AP Physics C: E&M · Topic 13.1

Topic 13.1: Magnetic Flux

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

Magnetic flux is how much magnetic field passes through a surface. In general it is the surface integral of the field over that area. When the field is constant across the area it collapses to a dot product of the field and the area vector, so the sign comes from the angle between them.

AP Physics: Unit 13 (topics 13.1 Magnetic Flux). AP Physics C: Electricity and Magnetism Unit 13, Topic 13.1. One learning objective, 13.1.A, describe the magnetic flux through an arbitrary area or geometric shape. Supported by 13.1.A.1 (for a magnetic field B that is constant across an area A, the magnetic flux through the area is defined as the dot product of B and A), 13.1.A.1.i (the area vector is defined as perpendicular to the plane of the surface and outward from a closed surface), 13.1.A.1.ii (the sign of flux is given by the dot product of the magnetic field vector and the area vector), and 13.1.A.2 (the total magnetic flux passing through a surface is defined by the surface integral of the magnetic field over the surface area, with the relevant equation Phi_B equals the integral of B dot dA). Topic 13.1 prints no boundary statement; the only boundary statement in Unit 13 sits under Topic 13.2 and concerns deriving the speed of light from Maxwell's equations. The equation sheet prints magnetic flux only as the surface integral, not as a plain dot product or in a B A cos theta form, which is a difference from the AP Physics 2 sheet. Suggested skills are 1.A, 2.A, 2.C and 3.B. Unit 13 is weighted 10 to 20% of the multiple-choice section over about 10 to 20 class periods. The CED's own sample free-response Question 2 aligns to 12.4.A, 13.1.A and 13.2.A and turns on using the solenoid's radius rather than the encircling loop's.

What Topic 13.1 requires

Topic 13.1 has one learning objective and four essential-knowledge statements under it.

13.1.A, describe the magnetic flux through an arbitrary area or geometric shape.

13.1.A.1 states that for a magnetic field $\vec{B}$ that is constant across an area $\vec{A}$ , the magnetic flux through the area is defined as $\Phi_B = \vec{B} \cdot \vec{A}$ .
13.1.A.1.i states that the area vector is defined as perpendicular to the plane of the surface and outward from a closed surface.
13.1.A.1.ii states that the sign of flux is given by the dot product of the magnetic field vector and the area vector.
13.1.A.2 states that the total magnetic flux passing through a surface is defined by the surface integral of the magnetic field over the surface area, with the relevant equation $\Phi_B = \int \vec{B} \cdot d\vec{A}$ .

Topic 13.1 prints no boundary statement. Neither do Topics 13.3, 13.4, 13.5 or 13.6. The whole of Unit 13 carries one boundary statement and it sits under Topic 13.2, restricting nothing about flux. So nothing in the framework fences off the shape of the surface or the angle you have to handle, and the objective's phrase "an arbitrary area or geometric shape" points the other way.

What does bound the topic is the sheet. The AP Physics C: E&M equation sheet prints magnetic flux exactly once, as the surface integral:

\Phi_B = \int \vec{B} \cdot d\vec{A}

The simpler dot-product form in 13.1.A.1 is in the framework and not on the sheet. That is a real difference from the algebra-based course, whose sheet prints both a vector form and a $\cos\theta$ component form. Collapsing the integral to a product is a step you take and, on a free-response question, a step you say out loud.

The CED lists four suggested skills for this topic: 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.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Unit 13 is weighted 10 to 20% of the multiple-choice section over a suggested 10 to 20 class periods.

The integral is the definition, and the product is a special case

Read the two statements in the order the CED puts them and the structure is backwards from how it is usually taught. Statement 13.1.A.1 gives the constant-field case first, then 13.1.A.2 gives the general one. Physically the general one comes first.

The surface integral says: chop the surface into patches small enough that the field does not vary across any one of them, give each patch an area vector $d\vec{A}$ perpendicular to it, take the dot product $\vec{B} \cdot d\vec{A}$ for each, and add up the results.

\Phi_B = \int \vec{B} \cdot d\vec{A}

Everything that makes flux problems easy is a reason that sum turns into something you can do without integrating.

Case 1: the field is constant across the surface. Then $\vec{B}$ comes out of the integral and $\int d\vec{A} = \vec{A}$ , so $\Phi_B = \vec{B} \cdot \vec{A}$ . This is 13.1.A.1, and in components it is $\Phi_B = BA\cos\theta$ with $\theta$ the angle between the field and the area vector. Most exam problems are this case, but the reason it works is the constancy, not the shape of the loop.

Case 2: the field is perpendicular to the surface everywhere but varies in magnitude. Then the dot product is just $B\,dA$ and you integrate a scalar. This is the wire-and-rectangle problem in the second worked example, where $B$ depends on distance from a long straight wire.

Case 3: the field is parallel to the surface everywhere. Then $\vec{B} \cdot d\vec{A} = 0$ patch by patch and the flux is zero without any work. A loop held edge-on to the field has no flux through it.

Case 4: the field is uniform but only crosses part of the surface. Split the integral. The flux is the field times the area it actually crosses, and the rest contributes nothing. This case does most of the damage on exam day and it gets its own section below.

The unit of flux is the weber, one tesla times one square metre. Hold onto that: a weber per second has to come out as a volt, which is the units check on everything in Topic 13.2.

The area vector, and the sign you have to declare

Statement 13.1.A.1.i defines the area vector as perpendicular to the plane of the surface, and outward from a closed surface. Two things follow, and the second one is the one that gets marked wrong.

For a closed surface the direction is fixed for you: outward. That is what makes the closed-surface statement of Gauss's law for magnetism meaningful. The C: E&M sheet prints it,

\oint \vec{B} \cdot d\vec{A} = 0

and it says the total magnetic flux through any closed surface is zero. Every field line that enters leaves, because field lines close on themselves and there are no magnetic monopoles. Put a closed box anywhere in any magnetic field and its net flux is zero, no calculation required. Compare the electric case on the same sheet, where the closed-surface integral equals the enclosed charge over $\varepsilon_0$ , and the contrast is the whole content of Gauss's law for magnetism.

For an open surface, such as a flat loop, the direction is your choice. Nothing in physics picks it, so you pick it, and statement 13.1.A.1.ii then makes that choice visible: the sign of the flux is given by the dot product of the field vector and the area vector. Parallel gives positive flux, antiparallel gives negative, and perpendicular gives zero.

Declare the choice before you calculate and keep it to the end of the problem. An axis that flips halfway through is where sign errors come from, not from Lenz's law.

Field relative to the area vector	Angle	Flux
Along it, straight through the loop	$0^\circ$	$\Phi_B = BA$ , the maximum
At sixty degrees to it	$60^\circ$	$\Phi_B = BA/2$
In the plane of the loop, skimming across	$90^\circ$	$\Phi_B = 0$
Straight through but the other way	$180^\circ$	$\Phi_B = -BA$

The reassuring part is that the physical answer never depends on your choice. Flip the convention and both the flux and its rate of change change sign, the two cancel in Faraday's law, and the induced current still runs the same way around the loop. The convention is bookkeeping. The current is physics.

One wording trap is worth naming, because the words look interchangeable and are not. "The plane of the loop is perpendicular to the field" means the area vector is along the field, angle zero, maximum flux. "The plane of the loop contains the field" means the area vector is perpendicular to the field, angle ninety degrees, zero flux. Read the sentence, draw the loop, then draw the arrow sticking out of it. Skill 1.A, creating diagrams to represent physical situations, is listed first for this topic for a reason.

The geometry that makes flux easy

Almost every flux calculation on this exam is easy for one of a small number of reasons. Learning the reasons is faster than learning the cases.

Uniform field, flat loop. The classic. $\Phi_B = BA\cos\theta$ and the only decisions are the area and the angle.

Inside a long solenoid. Statement 12.4.A.1.ii of the previous unit says that unless otherwise stated, all solenoids are assumed to be very long, with uniform magnetic fields inside and negligible magnetic fields outside. That single assumption turns solenoid flux into a product. Inside, $B_{\text{sol}} = \mu_0 n I$ , which the sheet prints, and it is the same everywhere across the cross-section, so any coaxial loop inside gets $\Phi_B = \mu_0 n I \cdot \pi r_{\text{loop}}^2$ .

Outside that same solenoid. Here is the trap, and it is the one the CED's own sample free-response question is built on. A loop encircling the solenoid from outside collects flux only where there is field, and the field outside is negligible. So its flux is $\mu_0 n I \cdot \pi r_{\text{sol}}^2$ , using the solenoid's radius, not the loop's. Widen the outer loop as much as you like and the flux does not change.

The rule underneath both cases: the area in a flux is the area the field actually crosses. A loop bigger than the field region takes the field region's area; a loop smaller than the field region takes its own.

Field varying with position. Then you integrate, and the setup is more than half the marks. Choose a strip over which the field is constant, write its area as a differential, and put the limits where the surface starts and stops. Near a long straight wire the field is $\mu_0 I / (2\pi r)$ , a derived result from Ampere's law in Unit 12, so a rectangle lying in the same plane as the wire needs a strip parallel to the wire and an integral in $r$ . That is the second worked example, and the logarithm it produces is the signature of this problem.

Several sources. Superposition still holds. Compute the field as a vector sum first, then take one flux, rather than computing separate fluxes and worrying about whether they add.

One more geometric fact saves time on multiple-choice questions. Because flux depends on area and a circle's area goes as $r^2$ , doubling a loop's radius quadruples its flux in a uniform field. The CED's sample Question 2 ends on exactly that point: it asks by what factor the vertical intercept of a flux-against-time graph changes when the loop radius doubles, and the scoring guideline awards the point for indicating that flux is proportional to the square of the radius and so increases by a factor of 4. Functional dependence is the currency here.

Where the calculus actually shows up

Topic 13.1 is a definition, so it can look like the one place calculus is decoration. It is not, for two reasons.

The integral is load-bearing whenever the field is not uniform. The algebra-based course never poses that problem, because it has no way to. A rectangle beside a current-carrying wire, a loop in the fringing field of a magnet, a triangular loop in a field that grows with distance: all of these are ordinary here and impossible there.

The flux is usually wanted as a function of something, not as a number. Topic 13.2 differentiates whatever you write down here, so a flux expressed as $\Phi_B(t)$ or $\Phi_B(x)$ is worth more than a flux expressed as 0.024 Wb. Keep the symbol you are going to differentiate.

That is the practical reason the CED's sample Question 2 asks for a derivation and a graph rather than a value. Its scoring guidelines award a point for substituting the small loop's area into an expression for the magnetic flux, and another for substituting a correct expression for the field inside a solenoid, before any differentiation happens. Two of the four points in that part are for building the flux expression.

A short symbolic habit that pays for itself: write the flux with every dependence visible, even the constants.

\Phi_B(t) = \mu_0 n_0 I(t) \, \pi r_0^2

Everything not named $t$ is a constant that will come out of the derivative in the next topic. If you have already collapsed the expression to a number, you cannot see which factor is doing the changing, and functional-dependence questions, skill 2.C for this topic, are asked in exactly those terms.

Reading and sketching flux graphs

Skill 1.A is listed first for Topic 13.1, and its partner in Topic 13.2 is graph creation. A flux-against-time graph is the usual meeting point, so read it as a flux graph before you read it as an emf graph.

What the situation does	What $\Phi_B$ against $t$ looks like
Loop at rest in a steady field	horizontal line at a non-zero value
Field ramping at a constant rate	straight line, constant slope
Field growing as $t^2$	parabola opening upward
Current in a source solenoid falling as $I_0(1 - bt^2)$	downward parabola, zero at $t = 1/\sqrt{b}$
Loop rotating steadily	cosine
Loop sliding fully out of a field region	ramp down to zero, then flat at zero

The fourth row is the CED's sample Question 2 again, and its scoring guidelines are specific about what a correct sketch shows: a curve concave downward everywhere, passing through the point $\left(1/\sqrt{b},\, 0\right)$ on the horizontal axis with that value clearly indicated, and a positive maximum on the vertical axis with zero slope at that maximum. Zero slope at the intercept is the graphical statement that the current, and so the flux, starts off flat.

The last row is worth drawing yourself. When a loop leaves a field region at constant speed the flux falls linearly, so the graph is a straight ramp, and the instant the loop clears the boundary the flux is zero and stays zero. The kink at that instant is a real feature: the emf is constant during the exit and zero afterwards, which is a step, not a ramp.

If you landed here from AP Physics 2

Magnetic flux appears in both courses and they are not the same page.

In the algebra-based course it is one essential-knowledge statement inside AP Physics 2 Topic 12.4, Electromagnetic Induction and Faraday's Law, where flux, Faraday's law, Lenz's law and the sliding rod share a single topic. There, flux is $\Phi_B = BA\cos\theta$ , printed twice on the sheet, and the field is always uniform across the loop.

In AP Physics C, flux gets a topic of its own, the sheet prints only the surface integral, and the objective says "arbitrary area or geometric shape".

That Physics 2 page is for students taking the algebra-based course. This page is for students taking AP Physics C: Electricity and Magnetism. If you are in Physics 2 you want the product form and the finite-difference version of Faraday's law, and you can stop there. If you are in Physics C you want the integral, because you will meet fields that vary across the surface, and you want the flux as a symbolic function of time, because the next topic differentiates it.

The definitions of the area vector are identical in the two courses, word for word in substance: perpendicular to the plane of the surface and outward from a closed surface. So the angle conventions and the sign rules carry over intact. It is only the integral and the derivative that are new.

How Topic 13.1 is tested

Flux almost never appears alone. It appears as the first two or three lines of an induction question, which is why learning objective 13.1.A shares billing with 12.4.A and 13.2.A on the CED's own sample Question 2, a 12-point Translation Between Representations question.

The patterns worth rehearsing:

Compute a flux from a field, an area, and an angle to the area vector (skills 2.C, 3.B). Check the angle is measured to the vector, not to the plane.
Write a flux as a symbolic function of time given a field or a current that varies (skill 2.A). Leave the constants as symbols.
Set up and evaluate a surface integral for a field that varies across the surface (skill 2.A). The wire-and-rectangle case is the standard one.
Decide which area the field actually crosses for a loop inside or outside a solenoid, or a loop larger than the field region (skill 3.B).
Predict a factor of change when the radius, area, field or angle changes (skill 2.C). Radius doubling gives flux times four in a uniform field.
Sketch or read a flux-against-time graph (skill 1.A, with 1.B and 1.C next door in Topic 13.2).

On the free-response section, skill 2.A carries a 40 to 45% weighting, the largest of any single skill, and it is listed for this topic. Symbolic setup is where the marks are. The scoring guidelines for the CED's sample question make that concrete: they award a point for substituting the loop's area into a flux expression and a point for substituting a correct field expression, before anything is differentiated, and they explicitly note on a later point that the expression being differentiated need not be correct for the derivative point to be earned. Show the pathway.

A loop inside a solenoid, and a loop around it

A long solenoid of radius 0.050 m has 2000 turns per metre and carries a current of 3.0 A. Find the magnetic flux through (a) a coaxial circular loop of radius 0.020 m inside the solenoid, (b) a coaxial circular loop of radius 0.040 m inside the solenoid, and (c) a coaxial circular loop of radius 0.10 m that lies outside the solenoid and encircles it.

Get the field first. Statement 12.4.A.1.iii of the previous unit gives $B_{\text{sol}} = \mu_0 n I$ as a derived result of Ampere's law, and the sheet prints it. With $\mu_0 = 4\pi \times 10^{-7} \ (\mathrm{T \cdot m})/\mathrm{A}$ from the Table of Information: $B = (4\pi \times 10^{-7})(2000)(3.0)$ .
Step by step: $(4\pi \times 10^{-7})(2000) = 2.5133 \times 10^{-3}$ , then $\times 3.0 = 7.5398 \times 10^{-3}$ T, so $B = 7.5 \times 10^{-3}$ T inside.
Declare the convention: take every area vector along the field, so all three fluxes are positive and $\cos\theta = 1$ .
(a) The field is uniform across the loop, so the surface integral collapses to $\Phi_B = BA$ . $A = \pi(0.020)^2 = 1.2566 \times 10^{-3} \ \mathrm{m^2}$ , so $\Phi_B = (7.5398 \times 10^{-3})(1.2566 \times 10^{-3}) = 9.475 \times 10^{-6}$ Wb, which is $9.5 \times 10^{-6}$ Wb.
(b) $A = \pi(0.040)^2 = 5.0265 \times 10^{-3} \ \mathrm{m^2}$ , so $\Phi_B = (7.5398 \times 10^{-3})(5.0265 \times 10^{-3}) = 3.790 \times 10^{-5}$ Wb, which is $3.8 \times 10^{-5}$ Wb. Exactly four times part (a), because the radius doubled and area goes as $r^2$ .
(c) Now the trap. The loop has radius 0.10 m but statement 12.4.A.1.ii says the field outside a long solenoid is negligible. Split the surface integral into the part inside the solenoid's cross-section, where the field is $7.5398 \times 10^{-3}$ T, and the annulus outside it, where the field is zero.
Only the first part contributes: $\Phi_B = B \cdot \pi r_{\text{sol}}^2 = (7.5398 \times 10^{-3}) \, \pi (0.050)^2 = (7.5398 \times 10^{-3})(7.8540 \times 10^{-3}) = 5.922 \times 10^{-5}$ Wb, so $5.9 \times 10^{-5}$ Wb.
Check the reasoning by pushing it: a loop of radius 1.0 m around the same solenoid has the same flux, $5.9 \times 10^{-5}$ Wb. Growing the loop outside the solenoid adds area but adds no field, so it adds no flux.

(a) $9.5 \times 10^{-6}$ Wb. (b) $3.8 \times 10^{-5}$ Wb, four times part (a) because area goes as the square of the radius. (c) $5.9 \times 10^{-5}$ Wb, computed with the solenoid's radius of 0.050 m rather than the loop's, because the field outside a long solenoid is negligible. The area in a flux is the area the field actually crosses.

A field that varies across the surface: rectangle beside a wire

A long straight wire carries a steady current of 8.0 A. A rectangular loop lies in the same plane as the wire, with its two long sides parallel to the wire. The rectangle is 0.15 m long in the direction parallel to the wire, its near side is 0.020 m from the wire, and its far side is 0.070 m from the wire. Find the magnetic flux through the rectangle.

Check whether the field is constant across the surface. It is not: the field of a long straight wire is $B = \dfrac{\mu_0 I}{2\pi r}$ , a derived result of Ampere's law from Unit 12, and $r$ runs from 0.020 m to 0.070 m across this rectangle. So statement 13.1.A.1 does not apply and 13.1.A.2 does.
Check the direction. The field vectors around a wire are tangent to concentric circles centred on it, so at every point of a rectangle lying in the plane of the wire the field is perpendicular to the rectangle. Take the area vector that way too, so $\vec{B} \cdot d\vec{A} = B \, dA$ with no cosine to carry.
Choose a strip over which the field is constant. A strip parallel to the wire at distance $r$ , of width $dr$ and length $\ell = 0.15$ m, has every point at the same distance from the wire, so $dA = \ell \, dr$ .
Set up the integral with the limits at the near and far sides: $\Phi_B = \displaystyle\int_{a}^{b} \frac{\mu_0 I}{2\pi r} \ell \, dr = \frac{\mu_0 I \ell}{2\pi} \int_{a}^{b} \frac{dr}{r}$ .
Integrate: $\displaystyle\int_a^b \frac{dr}{r} = \ln\frac{b}{a}$ , so $\Phi_B = \dfrac{\mu_0 I \ell}{2\pi} \ln\dfrac{b}{a}$ . Stop here if the question asks for a symbolic answer, which it usually does.
Now the numbers. $\dfrac{\mu_0}{2\pi} = 2.00 \times 10^{-7} \ (\mathrm{T \cdot m})/\mathrm{A}$ exactly, since $\mu_0 = 4\pi \times 10^{-7}$ . And $\ln(0.070/0.020) = \ln 3.5 = 1.2528$ .
$\Phi_B = (2.00 \times 10^{-7})(8.0)(0.15)(1.2528) = (2.40 \times 10^{-7})(1.2528) = 3.007 \times 10^{-7}$ Wb, so $3.0 \times 10^{-7}$ Wb.
Two sanity checks. The logarithm is dimensionless, so the units are $(\mathrm{T \cdot m/A})(\mathrm{A})(\mathrm{m}) = \mathrm{T \cdot m^2} = \mathrm{Wb}$ . And pushing the rectangle out to infinity makes $\ln(b/a)$ diverge slowly rather than converge, which is the honest behaviour of a field that only falls as $1/r$ .

$\Phi_B = \dfrac{\mu_0 I \ell}{2\pi} \ln\dfrac{b}{a} = 3.0 \times 10^{-7}$ Wb. The logarithm is the signature of a $1/r$ field integrated across a rectangle, and this is the kind of flux the algebra-based course cannot pose, because it has no way to add up a field that changes across the surface.

Angle, sign, and the wording that flips the answer

A square loop 0.25 m on a side sits in a uniform 0.60 T magnetic field. Find the flux when (a) the area vector is at 30 degrees to the field, (b) the loop is turned so the area vector is at 150 degrees to the field, and (c) the plane of the loop contains the field. (d) State what the flux would be through a closed cubical box placed in the same field.

Area: $A = (0.25)^2 = 0.0625 \ \mathrm{m^2}$ . The field is uniform across the loop, so 13.1.A.1 applies and $\Phi_B = BA\cos\theta$ with $\theta$ measured to the area vector, per 13.1.A.1.i.
(a) $\cos 30^\circ = 0.86603$ , so $\Phi_B = (0.60)(0.0625)(0.86603) = (0.0375)(0.86603) = 0.03248$ Wb, which is 0.032 Wb, positive.
(b) $\cos 150^\circ = -0.86603$ , so $\Phi_B = -0.032$ Wb. Same magnitude, opposite sign. Statement 13.1.A.1.ii is what makes the sign meaningful: it comes from the dot product, and it records that the field now has a component antiparallel to the area vector.
Note what did not happen. The physical situation in (b) is the same loop in the same field, just turned over. The sign is a statement about your chosen area vector, not about the field.
(c) "The plane of the loop contains the field" means the field lies in the plane, so it is perpendicular to the area vector: $\theta = 90^\circ$ , $\cos 90^\circ = 0$ , and $\Phi_B = 0$ . Read that phrasing carefully, because "the plane of the loop is perpendicular to the field" means the opposite: area vector along the field, $\theta = 0$ , and the maximum flux $BA = 0.0375$ Wb.
(d) Zero, with no calculation. The sheet prints $\oint \vec{B} \cdot d\vec{A} = 0$ : the net magnetic flux through any closed surface is zero, in any field, because magnetic field lines close on themselves. Every line entering the box leaves it.

(a) $+0.032$ Wb. (b) $-0.032$ Wb, the same magnitude with the sign flipped by the dot product. (c) zero, because the field lies in the plane of the loop and so has no component along the area vector. (d) exactly zero for any closed surface, by the printed result that the closed-surface integral of the magnetic field vanishes.

Frequently asked questions

What is magnetic flux in AP Physics C?

Magnetic flux is a measure of how much magnetic field passes through a surface. Essential knowledge 13.1.A.2 defines the total flux through a surface as the surface integral of the magnetic field over the surface area, and that integral form is the only one printed on the AP Physics C: Electricity and Magnetism equation sheet. Essential knowledge 13.1.A.1 gives the special case where the field is constant across the area, in which the flux is the dot product of the field vector and the area vector. The unit is the weber, equal to one tesla times one square metre.

Why is magnetic flux a surface integral rather than just B times A?

Because the field is not always the same everywhere on the surface. The product form assumes a single value of the field across the whole area, which is only true for a uniform field. When the field varies with position, for example the field near a long straight wire, which falls off as one over the distance, you have to add up the contribution from each patch of surface separately, and that sum is the surface integral. Learning objective 13.1.A asks students to describe the flux through an arbitrary area or geometric shape, which is why the general form is the one the course prints.

How do you find the direction of the area vector?

Essential knowledge 13.1.A.1.i says the area vector is perpendicular to the plane of the surface, and directed outward from a closed surface. For a closed surface the direction is therefore fixed for you. For an open surface such as a flat loop, either perpendicular direction is allowed and you choose one, then keep that choice for the whole problem. The choice fixes the sign of the flux by the dot product, per 13.1.A.1.ii, but not any physical result: reverse the choice and both the flux and its rate of change change sign, so the induced current still comes out running the same way.

What is the magnetic flux through a loop outside a solenoid?

It is the field inside the solenoid multiplied by the cross-sectional area of the solenoid, not of the loop. Essential knowledge 12.4.A.1.ii states that unless otherwise stated, solenoids are assumed to be very long, with uniform magnetic fields inside and negligible magnetic fields outside. So the part of the loop's area that lies outside the solenoid contributes nothing, and widening the loop further does not change the flux at all. The general rule is that the area in a flux calculation is the area the field actually crosses.

Is magnetic flux a vector?

No. Magnetic flux is a scalar, produced by a dot product of two vectors, so it has a sign but no direction. Fluxes through the same surface add as ordinary signed numbers. The sign records whether the field has a component along or against the area vector you chose, per essential knowledge 13.1.A.1.ii, with parallel giving positive flux, antiparallel giving negative flux, and perpendicular giving zero.

Why is the magnetic flux through a closed surface always zero?

Because magnetic field lines have no beginning or end. The AP Physics C: Electricity and Magnetism equation sheet prints the closed-surface integral of the magnetic field over area as exactly zero, which is Gauss's law for magnetism. Physically it says there are no magnetic monopoles, so no closed surface can enclose a net source of magnetic field. Every field line that enters a closed surface leaves it somewhere else. The electric case on the same sheet is different, because the closed-surface integral of the electric field equals the enclosed charge divided by the vacuum permittivity, and isolated charges do exist.

Does AP Physics C Topic 13.1 have a boundary statement?

No. Topic 13.1 prints no boundary statement, and neither do Topics 13.3, 13.4, 13.5 or 13.6. The whole of Unit 13 carries exactly one boundary statement, under Topic 13.2, and it concerns deriving the speed of light from Maxwell's equations, which is something students are told they are not expected to do. Nothing in the framework restricts the shape of the surface or the angles you may be asked to handle in a flux calculation.