AP Physics C: E&M · Topic 11.1

Topic 11.1: Electric Current

Unit 11: Electric Circuits15-25% of the multiple-choice section

Electric current is the rate at which charge passes through a cross-sectional area. AP Physics C defines it as a derivative, I = dq/dt, and adds current density: current through a surface is the integral of J over that area. Current is a scalar with a direction, never resolved into components.

AP Physics: Unit 11 (topics 11.1 Electric Current). Topic 11.1 of the current AP Physics C: Electricity and Magnetism course and exam description, in Unit 11 Electric Circuits (15 to 25% of the multiple-choice section, about 12 to 24 class periods). One learning objective, 11.1.A, describe the movement of electric charges through a medium, carrying twelve essential knowledge statements: 11.1.A.1 with sub-statements i to iii, 11.1.A.2 with sub-statements i to iii, 11.1.A.3, and 11.1.A.4 with sub-statements i and ii. The topic prints no boundary statement. Relevant equations are I = dq/dt (11.1.A.1), I = n q v_d A (11.1.A.1.i), I = the integral of J dotted with dA (11.1.A.2), J = n q v_d (11.1.A.2.i) and E = rho J (11.1.A.2.iii); 11.1.A.3 carries the Derived equation label for the total current as the integral of J of r dotted with dA. Of these, I = dq/dt, the flux integral for current and E = rho J are printed on the equation sheet; the two drift-velocity forms and the derived integral are not. Suggested skills 1.A, 2.A, 2.D and 3.B. Learning objective 11.1.A is one of five objectives aligned to sample free-response Question 4 in the course description.

What Topic 11.1 requires

Topic 11.1 of AP Physics C: Electricity and Magnetism Unit 11 has one learning objective and twelve essential-knowledge statements. It prints no boundary statement, so nothing in the framework fences off the treatment here.

11.1.A, describe the movement of electric charges through a medium.

11.1.A.1 Current is the rate at which charge passes through a cross-sectional area of a wire. Relevant equation $I = \dfrac{dq}{dt}$ .
11.1.A.1.i Current within a conductor consists of charge carriers traveling through the conductor with an average drift velocity. Relevant equation $I = n q v_d A$ .
11.1.A.1.ii Electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or emf $(\mathcal{E})$ .
11.1.A.1.iii If the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed.
11.1.A.2 Current density is the flow of charge per unit area. Relevant equation $I = \int \vec{J} \cdot d\vec{A}$ .
11.1.A.2.i Current density is related to the motion of the charge carriers within a conductor. Relevant equation $\vec{J} = n q \vec{v}_d$ .
11.1.A.2.ii Current density is a vector quantity.
11.1.A.2.iii A potential difference across a conductor creates an electric field within the conductor that is proportional to the resistivity of the conductor and the current density. Relevant equation $\vec{E} = \rho \vec{J}$ .
11.1.A.3 If a function of current density is given, the total current can be determined by integrating the current density over the area. Derived equation $I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$ .
11.1.A.4 Although current is a scalar quantity, it does have a direction. Because its direction is relative to the current carrier and not space, current does not obey the laws of vector addition and has no vector components.
11.1.A.4.i The direction of conventional current is chosen to be the direction in which positive charge would move.
11.1.A.4.ii In common circuits, the current is actually due to the movement of electrons (negative charge carriers).

The suggested skills are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Skill 2.A is the tell. Of the eight topics in Unit 11, only 11.1, 11.5, 11.6 and 11.8 list the derivation skill, and all four of those are places where the framework hands you an equation to build rather than a number to compute.

The definition is a derivative, and the equation sheet prints it that way

The single sharpest difference between this topic and its algebra-based twin is one character on the formula sheet.

Course	What the sheet prints
AP Physics 2	$I = \dfrac{\Delta q}{\Delta t}$
AP Physics C: E&M	$I = \dfrac{dq}{dt}$

Same physical idea, different question answered. $\Delta q / \Delta t$ answers "how much charge crossed, on average, over this interval". $dq/dt$ answers "what is the current right now", which is the only version that survives contact with a current that changes.

Two consequences follow immediately, and both get examined.

Going forward in time, charge is an integral. If you know $I(t)$ , then the charge that passes a cross-section between $t_1$ and $t_2$ is

q = \int_{t_1}^{t_2} I(t)\, dt

That is the area under a current-versus-time graph, and it is how a discharging capacitor's total charge, a nonconstant charging current, or a lab data set gets turned into a coulomb count. The relation is not printed on the sheet, because it is just the inverse of $I = dq/dt$ , but the sheet's calculus table prints the integral rules you need to evaluate it.

Going backward in time, current is a slope. Given $q(t)$ , the current at an instant is the tangent slope of the charge-versus-time graph. Skill 1.B is listed on Topics 11.3, 11.7 and 11.8 for exactly this reason: the Experimental Design and Analysis free-response question wants data plotted so that a slope means something.

Be careful with the symbol. The framework writes $I = dq/dt$ with a lowercase $q$ , and the sheet's symbol list defines both $q$ and $Q$ as charge. Nothing hangs on which one a question uses.

One quantity that is not the derivative of anything here: the average current. If a question gives you a graph and asks for an average, you want total charge over total time, not the average of the endpoint values, and those agree only when the current is linear in time.

Current density, the quantity AP Physics 2 never meets

Statements 11.1.A.2 through 11.1.A.3 are entirely absent from the AP Physics 2 framework. They are the largest single addition this topic makes, and they are worth more than the derivative notation.

Current density is charge flow per unit area, and it is a vector. Statement 11.1.A.2 defines it and 11.1.A.2.ii makes the vector claim explicit. Read those two beside 11.1.A.4, which says current is a scalar, and the pairing stops being confusing: $\vec{J}$ is a vector field defined at every point inside the conductor, and $I$ is the scalar you get after integrating that field over a surface. The same relationship holds between the electric field and electric flux.

Its microscopic form ties it to the carriers. Statement 11.1.A.2.i gives

\vec{J} = n q \vec{v}_d

where $n$ is the number of mobile carriers per unit volume, $q$ the charge on each, and $\vec{v}_d$ the average drift velocity from 11.1.A.1.i. Multiply both sides by a uniform cross-sectional area $A$ and you recover $I = n q v_d A$ , which is the form 11.1.A.1.i prints. They are the same statement, one written per unit area and one written for the whole wire.

Its macroscopic form ties it to the field inside the metal. Statement 11.1.A.2.iii is the one students skip and then need:

\vec{E} = \rho \vec{J}

A potential difference across a conductor creates a field inside it, proportional to the resistivity and the current density. This is the microscopic version of Ohm's law, and it is the reason a current-carrying conductor is not an equipotential the way an electrostatic conductor is. Unit 10 tells you the field inside a conductor in electrostatic equilibrium is zero; Unit 11 puts a current through it and the field is no longer zero. Nothing contradicts, because the conductor is no longer in electrostatic equilibrium.

Units are a useful check on $\vec{E} = \rho \vec{J}$ : an ohm metre times an amp per square metre is $(\Omega \cdot \mathrm{m}) (\mathrm{A}/\mathrm{m}^2) = \Omega \cdot \mathrm{A}/\mathrm{m} = \mathrm{V}/\mathrm{m}$ .

Integrating a current density (11.1.A.3)

Statement 11.1.A.3 says that if a function of current density is given, the total current can be determined by integrating the current density over the area, and it carries the Derived equation label:

I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}

The CED's Required Equations page defines that label. Not all equations in the framework appear on the equation sheet; many are provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam, and those are the Derived Equations. The general form $I = \int \vec{J} \cdot d\vec{A}$ from 11.1.A.2 is printed. The $\vec{J}(r)$ version is the same integral with the dependence made visible, and it is telling you that a radially varying current density is fair game.

The machinery is the machinery you already built in Unit 8 for Gauss's law. Three moves, every time:

Choose an area element that matches the symmetry. For a cylindrical wire with $J$ depending only on distance $r$ from the axis, the element is a thin annulus, $dA = 2\pi r\, dr$ , because $J$ is constant on it.
Take the dot product. If $\vec{J}$ is parallel to $d\vec{A}$ , which it is for current straight down a wire through a cross-section, the dot product is just $J\, dA$ and the vectors drop out.
Integrate over the range the surface covers, usually $0$ to $R$ .

The dot product is doing real work when the surface is tilted. A cross-section cut at an angle to the wire axis has a larger area but the same total current, because $\cos\theta$ in the dot product cancels the extra area exactly. That is the same argument that makes flux through a tilted surface independent of the tilt when nothing is enclosed.

A sanity check to apply every time: divide your answer by the cross-sectional area and compare with the largest and smallest values of $J$ on the surface. The average current density has to land between them. If it does not, you have an algebra error.

Drift velocity, and why the bulb lights immediately

Statement 11.1.A.1.i says current within a conductor consists of charge carriers traveling through the conductor with an average drift velocity. The word doing the work is average. Individual carriers in a metal move fast and in every direction; the drift velocity is the small net bias on top of that random motion.

Statement 11.1.A.1.iii makes the same point from the other side, and it is the most quotable line in the topic: if the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed. Zero current means zero net transport, not stillness.

Drift speeds in a copper wire carrying an ordinary current come out around $10^{-4}$ to $10^{-5}$ metres per second, which is slower than a walking pace by five orders of magnitude. The bulb still lights the instant you close the switch, because what propagates is the field, not the carriers. Statement 11.1.A.2.iii is the mechanism: the potential difference establishes $\vec{E} = \rho \vec{J}$ throughout the conductor, and every carrier along the whole length starts drifting at once.

11.1.A.1.ii names the driver: charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or emf. Two things to hold on to about that word. Emf is measured in volts and is not a force, despite the name. And Topic 11.5 sharpens the definition at 11.5.B.1.iii: the emf of a battery is the potential difference measured across its terminals when there is no current in the battery. Once current flows through a real battery, the terminal potential difference is smaller.

The voltage against current comparison is the place to go if the two quantities still blur together.

Current is a scalar with a direction, quoted whole

Statement 11.1.A.4 is one of the few places where the two courses use noticeably different words for the same idea, and the C version is stricter.

AP Physics C: "Although current is a scalar quantity, it does have a direction. Because its direction is relative to the current carrier and not space, current does not obey the laws of vector addition and has no vector components."

AP Physics 2 instead says current is not a vector quantity and that its direction is associated with what the motion of positive charge would be but not with any coordinate system in space.

The C wording adds the operational consequence, and it is the one that shows up in a wrong answer: current has no vector components. You never write $I_x$ and $I_y$ . You never add two currents at a junction with the parallelogram rule. At a junction you use Kirchhoff's junction rule, which is signed scalar bookkeeping along wires, and the geometry of the wires is irrelevant to it.

Current density, on the other hand, is a genuine vector by 11.1.A.2.ii, with components, direction in space, and a dot product. So in this topic one quantity resolves into components and the other does not, and which is which is worth a moment. The scalar against vector comparison covers the general distinction.

Two conventions close the topic. 11.1.A.4.i fixes conventional current as the direction positive charge would move. 11.1.A.4.ii admits the physical truth: in common circuits the current is actually electrons going the other way. The Topic 11.2 boundary statement then makes conventional current the default for every schematic on the exam unless a question says otherwise, and the exam's own reference information repeats it as a standing convention: the direction of current is the direction in which positive charges would drift.

What the equation sheet prints for Topic 11.1

Six equations appear in Topic 11.1's required content. Three are printed on the sheet and three are not. Here is each one checked against the AP Physics C: E&M Table of Information directly, rather than recalled.

Equation	CED statement	On the sheet
$I = \dfrac{dq}{dt}$	11.1.A.1, relevant	yes
$I = n q v_d A$	11.1.A.1.i, relevant	no
$I = \int \vec{J} \cdot d\vec{A}$	11.1.A.2, relevant	yes
$\vec{J} = n q \vec{v}_d$	11.1.A.2.i, relevant	no
$\vec{E} = \rho \vec{J}$	11.1.A.2.iii, relevant	yes
$I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$	11.1.A.3, derived	no

Two of the three that are missing are the microscopic ones, both involving $n$ and $v_d$ , and they are missing together because they are the same equation. If you can write either, you have both: multiply $\vec{J} = nq\vec{v}_d$ by a uniform area and you get $I = nqv_dA$ . The third absentee, the derived $I_{\text{tot}}$ integral, is the printed $I = \int \vec{J} \cdot d\vec{A}$ with the radial dependence written in, so nothing new has to be remembered for it either.

The sheet's symbol list is worth reading once for this topic. It defines $J$ as current density and $\rho$ as "resistivity or charge density", which is the one genuinely ambiguous symbol in the course. In $\vec{E} = \rho\vec{J}$ and $R = \rho\ell/A$ it is resistivity, measured in ohm metres. In $Q_{\text{total}} = \int \rho(r)\, dV$ , printed a few lines above on the same sheet, it is volume charge density in coulombs per cubic metre. The equation tells you which.

Because the C: E&M sheet reprints the whole C: Mechanics table, a few mechanics lines are available here too. The one worth remembering for Unit 11 is $P_{\text{inst}} = dW/dt$ , which is what lets electric power be integrated over time to get energy.

How Topic 11.1 is tested, and who each page is for

Learning objective 11.1.A appears in the CED's own sample free-response set, on Question 4, alongside 10.3.A, 11.7.A, 11.3.B and 11.8.B. That question is the Qualitative/Quantitative Translation type, worth 8 points, and it is built on an RC circuit driven by an ideal constant current source. Its part B rubric awards a point for substituting the current in the capacitor branch as $dq/dt$ into an equation expressing Kirchhoff's junction rule.

That is the shape to expect. Topic 11.1 rarely arrives as a question about current on its own. It arrives as the definition you need in the middle of something else, and the thing it most often unlocks is treating a capacitor branch current as a derivative of charge.

The science-practice weighting explains the rest. On the free-response section, Science Practice 2 carries 40 to 45%, the largest of the three, and skill 2.A alone carries 25 to 30% of the multiple-choice section. Symbolic answers are the norm, so expect $I = \pi J_0 R^2 / 3$ rather than 3.8 amperes.

The [AP Physics 2 Topic 11.1 page](/ap-physics-2/unit-11-electric-circuits/11-1-electric-current) is for AP Physics 2 students; this page is for AP Physics C students. The algebra-based course stops at $I = \Delta q / \Delta t$ , conventional current, and the zero-current statement. It never mentions current density, drift velocity as a defined quantity with its own equation, or the field inside a conductor. If your syllabus is Physics 2, that page covers your topic completely and this one adds material you will not be assessed on.

From here, Topic 11.3 turns $\vec{E} = \rho\vec{J}$ into $R = \rho\ell/A$ and then into Ohm's law, and Topic 11.8 is where $I = dq/dt$ stops being notation and becomes the term that makes a circuit equation differential. The full unit map is on the Unit 11 hub.

A current that changes with time, both ways

The current in a wire increases from zero as $I(t) = \alpha t^2$ , with $\alpha = 3.0\ \mathrm{A/s^2}$ . Find (a) the charge that passes a cross-section between $t = 0$ and $t = 4.0$ s, (b) the average current over that interval, and (c) the instant at which the instantaneous current equals that average.

(a) Statement 11.1.A.1 gives $I = dq/dt$ . Inverting it, the charge is the time integral of the current: $q = \int_0^{4.0} \alpha t^2\, dt$ . The sheet's calculus table prints $\int x^n dx = \dfrac{1}{n+1}x^{n+1}$ for $n \neq -1$ , so $q = \dfrac{\alpha t^3}{3}\bigg|_0^{4.0} = \dfrac{(3.0)(4.0)^3}{3} = \dfrac{(3.0)(64)}{3} = 64$ C.
(b) The average current is total charge over total time, which is what $\Delta q / \Delta t$ means: $I_{\text{avg}} = \dfrac{64\ \mathrm{C}}{4.0\ \mathrm{s}} = 16$ A. This is the quantity the AP Physics 2 sheet prints, and it is the only one that survives if you never learn the derivative form.
(c) Set the instantaneous current equal to it: $3.0 t^2 = 16$ , so $t^2 = 5.333\ \mathrm{s^2}$ and $t = 2.31$ s. Round to $2.3$ s.
Read the result. The instantaneous current reaches the average value at $t = 2.3$ s, not at the midpoint $t = 2.0$ s, because the current is not linear in time. Confirm the endpoints: at $t = 4.0$ s the instantaneous current is $(3.0)(16) = 48$ A, three times the average, which is the signature of a quadratic starting from zero.
Units check on $\alpha$ : amps per second squared times seconds squared gives amps, and $\alpha t^3/3$ gives amp seconds, which is coulombs.

(a) $q = 64$ C. (b) $I_{\text{avg}} = 16$ A. (c) $t = 2.3$ s, later than the midpoint of the interval because the current grows quadratically. The instantaneous current at the end of the interval is 48 A, three times the average.

Total current from a nonuniform current density

A long cylindrical conductor of radius $R = 2.0$ mm carries a current directed along its axis with current density $J(r) = J_0\left(1 - \dfrac{r}{R}\right)$ , where $r$ is the distance from the axis and $J_0 = 9.0 \times 10^5\ \mathrm{A/m^2}$ . Find (a) the total current, symbolically and then numerically, and (b) the uniform current density that would carry the same current.

(a) Statement 11.1.A.3 gives $I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$ . The current is along the axis and the cross-section is perpendicular to it, so $\vec{J}$ and $d\vec{A}$ are parallel and the dot product is $J\, dA$ .
Choose the area element to match the symmetry. $J$ depends only on $r$ , so use a thin annulus of radius $r$ and thickness $dr$ : $dA = 2\pi r\, dr$ . Every point on that annulus has the same current density, which is the whole reason for the choice.
$I = \displaystyle\int_0^R J_0\left(1 - \frac{r}{R}\right) 2\pi r\, dr = 2\pi J_0 \int_0^R \left(r - \frac{r^2}{R}\right) dr = 2\pi J_0 \left[\frac{R^2}{2} - \frac{R^3}{3R}\right] = 2\pi J_0 R^2\left(\frac{1}{2} - \frac{1}{3}\right)$
$\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$ , so $I = \dfrac{\pi J_0 R^2}{3}$ . Stop and read it: that is exactly one third of $J_0 \pi R^2$ , the current the wire would carry if the density were $J_0$ everywhere. Leaving the answer in this form is what skill 2.A asks for.
Numerically, $R^2 = (2.0 \times 10^{-3})^2 = 4.0 \times 10^{-6}\ \mathrm{m^2}$ , so $I = \dfrac{\pi (9.0 \times 10^5)(4.0 \times 10^{-6})}{3} = \dfrac{\pi (3.6)}{3} = 1.2\pi = 3.77$ A, or $3.8$ A to two figures.
(b) A uniform density carrying the same current satisfies $J_{\text{unif}} \pi R^2 = \pi J_0 R^2 / 3$ , so $J_{\text{unif}} = J_0/3 = 3.0 \times 10^5\ \mathrm{A/m^2}$ .
Sanity check, the one to run every time: $J$ ranges from $J_0 = 9.0 \times 10^5$ at the axis down to zero at the surface, and the average $3.0 \times 10^5$ sits between them. It is below the midpoint $4.5 \times 10^5$ because the annuli near the surface, where $J$ is small, carry the most area.

(a) $I = \dfrac{\pi J_0 R^2}{3} = 3.8$ A, one third of what a uniform $J_0$ would deliver. (b) The equivalent uniform current density is $J_0/3 = 3.0 \times 10^5\ \mathrm{A/m^2}$ .

Drift speed and the field inside the metal

A copper wire of cross-sectional area $A = 3.0 \times 10^{-6}\ \mathrm{m^2}$ carries a steady current of $1.8$ A. Copper has $n = 8.5 \times 10^{28}$ mobile electrons per cubic metre, each of charge magnitude $e = 1.60 \times 10^{-19}$ C, and resistivity $\rho = 1.7 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ . Find (a) the current density, (b) the drift speed, and (c) the electric field inside the metal, and check (c) against $R = \rho \ell / A$ .

(a) The current is uniform across the wire, so the integral of 11.1.A.2 collapses to a product: $J = \dfrac{I}{A} = \dfrac{1.8}{3.0 \times 10^{-6}} = 6.0 \times 10^{5}\ \mathrm{A/m^2}$ .
(b) Statement 11.1.A.2.i gives $\vec{J} = n q \vec{v}_d$ , so $v_d = \dfrac{J}{ne} = \dfrac{6.0 \times 10^{5}}{(8.5 \times 10^{28})(1.60 \times 10^{-19})}$ . The denominator is $1.36 \times 10^{10}\ \mathrm{C/m^3}$ , giving $v_d = 4.41 \times 10^{-5}$ m/s, or $4.4 \times 10^{-5}$ m/s to two figures.
That is about 0.044 millimetres per second. A carrier would take roughly six hours to travel one metre of this wire, and yet a switch at one end changes the current at the other end essentially at once. Statement 11.1.A.1.iii and the field mechanism of 11.1.A.2.iii are what resolve that.
(c) Statement 11.1.A.2.iii gives $\vec{E} = \rho \vec{J}$ , so $E = (1.7 \times 10^{-8})(6.0 \times 10^{5}) = 1.02 \times 10^{-2}$ V/m, about $1.0 \times 10^{-2}$ V/m.
Check it the long way round. One metre of this wire has $R = \dfrac{\rho \ell}{A} = \dfrac{(1.7 \times 10^{-8})(1.0)}{3.0 \times 10^{-6}} = 5.67 \times 10^{-3}\ \Omega$ . Ohm's law in the printed form $I = \Delta V / R$ gives $\Delta V = (1.8)(5.67 \times 10^{-3}) = 1.02 \times 10^{-2}$ V across that metre, and a uniform field over one metre is $E = \Delta V / \ell = 1.02 \times 10^{-2}$ V/m. The two routes agree, which is the point: $\vec{E} = \rho \vec{J}$ and $R = \rho \ell / A$ are the same physics at two scales.
Units check: $(\Omega \cdot \mathrm{m})(\mathrm{A}/\mathrm{m}^2) = \Omega \cdot \mathrm{A}/\mathrm{m} = \mathrm{V}/\mathrm{m}$ .

(a) $J = 6.0 \times 10^{5}\ \mathrm{A/m^2}$ . (b) $v_d = 4.4 \times 10^{-5}$ m/s. (c) $E = 1.0 \times 10^{-2}$ V/m inside the copper, confirmed independently by computing the resistance of a one-metre length and dividing its potential difference by the length.

Frequently asked questions

What is the definition of electric current in AP Physics C?

Current is the rate at which charge passes through a cross-sectional area of a wire, and the AP Physics C: Electricity and Magnetism course description defines it as a derivative, I = dq/dt. That form is printed on the C: E&M equation sheet. The algebra-based AP Physics 2 sheet prints the finite-difference version instead, I = the change in charge divided by the change in time, which gives an average rather than an instantaneous value. The derivative form is what lets a Physics C question use a current that varies with time, and inverting it means the charge delivered over an interval is the time integral of the current, which is the area under a current-versus-time graph.

What is current density and how is it different from current?

Current density is the flow of charge per unit area at a point inside a conductor, and unlike current it is a vector quantity. The AP Physics C course description states both facts, at essential knowledge 11.1.A.2 and 11.1.A.2.ii. Current is the scalar you get by integrating current density over a surface, through the flux integral I equals the integral of J dotted with dA, which is printed on the equation sheet. Two further relations pin current density down: it equals n q times the drift velocity, where n is the carrier number density, and the electric field inside a conductor equals the resistivity times the current density. Current density does not appear anywhere in the AP Physics 2 framework.

Is current a vector or a scalar?

Current is a scalar quantity that nonetheless has a direction. The AP Physics C course description says so explicitly at essential knowledge 11.1.A.4: because its direction is relative to the current carrier and not to space, current does not obey the laws of vector addition and has no vector components. The practical consequence is that you never resolve a current into x and y components and never combine currents with a parallelogram rule. At a junction you use Kirchhoff's junction rule, which is signed scalar bookkeeping and is indifferent to the geometry of the wires. Current density, by contrast, is a genuine vector with components and a direction in space.

How do you find the total current from a current density that varies with radius?

Integrate the current density over the cross-sectional area. The AP Physics C course description gives this as a derived equation at essential knowledge 11.1.A.3: the total current is the integral of J of r dotted with dA. For a cylindrical wire whose current density depends only on distance from the axis, choose a thin annular area element, dA equal to 2 pi r dr, so that the current density is constant across the element, then integrate from the axis to the outer radius. The dot product reduces to a simple product whenever the current is along the wire and the surface is a cross-section perpendicular to it. A useful check is that the total current divided by the cross-sectional area must lie between the smallest and largest values of the current density on that surface.

Why does a light turn on instantly if drift speed is so slow?

Because what travels quickly is the electric field, not the charge carriers. Drift speeds in an ordinary copper wire are of order ten to the minus four or minus five metres per second, so an individual electron takes hours to cross a metre of wire. Essential knowledge 11.1.A.2.iii in the AP Physics C course description supplies the mechanism: a potential difference across a conductor creates an electric field within the conductor, equal to the resistivity times the current density. That field is established along the whole length of the circuit almost immediately, so every carrier everywhere in the wire begins drifting at nearly the same moment, and the bulb receives energy at once.

Is I = dq/dt on the AP Physics C E&M equation sheet?

Yes. The AP Physics C: Electricity and Magnetism Table of Information prints I = dq/dt, and also prints I equals the integral of J dotted with dA and the field relation E equals rho times J. Two equations from this topic are not printed: the drift-velocity forms I = n q v sub d A and J = n q v sub d. Those two are the same statement written for a whole wire and for a point, so producing either one gives you the other. The derived-equation form for a varying current density, total current equals the integral of J of r dotted with dA, is also not printed, but the general version of that integral is.

What is the difference between AP Physics C Topic 11.1 and AP Physics 2 Topic 11.1?

Both courses have a Topic 11.1 called Electric Current with the same learning objective, describe the movement of electric charges through a medium. AP Physics C adds current density in full: what it is, that it is a vector, its relation to drift velocity, the electric field it creates inside the conductor, and the area integral that recovers the total current. None of that appears in the AP Physics 2 framework. AP Physics C also defines current as a derivative, dq over dt, where AP Physics 2 uses the change in charge over the change in time, and the two courses print those different forms on their respective equation sheets.