks inductance grows with the length of wire, as resistance does, picks this. Straight wires are typically modeled as having zero inductance; it is coiling the wire that gives significant inductance.
CA charged capacitor A student who thinks an inductor works by storing charge, as a capacitor does, takes a charged capacitor to be an inductor and picks this. A capacitor stores energy in an electric field and is modeled with capacitance, not inductance; an inductor stores no net charge.
DA high-value resistor A student who thinks inductance is a kind of resistance picks this. A resistor opposes the current itself and dissipates energy; it is modeled as having no inductance.
A length W of thin insulated wire is wound into a single-layer solenoid of radius r and length ℓ, with ℓ much greater than r. The core is air. What is the inductance of the solenoid?
Answer and reasoning
Aμ₀Wr/(2ℓ) A student who takes the inductance to be proportional to N rather than N² gets μ₀NA/ℓ = μ₀(W/2πr)(πr²)/ℓ = μ₀Wr/(2ℓ). The field inside and the number of linked turns are each proportional to N, so L ∝ N².
Bμ₀W²/(4πℓ)Correct Each turn uses a length 2πr of wire, so N = W/(2πr). With A = πr², L = μ₀N²A/ℓ = μ₀W²πr²/(4π²r²ℓ) = μ₀W²/(4πℓ): for a given length of wire, the radius cancels.
Cμ₀W²/(2πrℓ) A student who uses the circumference 2πr of a turn in place of its area gets μ₀(W/2πr)²(2πr)/ℓ = μ₀W²/(2πrℓ). The area in Lsol is the cross-sectional area πr².
Dμ₀πr²/ℓ A student who thinks the inductance depends only on the solenoid's shape leaves out N and gets μ₀A/ℓ. The number of turns, set here by the length of wire, enters as N².
Working N = W/(2πr); A = πr². L = μ₀N²A/ℓ = μ₀(W²/4π²r²)(πr²)/ℓ = μ₀W²/(4πℓ). Units: (T·m/A)(m²)/m = T·m²/A = Wb/A = H.
An air-core solenoid 0.20 m long has 400 turns, each of radius 1.0 cm. What is the inductance of the solenoid? (μ₀ = 4π × 10⁻⁷ T·m/A)
Answer and reasoning
A7.9 × 10⁻⁷ H A student who uses N instead of N² picks this: (4π × 10⁻⁷)(400)(3.14 × 10⁻⁴)/0.20 = 7.9 × 10⁻⁷ H. Both the field and the number of linked turns grow with N, so L ∝ N².
B6.3 × 10⁻² H A student who uses the circumference, 2π(0.010 m) = 0.063 m, in place of the area picks this. A is the cross-sectional area, π(0.010 m)².
C3.2 × 10⁻⁴ HCorrect A = π(0.010 m)² = 3.14 × 10⁻⁴ m². L = μ₀N²A/ℓ = (4π × 10⁻⁷ T·m/A)(400)²(3.14 × 10⁻⁴ m²)/(0.20 m) = 3.2 × 10⁻⁴ H.
D1.3 × 10⁻⁵ H A student who multiplies by the length, as for resistance, picks this: (4π × 10⁻⁷)(400)²(3.14 × 10⁻⁴)(0.20). Spreading the turns over a greater length weakens the field, so ℓ is in the denominator.
Working A = π(0.010)² = 3.142 × 10⁻⁴ m². L = 4π×10⁻⁷ × 400² × 3.142×10⁻⁴ / 0.20 = 3.16 × 10⁻⁴ H ≈ 3.2 × 10⁻⁴ H.
Three long air-core solenoids have these properties. Solenoid 1: N turns, length ℓ, cross-sectional area A. Solenoid 2: 2N turns, length 2ℓ, area A. Solenoid 3: N turns, length ℓ/2, area A/2. Which ranks their inductances L₁, L₂ and L₃?
Answer and reasoning
AL₁ = L₂ = L₃ A student who takes L ∝ NA/ℓ gets 1, 2/2 = 1 and (1/2)/(1/2) = 1 for all three. With N², doubling N quadruples the inductance, which outweighs doubling the length.
BL₂ > L₁ > L₃ A student who puts the length in the numerator, as for resistance, gets N²Aℓ: 1, 8 and 1/4. A longer solenoid with the same turns has a weaker field and less inductance, so ℓ belongs in the denominator.
CL₁ = L₃ > L₂ A student who ignores the number of turns ranks by A/ℓ: 1, 1/2 and 1. Solenoid 2's doubled number of turns multiplies its inductance by four.
The steady current in an inductor is doubled. By what factor is the energy stored in the inductor multiplied?
Answer and reasoning
A×4Correct UL = (1/2)LI². The inductance is a property of the inductor and does not change, so doubling I multiplies the energy by 2² = 4.
B×2 A student who takes the stored energy to be proportional to the current picks this. UL depends on I², so doubling the current quadruples the energy.
C×8 A student who thinks the inductance grows in proportion to the current doubles L as well as I: (2)(2²) = 8. L depends only on the inductor's construction, so it stays the same.
D×1 A student who thinks an inductor stores energy only while its current changes says a steady current stores none either way, so nothing changes. A steady current produces a steady field storing (1/2)LI².
A long air-core solenoid has N turns, length ℓ and radius r, and carries a steady current I. What is the energy stored in the solenoid?
Answer and reasoning
AUL = πμ₀NI²r²/(2ℓ) A student who takes the inductance to be proportional to N uses L = μ₀N(πr²)/ℓ. Both the field and the number of turns linked are proportional to N, so L ∝ N².
BUL = πμ₀N²I²r²/ℓ A student who takes the stored energy to be LI², leaving out the factor 1/2, picks this. The flux linkage grows from zero with the current, so the work done against the induced emf is (1/2)LI².
CUL = πμ₀N²I²r²/(2ℓ)Correct L = μ₀N²(πr²)/ℓ and UL = (1/2)LI² = πμ₀N²I²r²/(2ℓ).
DUL = πμ₀NI²r²/ℓ A student who takes the stored energy as the flux through the solenoid times the current, ΦB I with ΦB = μ₀(N/ℓ)I·πr², picks this. The energy is (1/2)LI² = (1/2)NΦB I: the flux must be multiplied by the number of turns it links, and the 1/2 comes from building the current up from zero.
Working L = μ₀N²πr²/ℓ; U = (1/2)LI² = πμ₀N²I²r²/(2ℓ). Units J. Distractors: m07 L ∝ N → πμ₀NI²r²/(2ℓ); m12 LI² → πμ₀N²I²r²/ℓ; m21 ΦB I → πμ₀NI²r²/ℓ.
An ideal inductor of inductance 0.50 H, connected only to a resistor, carries a current of 4.0 A. The current then decreases to 2.0 A. How much energy does the resistor dissipate while the current decreases?
Answer and reasoning
A2.0 J A student who takes the energy to be proportional to the current thinks halving the current releases half the initial 4.0 J. UL ∝ I², so at 2.0 A only one-quarter, 1.0 J, remains and 3.0 J has been released.
B3.0 JCorrect The resistor dissipates the energy released by the inductor: ΔUL = (1/2)L(I₁² − I₂²) = (1/2)(0.50 H)(16 A² − 4.0 A²) = 3.0 J.
C1.0 J A student who puts the change in current into the energy formula picks this: (1/2)(0.50 H)(2.0 A)² = 1.0 J. The energy released is the difference between the stored energies, (1/2)L(I₁² − I₂²).
D6.0 J A student who takes the stored energy to be LI², without the factor 1/2, picks this: (0.50 H)(16 A² − 4.0 A²) = 6.0 J. The stored energy is (1/2)LI².
Working ΔU = ½L(I₁² − I₂²) = ½(0.50)(4.0² − 2.0²) = 0.25 × 12 = 3.0 J.
The diagram shows an ideal inductor of inductance L between points X and Y. The current in the direction shown is I = I₀ + βt², where I₀ and β are positive constants. What is the potential difference VY − VX at time t?
Answer and reasoning
A2Lβt A student who thinks the induced emf acts with an increasing current has the potential rise from X to Y. The emf opposes the increase, so Y is at the lower potential.
B−L(I₀ + βt²) A student who treats the inductor like a resistor, with a potential drop LI in the direction of the current, picks this. The potential difference across an inductor depends on the rate of change of the current, L dI/dt, not on the current.
C−2LβtCorrect dI/dt = 2βt. The induced emf opposes the increase in current, so going from X to Y, in the direction of the current, the potential drops by L dI/dt: VY − VX = −L dI/dt = −2Lβt.
D0 A student who thinks an ideal inductor, having no resistance, has no potential difference across it picks this. While the current changes, the induced emf gives a potential difference of magnitude L dI/dt.
The graph shows the current I in an ideal inductor as a function of time t, with points P, Q and R marked on the curve. Which ranks the magnitudes of the emf induced across the inductor at the instants P, Q and R?
Answer and reasoning
AQ > R > P A student who takes the emf to be proportional to the current ranks by the height of the curve: Q highest, then R, then P. The emf depends on the slope, and at Q the slope is zero.
BP > R > QCorrect |ε| = L|dI/dt|, so the ranking follows the steepness of the curve. It is steepest at P, where the current is rising quickly; less steep at R, where it is falling slowly; and level at Q, where the current is momentarily constant, so the emf there is zero.
CP > Q = R A student who thinks an inductor has an emf only while its current rises gives Q and R zero emf. At R the current is decreasing, and the inductor opposes the decrease with a nonzero emf.
DP > Q > R A student who uses the current divided by the elapsed time, I/t, for the rate of change picks this: I/t falls from P to Q to R. The emf depends on the slope at each instant, which is zero at Q.
Working |ε| ∝ |slope|: steep at P, zero at Q (flat maximum), small at R (gentle decline).
A long air-core solenoid has N turns, length ℓ and radius r. The current in the solenoid produces a uniform magnetic field of magnitude B inside it. Which expression gives the energy stored in the solenoid?
Answer and reasoning
AπB²r²ℓ/μ₀ A student who takes the stored energy to be LI², with no factor 1/2, gets (μ₀N²πr²/ℓ)(Bℓ/(μ₀N))² and picks this, twice the stored energy. UL = (1/2)LI², because the flux linkage grows from zero as the current is built up.
BπB²rℓ/μ₀ A student who puts the circumference 2πr in place of the cross-sectional area in Lsol gets L = 2πμ₀N²r/ℓ and picks this. The flux through each turn is B times the area πr² that the turn encloses, and this expression does not even have the units of energy.
CπNBr²/2 A student who takes the stored energy to be proportional to the current, U = (1/2)LI, gets (1/2)(μ₀N²πr²/ℓ)(Bℓ/(μ₀N)) and picks this. The stored energy is (1/2)LI², proportional to the square of the current; this expression is not even an energy (its unit is the weber).
DπB²r²ℓ/(2μ₀)Correct B = μ₀NI/ℓ gives I = Bℓ/(μ₀N). With L = μ₀N²πr²/ℓ, U = (1/2)LI² = (1/2)(μ₀N²πr²/ℓ)(Bℓ/(μ₀N))² = πB²r²ℓ/(2μ₀). For a given field the stored energy depends only on B and the volume πr²ℓ the field fills, not on N.
Working Field inside a long solenoid: B = μ₀NI/ℓ, so I = Bℓ/(μ₀N). Inductance: L = μ₀N²A/ℓ with A = πr², L = μ₀N²πr²/ℓ. UL = (1/2)LI² = (1/2)(μ₀N²πr²/ℓ)(B²ℓ²/(μ₀²N²)) = πB²r²ℓ/(2μ₀), that is, B²/(2μ₀) times the volume πr²ℓ that the field fills; N cancels. Units: T²·m³/(T·m/A) = T·m²·A = Wb·A = J. Distractors (checked with sympy; B = 0.2 T, r = 0.01 m, ℓ = 0.3 m, N = 500 give key 1.5 J): U = LI² gives πB²r²ℓ/μ₀ = 3.0 J; circumference 2πr in place of the area gives πB²rℓ/μ₀ (units J/m); energy taken ∝ current, U = (1/2)LI, gives πNBr²/2 (units Wb).
An ideal inductor of inductance L carries a current I₀. At t = 0 it is connected across an uncharged capacitor of capacitance C, with no resistance and no other elements in the circuit. What is the magnitude of the charge on the capacitor at the instant when the current in the inductor has decreased to I₀/3?
Answer and reasoning
A0.82I₀√(LC) A student who takes the inductor’s energy to be proportional to its current thinks one-third of the energy remains at I₀/3, gives the capacitor q²/(2C) = (2/3)(1/2)LI₀², and picks this. The energy depends on I², so at I₀/3 the inductor keeps only 1/9 of it.
B1.33I₀√(LC) A student who takes the inductor’s energy to be LI², with no factor 1/2, finds that the capacitor receives (8/9)LI₀² and picks this. The inductor stores (1/2)LI₀², so the capacitor receives (4/9)LI₀².
C0.94I₀√(LC)Correct With no resistance, the energy the inductor gives up goes to the capacitor: (1/2)LI₀² − (1/2)L(I₀/3)² = q²/(2C). So q²/(2C) = (4/9)LI₀², q² = (8/9)LCI₀², and q ≈ 0.94I₀√(LC).
D0.67I₀√(LC) A student who finds the energy given up by putting the change in current, ΔI = 2I₀/3, into (1/2)L(ΔI)² gives the capacitor (2/9)LI₀² and picks this. The energy given up is the difference of the stored energies, (1/2)LI₀² − (1/2)L(I₀/3)².
Working With no resistance, the energy the inductor gives up is stored in the capacitor: (1/2)LI₀² = (1/2)L(I₀/3)² + q²/(2C). So q²/(2C) = (1/2)LI₀²(1 − 1/9) = (4/9)LI₀², q² = (8/9)LCI₀², q = (2√2/3)I₀√(LC) = 0.943I₀√(LC) ≈ 0.94I₀√(LC). Units: A·√(H·F) = A·s = C. Distractors (sympy): energy taken ∝ current, capacitor gets (2/3)(1/2)LI₀², q = √(2/3) = 0.816 → 0.82; UL = LI², capacitor gets (8/9)LI₀², q = 4/3 → 1.33; energy given up taken as (1/2)L(ΔI)² with ΔI = 2I₀/3, capacitor gets (2/9)LI₀², q = 2/3 → 0.67.
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
