4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The emf ε of a battery is sometimes called its electromotive force. Which of the following correctly describes the emf?
Answer and reasoning
AA force, in newtons, that the battery exerts on each charge carrier to push it along the wires of the circuit A student who takes the word 'force' literally picks this. The emf is measured in volts (joules per coulomb): it is a potential difference, not a force.
BA current, in amperes, that the battery delivers unchanged to any circuit that is connected to it A student who thinks a battery is a source of a fixed current picks this. A battery maintains a potential difference; the current it delivers depends on the circuit connected to it.
CAn amount of charge, in coulombs, that the battery stores and then releases into the connected circuit A student who thinks a battery stores the charge that flows picks this. The charge carriers are already in the wires and other elements; the battery moves them, and its emf is a potential difference, in volts, not an amount of charge.
DA potential difference, in volts, that the battery maintains and that drives charge around a circuitCorrect Despite its name, emf is a potential difference: the energy per unit charge, in volts, that the battery supplies by maintaining a potential difference between its terminals. The charge carriers already in the wires move around the circuit in response to it.
A long copper wire connected across a battery carries a steady current. A student recalls that the electric field inside a conductor is zero. Which statement about the electric field inside this wire is correct?
Answer and reasoning
AIt is zero, as inside every conductor; the charge carriers keep on drifting because nothing acts to stop them. A student who applies the electrostatic result to a conductor carrying a current picks this. The steady current requires a field E = ρJ throughout the wire; without it the current would not continue.
BIt is zero in the interior; there is a field just at the surface, which is where the current flows. A student who carries over the idea that charge on a conductor resides on its surface picks this. In a uniform wire the current is spread over the whole cross section, and so is the field that drives it.
CIt is not zero: charge is moving, so the wire is not in electrostatic equilibrium, and a field drives it.Correct The field inside a conductor is zero only in electrostatic equilibrium, when no charge moves. Here the battery maintains a potential difference along the wire, and the field E = ρJ acts on the carriers everywhere in the wire, keeping them drifting.
DIt is not zero near the battery alone, where it pushes carriers that then push the rest along the wire. A student who pictures the battery pushing charges only at its terminals picks this. There is a field all along the wire: E = ρJ, and J is the same in every part of a uniform wire.
A long, straight wire of radius R carries a current whose density is directed along the wire and has magnitude J(r) = J₀r/R, where r is the distance from the wire's axis and J₀ is a positive constant. What is the current in the wire?
Answer and reasoning
A2πJ₀R²/3Correct Divide the cross section into thin rings of radius r and width dr, each of area 2πr dr, across which J is constant. I = ∫₀R (J₀r/R)2πr dr = (2πJ₀/R)(R³/3) = 2πJ₀R²/3.
BπJ₀R² A student who multiplies the current density J₀ by the whole cross-sectional area picks this. J₀ is the density only at the surface; everywhere inside J is smaller, so the current must be found by integrating.
CπJ₀R²/2 A student who averages J over the radius, getting J₀/2, and multiplies by πR² picks this. The outer rings, where J is largest, have the most area, so the average over the area is 2J₀/3, not J₀/2.
DJ₀R²/3 A student who takes the area of a thin ring as r dr picks this: ∫₀R (J₀r/R) r dr = J₀R²/3. A ring of radius r and width dr has circumference 2πr, so its area is 2πr dr.
Working dA = 2πr dr. I = ∫₀R (J₀r/R)(2πr) dr = (2πJ₀/R)(R³/3) = 2πJ₀R²/3 (coefficient 2 in the key). Distractors (sympy-checked): J₀πR²; (1/R)∫₀R J dr = J₀/2, times πR² = πJ₀R²/2; ∫₀R (J₀r/R) r dr = J₀R²/3.
A copper wire carries a current directed toward the right. Which statement describes the motion of the charge carriers that make up this current?
Answer and reasoning
APositive ions drift toward the right, while the free electrons stay in their places. A student who thinks the current in a metal is carried by positive charges, because conventional current is defined by positive charge, picks this. In a metal the positive ions are fixed; the free electrons are what move through it.
BFree electrons drift toward the left, while the positive ions drift toward the right. A student who carries over the picture of a current in a solution, where positive and negative ions both move, picks this. In a solid metal the positive ions stay in the lattice; the free electrons alone carry the current.
CFree electrons drift toward the left, while the positive ions stay in their places.Correct In a metal the current is carried by free electrons. Being negative, they drift opposite to the conventional current, to the left. The positive ions are held in the metal's lattice and only vibrate about their positions.
DFree electrons drift toward the right, the same direction as the conventional current. A student who takes the direction of the current to be the direction in which the electrons move picks this. The electrons are negative, so they drift opposite to the conventional current.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.1.A.1 Electric current, I Fix
Electric current, I
The rate at which charge passes through a cross section of a conductor: I = dq/dt. For a constant current, the charge passing in a time Δt is IΔt. Unit: ampere (A = C/s).
Charge and current graphs
Because I = dq/dt, the current at an instant is the slope of the graph of charge passed against time, and the charge that passes between t₁ and t₂ is q = ∫I dt, the area under the current–time graph between those times.
Drift velocity, v⃗d
The small average velocity, along the conductor, of the charge carriers in a conductor carrying a current; it is superposed on their much faster random motion. In metal wires it is typically of order 10⁻⁴ m/s. Unit: m/s.
I = nqvdA
The current carried by charge carriers of charge q and number density n (carriers per unit volume, m⁻³) drifting at speed vd through a cross section of area A. In a time Δt the carriers in a length vdΔt of the conductor, with total charge nqAvdΔt, pass the cross section.
Electromotive force (emf), ε
The potential difference that a source such as a battery provides: for an ideal battery, the potential difference it maintains between its terminals, which makes the charge already in a circuit move around it. Despite its name it is a potential difference, not a force. Unit: volt (V = J/C).
Random motion of charge carriers
Even when there is no current, the free electrons in a metal move in random directions at high speeds, of order 10⁶ m/s. Because their directions are random, their average velocity is zero and no net charge crosses any cross section, so the current is zero.
Students often think The current at any instant is q/t, the total charge that has passed divided by the time elapsed, as in the relation for a constant current. In fact Not in general. The current is the rate at which charge passes, I = dq/dt: the slope of the q–t graph at that instant. The ratio q/t of a point's coordinates equals that slope only when the graph is a straight line through the origin, that is, for a constant current that began at t = 0.
Students often think Current and charge are the same kind of quantity, so the charge that has passed at an instant, the height of the q–t graph, gives the current. In fact No. The value of q is the total charge that has passed, in coulombs. The current is the rate at which charge passes, in amperes (C/s), given by the slope of the q–t graph.
11.1.A.2 Current density, J⃗ Fix
Current density, J⃗
The current per unit area at a point in a conductor, directed along the motion of positive charge. The current through a surface is I = ∫J⃗·dA⃗, which reduces to I = JA for a uniform J⃗ perpendicular to an area A. Unit: A/m².
J⃗ = nqv⃗d
The current density produced by carriers of number density n, charge q (sign included) and drift velocity v⃗d. For negative carriers such as electrons, J⃗ points opposite to v⃗d.
Current density as a vector
At each point in a conductor, J⃗ has a magnitude and a direction in space. The contributions of different kinds of carriers at a point add as vectors.
E⃗ = ρJ⃗
In a conductor carrying a current, the electric field at a point equals the resistivity times the current density, so the field points along J⃗. A conductor carrying a current is not in electrostatic equilibrium, and the field inside it is not zero.
Resistivity, ρ
A property of a conducting material: the ratio E/J of the electric field to the current density in it. It is larger for poorer conductors. Unit: Ω·m.
Students often think Current density and current are the same quantity, so the current can be used in place of the current density (in J = nqvd or E = ρJ), and the reverse. In fact No. Current density J⃗ is the current per unit area at a point, in A/m², and is a vector. Current I is the total through a surface, in A: I = ∫J⃗·dA⃗, which is JA for a uniform J⃗ perpendicular to an area A.
Students often think Because current density means 'current per area', the current is found by dividing the current density by the area (or the current density by multiplying the current by the area). In fact No. Current density is current per unit area, so for a uniform J⃗ perpendicular to an area A, the current is I = JA: multiply by the area. Conversely J = I/A.
11.1.A.3 Current from a non-uniform current density Fix
Current from a non-uniform current density
When the current density along a wire depends on the distance r from its axis, the current is I = ∫J(r)2πr dr: each thin ring of radius r and width dr has area 2πr dr and carries current J(r)2πr dr.
Students often think The current in a wire is its current density times its cross-sectional area, I = JA, even when J varies across the section, using one value of J (such as the value given at the wire's surface). In fact No. I = JA holds only for a uniform current density. When J depends on the distance r from the axis, the current is I = ∫J(r)2πr dr, adding the currents through thin rings.
Students often think The average current density over a wire's cross section is the average of J over the radius, so the current is that average times πR². In fact No. Outer rings of a cross section have more area than inner rings of the same width (area 2πr dr), so the current density must be averaged over the area, weighting each value by its ring's area; averaging over the radius gives the wrong result unless J is uniform.
11.1.A.4 Current as a scalar with a direction Fix
Current as a scalar with a direction
A current has a size and a direction along its conductor, but the direction is relative to the conductor rather than a fixed direction in space. Currents therefore have no vector components, and currents meeting at a junction are added as signed numbers (into or out of the junction), not as vectors.
Conventional current
The direction in which positive charge would move. In a metal wire it is opposite to the drift velocity of the electrons and in the direction of E⃗ and J⃗.
Charge carriers in metals
In metal wires the current is carried by free (conduction) electrons, which are negative. The positive ions of the metal remain at their positions in the lattice, vibrating about them.
Students often think Charge carriers move in the direction of the electric field whatever the sign of their charge. In fact No. The force on a charge is qE⃗, so a negative charge is pushed opposite to the field. Electrons drift opposite to E⃗, while the conventional current and the current density point along E⃗.
Students often think The direction of the current is the direction in which the electrons in the wire move. In fact No. Conventional current is defined as the direction in which positive charge would move. In a metal wire the moving charges are electrons, which are negative, so they drift opposite to the direction of the conventional current.
20 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 20
The graph shows the total charge q that has passed through a cross section of a wire as a function of time t. What is the current in the wire at t = 4.0 s?
Answer and reasoning
A1.0 A A student who divides the charge at t = 4.0 s by the time, 4.0 C/4.0 s = 1.0 A, picks this. That ratio of coordinates is not the slope of this graph, because the graph is not a single straight line from the origin; the current is dq/dt, the slope of the segment containing t = 4.0 s.
B4.0 A A student who reads the charge that has passed at t = 4.0 s, 4.0 C, as the current picks this. The graph's height is a charge; the current is how fast that charge is increasing, the slope.
C6.0 A A student who takes the area under the q–t graph from 0 to 4.0 s, 1.0 C·s + 5.0 C·s = 6.0 C·s, as the current picks this. That area has units C·s and means nothing here; the current is the slope of the q–t graph.
D1.5 ACorrect Between t = 2.0 s and t = 6.0 s the graph is a straight line, so its slope is the current at every instant in that interval: I = dq/dt = (7.0 C − 1.0 C)/(6.0 s − 2.0 s) = 1.5 A.
Working The segment from (2.0 s, 1.0 C) to (6.0 s, 7.0 C) contains t = 4.0 s: I = dq/dt = 6.0 C/4.0 s = 1.5 A. (q at 4.0 s = 4.0 C; q/t = 1.0 A; area under q–t from 0 to 4.0 s = ½(2.0 s)(1.0 C) + ½(1.0 C + 4.0 C)(2.0 s) = 6.0 C·s.)
The current in a wire decreases with time t as I(t) = I₀e−t/T, where I₀ and T are positive constants. How much charge passes through a cross section of the wire from t = 0 to t = T?
Answer and reasoning
AI₀T(1+e⁻¹)/2 A student who averages the currents at the start and end of the interval, (I₀ + I₀e⁻¹)/2, and multiplies by T picks this. That average is correct only for a current that changes linearly; the decaying exponential lies below the straight line joining its end points, so the area under it is smaller.
BI₀T(1−e⁻¹)Correct Charge is the integral of current over time: q = ∫₀T I₀e−t/T dt = I₀[−Te−t/T]₀T = I₀T(1 − e⁻¹), about 0.63I₀T. This is the area under the I–t curve, less than I₀T because the current falls throughout the interval.
CI₀Te⁻¹ A student who uses q = It with the current at the end of the interval, I₀e⁻¹, picks this. q = It holds only for a constant current; here the current is larger than I₀e⁻¹ throughout the interval, so more charge passes.
DI₀e⁻¹/T A student who differentiates the current instead of integrating it picks this, taking the magnitude of dI/dt = −(I₀/T)e−t/T at t = T. Its units, A/s, show that it is a rate of change of current, not a charge; the charge is the area under the I–t graph.
Working q = ∫₀T I₀e−t/T dt = I₀T(1 − e⁻¹) ≈ 0.632I₀T. Distractors (sympy-checked): [(I₀ + I₀e⁻¹)/2]T = I₀T(1 + e⁻¹)/2 ≈ 0.684I₀T; I(T)·T = I₀Te⁻¹ ≈ 0.368I₀T; |dI/dt| at t = T = I₀e⁻¹/T (units A/s).
A copper wire of diameter 1.6 mm carries a steady current of 3.0 A. Measurements show that its free electrons, each with a charge of magnitude e = 1.60 × 10⁻¹⁹ C, drift at 1.1 × 10⁻⁴ m/s. What is the number of free electrons per unit volume in the copper?
Answer and reasoning
A8.5 × 10²⁸ m⁻³Correct The cross-sectional area is A = π(d/2)² = π(0.80 × 10⁻³ m)² ≈ 2.0 × 10⁻⁶ m². From I = nevdA, n = I/(evdA) = (3.0 A)/[(1.60 × 10⁻¹⁹ C)(1.1 × 10⁻⁴ m/s)(2.0 × 10⁻⁶ m²)] ≈ 8.5 × 10²⁸ m⁻³.
B1.7 × 10²³ m⁻³ A student who uses the current in place of the current density, n = I/(evd), picks this, leaving out the area. The 3.0 A is spread over the cross section, so the current per unit area, I/A, must be used; the units of I/(evd) are m⁻¹, not m⁻³.
C1.4 × 10¹⁰ m⁻³ A student who treats the current as the number of electrons passing per second, n = I/(vdA), picks this, leaving out the electron's charge. Each electron carries only 1.60 × 10⁻¹⁹ C, so the number of electrons needed to carry 3.0 A is far larger.
D3.4 × 10¹⁷ m⁻³ A student who multiplies the current by the area to find the current density, n = IA/(evd), picks this. Current density is current per unit area, J = I/A, so the area divides: n = I/(evdA).
Working A = π(0.80 × 10⁻³ m)² = 2.01 × 10⁻⁶ m². n = I/(evdA) = 3.0/[(1.60 × 10⁻¹⁹)(1.1 × 10⁻⁴)(2.01 × 10⁻⁶)] m⁻³ = 8.5 × 10²⁸ m⁻³.
A wire consists of two sections made of the same metal and joined end to end. The diameter of section 2 is half the diameter of section 1. The wire carries a steady current, which is the same in both sections. What is the ratio v₂/v₁ of the drift speed of the free electrons in section 2 to that in section 1?
Answer and reasoning
Av₂/v₁ = 1 A student who thinks equal currents mean equal drift speeds picks this. The same current through a smaller area requires faster drift: vd = I/(nqA).
Bv₂/v₁ = 2 A student who takes the area to be proportional to the diameter picks this: half the diameter, half the area, twice the drift speed. The area is proportional to the square of the diameter, so it falls to 1/4.
Cv₂/v₁ = 4Correct I = nqvdA, and I, n and q are the same in both sections, so the drift speed is inversely proportional to the cross-sectional area. Halving the diameter makes the area (1/2)² = 1/4 as large, so the electrons drift 4 times as fast in section 2.
Dv₂/v₁ = ¼ A student who thinks the narrower section holds the electrons back, slowing them in proportion to its smaller area, picks this. For the same current through 1/4 of the area, the electrons must drift faster, not slower.
Working I = nqvdA with I, n, q the same: v₂/v₁ = A₁/A₂. A ∝ d², so A₂ = A₁/4 and v₂/v₁ = 4.
A lamp is connected to a battery and a switch by copper wires several meters long. The drift speed of the free electrons in the wires is about 10⁻⁴ m/s, yet the lamp lights almost as soon as the switch is closed. Which reasoning explains this?
Answer and reasoning
AThe electrons from the battery travel to the lamp at nearly the speed of light, and they slow down to the small drift speed only after they have reached it. A student who thinks the charge carriers themselves travel at nearly the speed of light picks this. What spreads through the circuit that quickly is the electric field; each carrier only drifts, at about 10⁻⁴ m/s.
BFree electrons are already present all around the circuit, and closing the switch sets up a field in every part of it almost at once, so they all start drifting together.Correct The wires and the lamp's filament contain free electrons before the switch is closed. Closing it lets the battery's potential difference set up an electric field throughout the circuit almost instantly, so carriers everywhere, including those in the filament, begin drifting at once. No electron has to travel from the battery to the lamp.
CThe electrons' random speeds, about 10⁶ m/s, carry the first of them from the battery all the way to the lamp in a very short time after the switch is closed. A student who thinks the electrons' high random speeds carry them along the wire picks this. Those motions are in random directions and average to zero; they produce no net motion from the battery toward the lamp.
DClosing the switch lets the charge that is stored in the battery flood out into the wires at once, filling them up as far as the lamp almost instantly. A student who thinks the battery stores the charge that flows picks this. The charge carriers are already in the wires; the battery maintains a potential difference that makes them drift, and it releases no store of charge.
Wire X is not connected to anything. Wire Y, made of the same metal and at the same temperature, carries a steady current. How does the motion of the free electrons in X compare with that in Y?
Answer and reasoning
ATheir average speeds are about equal, but only in Y is their average velocity not zero.Correct In both wires the free electrons move in random directions at high speeds, of order 10⁶ m/s, so their average speeds are about the same. In X the random velocities average to zero; in Y the current adds a small drift velocity, of order 10⁻⁴ m/s, along the wire. The current changes the average velocity, not noticeably the average speed.
BThey are at rest in X, while in Y they all move along the wire at the small drift speed. A student who thinks free electrons are at rest when there is no current picks this. Zero current means zero net motion of charge, not zero speed: the electrons in X move rapidly in random directions, as do those in Y.
CTheir average speeds are about equal, and in Y those high speeds are directed along the wire. A student who thinks the electrons' high speeds are directed along the wire when there is a current picks this. Even in Y the motion is almost entirely random; only a tiny drift velocity, of order 10⁻⁴ m/s, is along the wire.
DNo electrons in wire X are free to move; in wire Y, electrons supplied by a battery drift. A student who thinks a wire has no mobile charge until a battery supplies it picks this. Free electrons are present in any metal, connected or not; a battery only makes them drift.
A student notes that the free electrons in a copper wire move at speeds of about 10⁶ m/s even when the wire is not connected to anything, and concludes that such a wire must carry a large current. Which reasoning shows that the conclusion is wrong?
Answer and reasoning
AThe electrons in a wire with no connections are at rest, so the speed quoted applies only when a current is present. A student who thinks free electrons stop when there is no current picks this. The student's premise is correct: the electrons do move at about 10⁶ m/s with no current. The conclusion fails because their directions are random.
BThe electrons in any wire move at the drift speed, about 10⁻⁴ m/s, far too slowly to give the wire a large current. A student who thinks the drift speed is the actual speed of the electrons picks this. The electrons do move at about 10⁶ m/s; the drift speed is only the small average velocity that a current adds to that motion.
CTheir directions are random, so each second as much charge crosses any cross section one way as the other way.Correct Current is the net rate at which charge crosses a cross section. Because the velocities are in random directions, electrons cross any cross section in both directions at the same average rate, so the net charge crossing per second, and so the current, is zero despite the high speeds.
DA current needs charge that a battery supplies, and the electrons that are already in the wire do not form a current. A student who thinks the charge in a current must come from a battery picks this. In a circuit the current is carried by the free electrons already in the wire; here the current is zero because their motion is random, not because they are the wrong charges.
A wire carries a current I. It is replaced by a wire of twice the diameter that carries a current 2I. In each wire the current is spread uniformly over the cross section. What is the ratio J₂/J₁ of the magnitude of the current density in the second wire to that in the first?
Answer and reasoning
AJ₂/J₁ = 1.0 A student who takes the area to be proportional to the diameter picks this: twice the current over twice the area. The area is proportional to the square of the diameter, so it is 4 times as large.
BJ₂/J₁ = 0.5Correct For a uniform current density perpendicular to the cross section, I = ∫J⃗·dA⃗ = JA, so J = I/A. Doubling the diameter makes the area 4 times as large, and the current doubles, so J₂/J₁ = (2I/4A)/(I/A) = 0.5.
CJ₂/J₁ = 2.0 A student who treats the current density as the current itself picks this: the current doubles, so J doubles. Current density is current per unit area, so the larger area must be taken into account.
DJ₂/J₁ = 8.0 A student who multiplies the current by the area to find the current density picks this: (2I)(4A)/(IA) = 8. Current density is current per unit area, J = I/A, so the area divides.
Working J = I/A for uniform J⃗ perpendicular to A. A ∝ d², so A₂ = 4A₁. J₂/J₁ = (2I/4A₁)/(I/A₁) = 1/2.
The diagram shows a wire carrying a steady current and three surfaces, 1, 2 and 3, each of which cuts completely across the wire. I₁, I₂ and I₃ are the currents through the surfaces. Which ranking of the currents is correct?
Answer and reasoning
AI₂ > I₁ > I₃ A student who takes the current through a surface to be proportional to its area picks this, ranking the surfaces by size. The current through a surface is ∫J⃗·dA⃗: only the component of J⃗ along the normal counts, and J is larger in the narrow section.
BI₁ = I₃ > I₂ A student who thinks a slanted surface intercepts less current, by the factor cos θ, picks this. The slanted surface is larger by the factor 1/cos θ, which exactly makes up for the smaller normal component, J cos θ.
CI₁ = I₂ > I₃ A student who thinks the narrow section lets less current through picks this. In a steady state charge cannot pile up where the wire narrows, so the same current passes through every cross section; it is the current density that is larger in the narrow section.
DI₁ = I₂ = I₃Correct The current is steady, so charge does not accumulate anywhere in the wire, and the charge that crosses one surface each second must cross the others. For the slanted surface 2, its larger area is offset by J⃗ making an angle with the surface's normal, so ∫J⃗·dA⃗ is unchanged; for surface 3, the smaller area is offset by the larger current density in the narrow section.
Working Steady state: no accumulation of charge between surfaces, so I₁ = I₂ = I₃. Surface 2 (normal at θ to J⃗): area A/cos θ, J⃗·n̂ = J cos θ, product JA. Surface 3: area A₃ < A, current density I/A₃ > I/A.
A beam of protons moving at 3.0 × 10⁶ m/s has a circular cross section of radius 5.0 cm, over which the protons are spread uniformly, with 2.0 × 10⁹ protons per cubic meter. What is the current in the beam? (e = 1.60 × 10⁻¹⁹ C)
Answer and reasoning
A7.5 × 10⁻⁶ ACorrect The current density is J = nqv = (2.0 × 10⁹ m⁻³)(1.60 × 10⁻¹⁹ C)(3.0 × 10⁶ m/s) = 9.6 × 10⁻⁴ A/m², uniform over the cross section. The current is J times the area: (9.6 × 10⁻⁴ A/m²)π(0.050 m)² ≈ 7.5 × 10⁻⁶ A.
B9.6 × 10⁻⁴ A A student who takes the current density as the current picks this, stopping at nqv = 9.6 × 10⁻⁴. That is the current per square meter, in A/m²; the beam's cross section is only about 7.9 × 10⁻³ m², so the current is smaller.
C4.7 × 10¹³ A A student who takes the current to be the number of protons passing per second, nvA ≈ 4.7 × 10¹³ s⁻¹, picks this. Each proton carries 1.60 × 10⁻¹⁹ C, and the current is the charge passing per second.
D1.2 × 10⁻¹ A A student who divides the current density by the area, reading 'per unit area' as a division to be done, picks this. J is current per unit area, so the current is J multiplied by the area.
Working J = nqv = (2.0 × 10⁹ m⁻³)(1.60 × 10⁻¹⁹ C)(3.0 × 10⁶ m/s) = 9.6 × 10⁻⁴ A/m². A = π(0.050 m)² = 7.85 × 10⁻³ m². I = JA = 7.5 × 10⁻⁶ A.
The free electrons in a metal wire of resistivity ρ have number density n and charge −e, and they drift with velocity v⃗d. What is the electric field E⃗ in the wire?
Answer and reasoning
Aρnev⃗d A student who thinks charge carriers move in the direction of the field, whatever their sign, picks this. The force on an electron is −eE⃗, opposite to the field, so electrons drift opposite to E⃗.
B−nev⃗d/ρ A student who divides by the resistivity, as if a material of larger resistivity needed a smaller field, picks this. E⃗ = ρJ⃗: for a given current density, the field is larger in a poorer conductor.
C−ρnev⃗dCorrect The current density is J⃗ = nqv⃗d = −nev⃗d, opposite to the electrons' drift because their charge is negative. In a conductor carrying a current, E⃗ = ρJ⃗, so E⃗ = −ρnev⃗d: the field points opposite to the electrons' drift velocity, and the force −eE⃗ on each electron points along it.
Dρnv⃗d A student who treats the current density as the number of carriers crossing unit area per second, leaving out each carrier's charge, picks this: J⃗ = nv⃗d, so E⃗ = ρnv⃗d. Each electron carries charge −e, so J⃗ = −nev⃗d and E⃗ = −ρnev⃗d; without the charge the expression has units of Ω/(m·s), not V/m.
Working J⃗ = nqv⃗d with q = −e: J⃗ = −nev⃗d. E⃗ = ρJ⃗ = −ρnev⃗d. Units: (Ω·m)(m⁻³)(C)(m/s) = Ω·A/m = V/m. Distractors: sign of q dropped, +ρnev⃗d (V/m); E⃗ = J⃗/ρ, −nev⃗d/ρ; charge omitted (m07): ρnv⃗d, units Ω/(m·s).
In a gas discharge tube, singly charged positive ions drift in the +x-direction and free electrons drift in the −x-direction. Which statement about the current density J⃗ in the tube is correct?
Answer and reasoning
AJ⃗ is in the +x-direction, but the electrons' motion makes it smaller than the ions' part. A student who thinks carriers moving in opposite directions produce opposing current densities picks this. For the electrons both q and v⃗d are reversed, so their contribution nqv⃗d is in +x and adds to the ions'.
BJ⃗ is in the −x-direction, which is the direction in which the electrons are drifting. A student who takes the direction of the current to be the direction in which the electrons move picks this. J⃗ points in the direction of qv⃗d, which for negative carriers is opposite to their motion.
CJ⃗ is in the +x-direction and is due to the ions alone; electrons do not add to it. A student who thinks conventional current counts only the motion of positive charges picks this. The convention fixes the direction of J⃗; negative carriers moving in −x contribute to J⃗ in +x just as positive carriers moving in +x do.
DJ⃗ is in the +x-direction; ion and electron contributions reinforce each other.Correct Each kind of carrier contributes nqv⃗d. The ions have q > 0 and v⃗d in +x, giving a contribution in +x. The electrons have q < 0 and v⃗d in −x, so nqv⃗d is also in +x. The two contributions point the same way and add.
A long wire of length ℓ and diameter d is made of a metal of resistivity ρ. It carries a current I spread uniformly over its cross section. What is the magnitude of the potential difference between the ends of the wire?
Answer and reasoning
AρIℓ/(πd²) A student who uses the diameter d as the radius in A = πr² picks this: A = πd², so J = I/(πd²), E = ρI/(πd²) and ΔV = Eℓ = ρIℓ/(πd²). The radius is d/2, so A = πd²/4, and the field and the potential difference are 4 times as large.
B4ρIℓ/(πd²)Correct The cross-sectional area is π(d/2)² = πd²/4, so the current density is J = 4I/(πd²), and the field in the wire is E = ρJ = 4ρI/(πd²), the same all along it. The potential difference between the ends is the field times the length: ΔV = Eℓ = 4ρIℓ/(πd²).
CρIℓ A student who uses the current in place of the current density, E = ρI, picks this. The field depends on the current per unit area, 4I/(πd²); the units of ρIℓ, V·m², show that something is missing.
D0 A student who applies the electrostatic result that there is no field inside a conductor picks this. A wire carrying a current is not in electrostatic equilibrium; the field E = ρJ along it produces a potential difference between its ends.
Working A = πd²/4. J = I/A = 4I/(πd²). E = ρJ = 4ρI/(πd²), uniform along the wire. |ΔV| = |−∫E⃗·dℓ⃗| = Eℓ = 4ρIℓ/(πd²) (coefficient 4 in the key). Units: (Ω·m)(A)(m)/m² = V. Distractors: A = πd² (m41) gives ρIℓ/(πd²) (units V); J replaced by I gives ρIℓ (V·m²); electrostatic E = 0 gives 0.
A wire made of two segments of different metals joined end to end has the same cross-sectional area throughout and carries a steady current. The graph shows the electric potential V as a function of position x along the wire. What is the resistivity ρ₂ of segment 2 in terms of the resistivity ρ₁ of segment 1?
Answer and reasoning
Aρ₂ = 12ρ₁Correct The field magnitude in each segment is the magnitude of the slope of V(x): 1.0 V over 0.40 m, or 2.5 V/m, in segment 1, and 6.0 V over 0.20 m, or 30 V/m, in segment 2. The current and the area are the same, so J is the same in both segments, and E = ρJ gives ρ₂/ρ₁ = (30 V/m)/(2.5 V/m) = 12.
Bρ₂ = 6.0ρ₁ A student who takes the potential difference across each segment, 6.0 V and 1.0 V, as the field in it picks this. The field is the potential difference per unit length; segment 2 is half as long as segment 1, so its field is 12 times, not 6 times, as great.
Cρ₂ = 0.083ρ₁ A student who thinks a larger field means a better conductor, as if E = J/ρ, picks this: (2.5 V/m)/(30 V/m) ≈ 0.083. E = ρJ: for the same current density, the segment with the steeper potential graph has the larger resistivity.
Dρ₂ = 0.63ρ₁ A student who takes the field to be large where the potential is large picks this, comparing the average potentials of the segments, 6.0 V and 9.5 V: 6.0/9.5 ≈ 0.63. The field depends on how fast V changes with x, the slope, not on the value of V.
Working E = |dV/dx|. Segment 1: (10.0 V − 9.0 V)/0.40 m = 2.5 V/m. Segment 2: (9.0 V − 3.0 V)/0.20 m = 30 V/m. Same I and A ⇒ same J; E = ρJ ⇒ ρ₂/ρ₁ = 30/2.5 = 12. (ΔV ratio 6.0/1.0 = 6.0; inverse of key 1/12 = 0.083; ratio of mean potentials 6.0/9.5 = 0.63.)
A copper wire carries a current, and the electric field in it has magnitude E₀. It is replaced by an aluminum wire whose diameter is 1.5 times as great, carrying 0.50 times the current. The resistivity of aluminum is 1.6 times that of copper. What is the magnitude of the electric field in the aluminum wire?
Answer and reasoning
A0.53E₀ A student who takes the area to be proportional to the diameter picks this: E₀(1.6)(0.50)/1.5 ≈ 0.53E₀. The area is proportional to the square of the diameter, so it is 2.25 times as great.
B0.14E₀ A student who divides by the resistivity, as if E = J/ρ, picks this: E₀(0.50)/[(1.6)(2.25)] ≈ 0.14E₀. E = ρJ: aluminum's larger resistivity increases the field for a given current density.
C0.80E₀ A student who uses the current in place of the current density picks this: E₀(1.6)(0.50) = 0.80E₀, ignoring the change in area. The field depends on the current per unit area, J = I/A.
D0.36E₀Correct E = ρJ = ρI/A. The resistivity is 1.6 times as great and the current 0.50 times as great, while the area is 1.5² = 2.25 times as great. So E = E₀(1.6)(0.50)/2.25 ≈ 0.36E₀.
Working E = ρI/A. Factors: ρ × 1.6, I × 0.50, A × 1.5² = 2.25. E = E₀(1.6)(0.50)/2.25 = 0.356E₀ ≈ 0.36E₀. (Area ∝ d: 0.53; E = J/ρ: 0.139; area ignored: 0.80.)
A long wire consists of a core of one metal surrounded by a layer of a different metal. The graph shows the current density J, which is directed along the wire, as a function of distance r from the wire's axis. What is the current in the wire?
Answer and reasoning
A31 A A student who averages J over the radius, (4.0 + 1.0)/2 × 10⁶ A/m², and multiplies by the whole area picks this. The outer layer has three times the core's area, so its smaller density counts for more; J must be weighted by area, not by radius.
B22 ACorrect J is constant in each region, so each region's current is J times its area. Core: (4.0 × 10⁶ A/m²)π(1.0 × 10⁻³ m)² ≈ 12.6 A. Outer layer, an annulus from 1.0 mm to 2.0 mm: (1.0 × 10⁶ A/m²)π[(2.0 × 10⁻³ m)² − (1.0 × 10⁻³ m)²] ≈ 9.4 A. Total ≈ 22 A.
C25 A A student who takes the outer layer's area to be the full disk, π(2.0 mm)², picks this, counting the core's area twice. The layer is an annulus of area π[(2.0 mm)² − (1.0 mm)²].
D13 A A student who uses the current density at the wire's surface, 1.0 × 10⁶ A/m², as if it applied over the whole cross section picks this. The core carries a larger current density and its current must be found separately.
Working I = ∫J dA. Core: (4.0 × 10⁶ A/m²)π(1.0 × 10⁻³ m)² = 12.57 A. Layer: (1.0 × 10⁶ A/m²)π[(2.0 × 10⁻³ m)² − (1.0 × 10⁻³ m)²] = 9.42 A. Total 22.0 A ≈ 22 A.
The diagram shows three wires joined at point P and the currents in wires 1 and 2. What is the magnitude of the current in wire 3?
Answer and reasoning
A5.8 A A student who adds currents as vectors picks this: setting 5.0 A to the right equal to the vector sum of 3.0 A downward and the current in wire 3 gives √(5.0² + 3.0²) ≈ 5.8 A. A current's direction is along its wire, not in space; currents at a junction add as signed numbers.
B5.0 A A student who thinks only the part of each current directed along wire 3 can enter it picks this: all 5.0 A of wire 1's current and none of wire 2's perpendicular current. Currents have no components; what matters is how much charge per second arrives at and leaves P.
C2.0 ACorrect Charge does not build up at P, so the charge arriving each second equals the charge leaving: 5.0 A arrives through wire 1 and 3.0 A leaves through wire 2, so 2.0 A leaves through wire 3. The directions of the wires in space play no part.
D8.0 A A student who thinks a scalar current has no direction adds the two currents, 5.0 A + 3.0 A. Each current has a direction along its wire: the 5.0 A flows into P but the 3.0 A flows out, so they subtract.
Working Conservation of charge at P (no accumulation): I₁(in) = I₂(out) + I₃(out), so I₃ = 5.0 A − 3.0 A = 2.0 A, leaving P. Vector (wrong): √(5.0² + 3.0²) = 5.8 A; component along wire 3: 5.0 A; no direction: 8.0 A.
The diagram shows the direction of the drift velocity of the free electrons in a section of copper wire. Which statement correctly gives the directions of the conventional current and of the electric field in the wire?
Answer and reasoning
AThe current and the field both point toward the left. A student who takes the direction of the current to be the direction in which the electrons move picks this. By convention the current points the way positive charge would move, opposite to the electrons' drift, and the field points the same way as the current.
BThe current points right; the field points toward the left. A student who thinks charges move along the field whatever their sign picks this, putting the field in the direction of the electrons' drift. The force on a negative charge, qE⃗, is opposite to E⃗, so electrons drifting to the left mean a field pointing to the right.
CThe current points right, and the field in the wire is zero. A student who applies the electrostatic result that the field inside a conductor is zero picks this. A wire carrying a current is not in electrostatic equilibrium; the field E⃗ = ρJ⃗ points along the current.
DThe current and the field both point toward the right.Correct Conventional current is the direction in which positive charge would move, opposite to the electrons' drift: to the right. In a conductor carrying a current the field is E⃗ = ρJ⃗, parallel to the current density, so it also points to the right; it exerts a force to the left on the negative electrons, the direction in which they drift.
A copper wire of length ℓ and diameter d carries a steady current I, spread uniformly over its cross section. The wire contains n free electrons per unit volume, each with a charge of magnitude e. How long does a free electron, moving at the drift speed, take to travel the length of the wire?
Answer and reasoning
Anπed²ℓ/(4I)Correct From I = nevdA with A = π(d/2)² = πd²/4, the drift speed is vd = 4I/(nπed²). At this constant speed the electron takes t = ℓ/vd = nπed²ℓ/(4I). Equivalently, t is the free charge in the wire, neAℓ, divided by the rate I at which charge passes a cross section.
Bnπd²ℓ/(4I) A student who treats the current as the number of electrons passing per second, I = nvdA, picks this. Each electron carries charge e, so I = nevdA; without e the expression has units of s/C, not seconds.
Cnπed²ℓ/I A student who uses the diameter d as the radius in A = πr² picks this. The radius is d/2, so A = πd²/4; using πd² makes the area, and so the time, four times too large.
Dneℓ/I A student who uses the current I in place of the current density in J = nevd picks this, taking vd = I/(ne). The current density is I/A, so the area πd²/4 is missing, and the expression has units of s/m², not seconds.
Working A = π(d/2)² = πd²/4. From I = nevdA, vd = I/(neA) = 4I/(nπed²). t = ℓ/vd = nπed²ℓ/(4I), which is also the free charge in the wire, neAℓ, divided by the current. Units: (m⁻³)(C)(m²)(m)/(C/s) = s. Distractors (sympy-checked, each different from the key at test values): A = πd² gives nπed²ℓ/I (s); omitting e gives nπd²ℓ/(4I) (s/C); I in place of J in J = nevd gives vd = I/(ne) and t = neℓ/I (s/m²).
A long wire of length ℓ consists of a solid core of radius a and resistivity ρ, surrounded by a tightly fitting layer of a different metal, of resistivity ρ/4, that extends from r = a to r = 2a, where r is the distance from the wire's axis. Each end of the wire is attached to a thick metal plate, and the two plates are held at a potential difference ΔV. What is the current in the wire?
Answer and reasoning
A17πa²ΔV/(ρℓ) A student who takes the layer's area as the full disk π(2a)² picks this, counting the core's area a second time. The layer is an annulus of area π[(2a)² − a²] = 3πa².
B13πa²ΔV/(ρℓ)Correct Both metals span the same two equipotential plates, so each has the same field E = ΔV/ℓ along the wire. From E = ρJ, J = ΔV/(ρℓ) in the core and 4ΔV/(ρℓ) in the layer. Adding J times the area of each region: [ΔV/(ρℓ)]πa² + [4ΔV/(ρℓ)]π(4a² − a²) = 13πa²ΔV/(ρℓ).
C16πa²ΔV/(ρℓ) A student who multiplies the current density at the wire's surface, 4ΔV/(ρℓ), by the whole cross section π(2a)² picks this. The core has only a quarter of that current density, so its current must be found separately.
D7πa²ρΔV/(4ℓ) A student who writes J = ρE, as if E = J/ρ, picks this, giving the layer of smaller resistivity the smaller current density. From E = ρJ, J = E/ρ, so the layer of resistivity ρ/4 carries four times the core's current density; the expression also has units of V·Ω, not A.
Working Each end face is at a single potential, so both metals have the potential difference ΔV across the same length ℓ and the same uniform field E = ΔV/ℓ along the wire. E = ρJ gives J = ΔV/(ρℓ) in the core and J = 4ΔV/(ρℓ) in the layer. I = ∫J dA = [ΔV/(ρℓ)]πa² + [4ΔV/(ρℓ)]π[(2a)² − a²] = πa²ΔV/(ρℓ) + 12πa²ΔV/(ρℓ) = 13πa²ΔV/(ρℓ) (coefficient 13 in the key). Units: m²·V/(Ω·m·m) = A. Distractors (sympy-checked): layer area taken as π(2a)² gives (1 + 16)πa²ΔV/(ρℓ) = 17πa²ΔV/(ρℓ); the surface value J = 4ΔV/(ρℓ) times π(2a)² gives 16πa²ΔV/(ρℓ); J = ρE gives (1 + 3/4)πa²ρΔV/ℓ = 7πa²ρΔV/(4ℓ) (units V·Ω, not A).
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
