Gauss's Law vs Ampere's Law: The Difference

Q: Is Ampere's law the magnetic version of Gauss's law?

No. The magnetic version of Gauss's law is the closed surface integral of B dot dA equals zero, which is printed on the AP Physics C: Electricity and Magnetism equation sheet a few lines above Ampere's law. It says the magnetic flux through any closed surface is zero, because there is no magnetic charge to enclose. Ampere's law is the magnetic counterpart of Faraday's law in structure, since both are loop integrals of a field. Gauss's law and Ampere's law resemble each other only in that both close an integral and count an enclosed source.

Q: Why does Gauss's law divide by epsilon zero while Ampere's law multiplies by mu zero?

It follows from how the two constants are defined. Epsilon zero, the vacuum permittivity, appears in the denominator of Coulomb's law through k equals one over four pi epsilon zero, so a larger permittivity means a weaker electric field for the same charge. Mu zero, the vacuum permeability, appears in the numerator of the Biot-Savart law, so a larger permeability means a stronger magnetic field for the same current. The AP sheet prints epsilon zero as 8.85 times ten to the minus twelve and mu zero as four pi times ten to the minus seven.

Q: Does an Amperian loop have to be a circle?

No, but it has to be a path on which the calculation actually simplifies. For a long straight wire or a cylindrical conductor the right choice is a circle centred on the axis, because the field magnitude is constant along it and points along the path. For a long solenoid the right choice is a rectangle with one side inside the coil and one side outside, because the field is uniform inside and negligible outside, so only one side of the rectangle contributes. Any closed path is legal; only some make the integral collapse.

Q: Are Gauss's law and Ampere's law on the AP Physics 2 equation sheet?

Neither one is. The AP Physics 2 CED contains no Gauss's law, no Ampere's law, no electric flux and no Gaussian surfaces. What the Physics 2 sheet prints instead is the finished result B equals mu zero I over two pi r for the field of a long straight wire, which is what Ampere's law produces for that case. Both closed-integral laws are AP Physics C: Electricity and Magnetism content, appearing as Topic 8.6 and Topic 12.4 respectively.

Q: What does it mean when the circulation of B around a loop is zero?

It means the net current threading the loop is zero, with currents counted positive or negative according to the right-hand rule applied to the direction you traverse the loop. It does not mean the magnetic field is zero anywhere. Two parallel wires carrying equal and opposite currents give zero circulation around any loop enclosing both, while the field between them is twice what a single wire would produce. This mirrors the electrostatic case where a Gaussian surface around a dipole has zero net flux and a large field on it.

Q: Which Maxwell's equations are Gauss's law and Ampere's law?

The AP Physics C: Electricity and Magnetism CED names both. Essential knowledge 8.6.A.6 states that Gauss's law is Maxwell's first equation. Essential knowledge 12.4.A.4 states that Maxwell's fourth equation is Ampere's law with Maxwell's addition, which adds that a changing electric field creates a magnetic field alongside the contribution from current. A boundary statement on that topic says the course does not expect students to use the fourth equation with a changing electric field, only to understand that a changing electric field generates a magnetic field.

Gauss's law integrates the electric field over a closed surface and equals the enclosed charge over epsilon zero. Ampere's law integrates the magnetic field around a closed loop and equals mu zero times the enclosed current. Surface against loop, charge against current, symmetry needed by both.

AP Physics: Unit 12 (topics 8.6 Gauss's Law, 12.4 Ampere's Law). Gauss's law is Topic 8.6 in AP Physics C: Electricity and Magnetism Unit 8, weighted at 15 to 25 percent of the multiple-choice section over approximately 12 to 24 class periods. Ampere's law is Topic 12.4 in Unit 12, Magnetic Fields and Electromagnetism, weighted at 10 to 20 percent over approximately 10 to 20 class periods. Essential knowledge 8.6.A.1 relates electric flux through a Gaussian surface to the enclosed charge; the Topic 8.6 boundary statement limits quantitative application to point charges and distributions with spherical, cylindrical, or planar symmetry. Essential knowledge 12.4.A.1 relates the magnitude of the magnetic field to the current enclosed by a closed imaginary path called an Amperian loop, with 12.4.A.1.i and 12.4.A.1.iii giving the long straight wire and long solenoid as derived cases; the Topic 12.4 boundary statement limits quantitative application to symmetrical magnetic fields and names long straight wires, long solenoids, and conductive slabs or cylindrical conductors carrying a current density. Essential knowledge 8.6.A.6 names Gauss's law as Maxwell's first equation and 12.4.A.4 names Ampere's law with Maxwell's addition as the fourth. Neither law appears in the AP Physics 2 CED.

The parallel, stated once

Set the two printed equations next to each other and the resemblance is immediate:

\oint \vec{E} \cdot d\vec{A} = \frac{q_{\mathrm{enc}}}{\varepsilon_0} \qquad\qquad \oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\mathrm{enc}}

Each one closes an integral around something, counts what is inside, and multiplies by a constant of the vacuum. Each one is exactly true always and useful only under symmetry. Each one is a member of Maxwell's equations. Each one gets you a field in two lines when the geometry cooperates and nothing at all when it does not.

The resemblance is real and it is worth using, because a student who has understood Gauss's law already understands the shape of Ampere's law and only has to learn the substitutions. But the resemblance is not an identity, and three things are genuinely different:

The object you integrate over. Gauss uses a two-dimensional closed surface. Ampere uses a one-dimensional closed loop, a curve. That is a difference in kind, not in flavour.
What "enclosed" means. For Gauss, enclosed means inside the volume the surface bounds. For Ampere, enclosed means threading through the open surface the loop bounds, and which way it threads decides its sign.
The constant's position. $\varepsilon_0$ divides, $\mu_0$ multiplies. That is not cosmetic, and it comes out of how the two constants are defined.

And there is a fourth difference that reframes the whole comparison, which the next section but one is about: Ampere's law is not the magnetic counterpart of Gauss's law. It has one, it is printed on the same sheet, and it is not this.

Side by side

	Gauss's law	Ampere's law
Printed form, C: E&M sheet	$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\mathrm{enc}}}{\varepsilon_0}$	$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\mathrm{enc}}$
You integrate over	A closed surface, two dimensional	A closed loop, one dimensional
The integrand	Flux, field dotted with an area element	Circulation, field dotted with a length element
What is counted inside	Charge, in coulombs	Current, in amperes
Vacuum constant	$\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)}$ , dividing	$\mu_0 = 4\pi \times 10^{-7} \ \mathrm{(T \cdot m)/A}$ , multiplying
Direction convention	Outward normal on the closed surface	Right-hand rule: curl fingers along the loop, thumb gives positive current
Symmetries the CED uses	Spherical, cylindrical, planar	Long straight wires, long solenoids, conductive slabs, cylindrical conductors with a current density
Which Maxwell equation	The first	The fourth, once Maxwell's addition is included
Complete as printed?	Yes	No, the full fourth equation adds a changing electric flux term
Units of the left side	$\mathrm{N \cdot m^2/C}$	$\mathrm{T \cdot m}$
In AP Physics 2?	No	No

The last row is worth pausing on. Neither law appears anywhere in the AP Physics 2 CED or on its equation sheet. What that course gets instead is the answer Ampere's law produces for the one case it needs: $B = \frac{\mu_0 I}{2\pi r}$ for a long straight wire, printed on the Physics 2 sheet as a finished result with no derivation attached. So a Physics 2 student uses the fruit of Ampere's law without ever meeting the law, which is exactly the design decision that makes these two laws a Physics C comparison.

The parallel that is actually correct: Gauss's law for magnetism

Pair Gauss's law with Ampere's law and you have paired an electric law with a magnetic one, which feels right and is structurally wrong. Look at what a closed surface integral of $\vec{B}$ gives, because the C: E&M sheet prints that too, three lines above Ampere's law:

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

That is Gauss's law for magnetism, and it is the true twin. Same surface, same flux integral, same question asked of a different field. The answer is zero, and the reason is that there is no magnetic charge to enclose: field lines that enter a closed surface always leave it, because they close on themselves rather than starting and ending on sources.

So the honest map of the four is a two-by-two grid, and it is worth drawing once:

	Closed surface, flux	Closed loop, circulation
Electric field	$\oint \vec{E} \cdot d\vec{A} = q_{\mathrm{enc}}/\varepsilon_0$ (Gauss's law)	$\oint \vec{E} \cdot d\vec{\ell} = -d\Phi_B/dt$ (Faraday's law)
Magnetic field	$\oint \vec{B} \cdot d\vec{A} = 0$ (Gauss's law for magnetism)	$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\mathrm{enc}}$ (Ampere's law)

All four cells are printed on the C: E&M sheet, and together they are Maxwell's equations in the form the course uses them. Gauss's law and Ampere's law sit on the diagonal of that table, which is why the analogy between them works at all and why it is incomplete. The row-mate of Gauss's law is the zero on the right of the magnetic flux equation; the column-mate of Ampere's law is Faraday's law, which is the subject of a comparison of its own.

One consequence is worth carrying into the exam. Because $\oint \vec{B} \cdot d\vec{A} = 0$ has no source term, the magnetic flux through any closed surface is zero no matter what is inside it: a bar magnet, a solenoid, a coil carrying any current at all. There is no magnetic analogue of "enclosed charge", so there is nothing for a magnetic Gauss's law to solve for. That is exactly why the magnetic field calculation had to be pushed onto a loop integral instead.

Where the parallel breaks down

Four places, in rising order of how much trouble they cause.

A loop bounds infinitely many surfaces, and it does not matter. When you count $I_{\mathrm{enc}}$ for an Amperian loop, you count the current through any surface bounded by that loop. For a steady current, flat disc, hemisphere and drooping bag shape all give the same enclosed current, because a steady current cannot pile up between two surfaces sharing a rim. Gauss's law has no equivalent freedom, because a closed surface bounds exactly one volume. This is the piece of Ampere's law that has no counterpart, and it is also where Maxwell's addition enters: for a circuit with a capacitor gap, a bulging surface through the gap carries no current at all, which is the defect the changing-electric-flux term repairs.

Signs work differently. On a Gaussian surface, the outward normal is fixed by the surface itself, and a negative enclosed charge simply makes the flux negative. On an Amperian loop, you choose which way to walk around the loop, and the right-hand rule then fixes which direction of current counts as positive. Two currents through the same loop in opposite directions cancel in $I_{\mathrm{enc}}$ . A loop enclosing $+5.0 \ \mathrm{A}$ and $-5.0 \ \mathrm{A}$ gives $\oint \vec{B} \cdot d\vec{\ell} = 0$ with a strong field everywhere on the path, which is worked example three.

Ampere's law as printed is incomplete. Essential knowledge 12.4.A.4 states that Maxwell's fourth equation is Ampere's law with Maxwell's addition: magnetic fields are generated by electric current, and a changing electric field also creates a magnetic field. The CED prints that full version, $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I + \mu_0\varepsilon_0 \frac{d\Phi_E}{dt}$ , and then a boundary statement says the course does not expect students to use the fourth equation with a changing electric field, though students should understand that a changing electric field generates a magnetic field. Gauss's law needs no such repair; it is complete as printed.

The symmetry menus do not match. Both laws need symmetry, but not the same list. The Topic 8.6 boundary statement gives Gauss's law point charges and spherical, cylindrical, or planar symmetry. The Topic 12.4 boundary statement limits quantitative application of Ampere's law to situations involving symmetrical magnetic fields, and names the shapes: long straight wires, long solenoids carrying currents, and conductive slabs or cylindrical conductors carrying a current density. Note what is absent from the magnetic list: nothing spherical. A uniformly charged sphere is the first thing Gauss's law does. Nothing on Ampere's list is a sphere, because a magnetic field circulates rather than radiating, so it never has one magnitude and one orientation all over a spherical surface the way a radial electric field does.

The case where the parallel is exact

There is one geometry where the two laws run in step so closely that you can write the solutions in the same handwriting: the long solid cylinder.

Take a cylinder of radius $R$ filled with uniform charge density $\rho$ , and a cylinder of the same radius carrying a current $I$ spread uniformly across its cross-section. Both problems have cylindrical symmetry. Both use a concentric surface or loop of radius $r$ . Both count the fraction of the source inside, which goes as $r^2/R^2$ in each case. And both give a field that rises linearly with $r$ inside the material and falls as $1/r$ outside it.

	Charged cylinder	Current-carrying cylinder
Object used	Coaxial Gaussian cylinder, radius $r$ , length $L$	Circular Amperian loop, radius $r$
Left side collapses to	$E(2\pi r L)$	$B(2\pi r)$
Source enclosed, $r < R$	$\rho \pi r^2 L$	$I r^2/R^2$
Field inside	$E = \dfrac{\rho r}{2\varepsilon_0}$	$B = \dfrac{\mu_0 I r}{2\pi R^2}$
Field outside	$E = \dfrac{\rho R^2}{2\varepsilon_0 r}$	$B = \dfrac{\mu_0 I}{2\pi r}$
Direction	Radially outward from the axis	Circling the axis, right-hand rule

The two rows of results have the same algebra and describe completely different pictures: one field points away from the axis, the other wraps around it. Worked examples one and two run the numbers on both with the same radii, so you can watch the identical structure produce unrelated answers.

That difference in direction is not decoration. It is why the electric field can be extracted from a surface integral (the field pierces the surface everywhere at right angles) and the magnetic field cannot (it never pierces anything; it runs along a circle, which is exactly what a loop integral is built to measure).

When it costs a mark

Putting $\varepsilon_0$ in the wrong place. $q_{\mathrm{enc}}/\varepsilon_0$ divides; $\mu_0 I_{\mathrm{enc}}$ multiplies. Since $\varepsilon_0$ is small and $\mu_0$ is small, flipping either one moves the answer by more than twenty orders of magnitude, and the mistake survives the sanity check of "does this look like physics" only because nobody has an intuition for teslas.

Counting the current that runs along the loop rather than through it. A wire that lies in the plane of the Amperian loop but does not pass through the surface it bounds contributes nothing to $I_{\mathrm{enc}}$ . It still contributes to $\vec{B}$ at points on the loop.

Assuming $\oint \vec{B} \cdot d\vec{\ell} = 0$ means $\vec{B} = 0$ . The exact counterpart of the dipole trap in electrostatics, and it appears on exams for the same reason. Zero circulation means the positive and negative contributions around the path cancelled, not that the field vanished.

Using a rectangular Amperian loop for a straight wire. The loop must be a path along which $B$ has constant magnitude and a fixed angle to the path. For a straight wire that path is a circle centred on the wire. A rectangle samples several field strengths and several angles, and the integral will not collapse. For a long solenoid, by contrast, the rectangle is exactly right, because the field is uniform inside and negligible outside.

Reaching for a sphere. Spherical symmetry is on Gauss's list and not on Ampere's. If a magnetic problem seems to want a spherical surface, the intended tool is the Biot-Savart law, not Ampere's law.

Which one a problem is asking for

The two laws almost never compete, because they act on different fields. The real decision on the exam is not Gauss against Ampere but, within each field, the symmetric route against the integrating route.

Electric field, symmetric charge: Gauss's law. Electric field, no symmetry: Coulomb's law in integral form. That fork is Coulomb's law vs Gauss's law.
Magnetic field, symmetric current: Ampere's law. Magnetic field, no symmetry: the Biot-Savart law. That fork is Ampere's law vs the Biot-Savart law.

Said together, the four laws form a rectangle: Coulomb is to Gauss as Biot-Savart is to Ampere. The integrating law in each pair works everywhere and costs you an integral; the symmetric law in each pair costs almost nothing and works on a short list of shapes. Learning that once covers both halves of the course, which is why it is worth stating as a structure rather than as four separate laws.

The pairing that people build instead, Gauss with Ampere because both use $\oint$ , is not wrong so much as sideways. It links the two shortcuts to each other and leaves the two workhorses unconnected.

Where these sit in the course

Gauss's law is Topic 8.6, the closing topic of Unit 8, which the CED weights at 15 to 25 percent of the multiple-choice section over roughly 12 to 24 class periods. Ampere's law is Topic 12.4, the closing topic of Unit 12, weighted at 10 to 20 percent over roughly 10 to 20 class periods.

They occupy the same position in their units, and that is not a coincidence: in each case the unit builds the field from its sources first, defines the flux or the geometry second, and only then introduces the closed integral as the payoff. Reading Unit 12 with Unit 8 fresh in mind is the cheapest way to make the magnetic half of the course feel less like a new subject.

Both units end with an essential knowledge statement placing the law inside Maxwell's equations: 8.6.A.6 for Gauss's law as the first, and 12.4.A.4 for Ampere's law with Maxwell's addition as the fourth. Neither statement is decoration on the exam, because free-response prompts do ask students to identify which law applies to a described situation, and the Maxwell framing is the CED's way of signalling that these are four coordinates of one theory rather than four tricks.

Gauss on a charged cylinder: the field inside

An infinitely long solid insulating cylinder of radius $R = 0.040 \ \mathrm{m}$ carries a uniform volume charge density $\rho = 2.0 \times 10^{-6} \ \mathrm{C/m^3}$ . Find the electric field at $r = 0.020 \ \mathrm{m}$ from the axis, inside the material. Use $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)}$ .

Symmetry: the cylinder is infinite and uniform, so the field can only point radially outward and can only depend on $r$ . That licenses a coaxial Gaussian cylinder of radius $r$ and length $L$ .
The two flat end caps contribute nothing, because the field is parallel to them. The curved wall has $\vec{E}$ perpendicular to it with constant magnitude, so $\oint \vec{E} \cdot d\vec{A} = E(2\pi r L)$ .
The enclosed charge is the volume inside the Gaussian cylinder times the density: $q_{\mathrm{enc}} = \rho(\pi r^2 L)$ . Note that only the charge within radius $r$ counts, not the whole cylinder.
Gauss's law: $E(2\pi r L) = \rho \pi r^2 L/\varepsilon_0$ . The $L$ and one power of $\pi r$ cancel, leaving $E = \dfrac{\rho r}{2\varepsilon_0}$ .
Numerically: $\rho r = (2.0 \times 10^{-6})(0.020) = 4.0 \times 10^{-8}$ , and $2\varepsilon_0 = 1.77 \times 10^{-11}$ .
$E = (4.0 \times 10^{-8})/(1.77 \times 10^{-11}) = 2.26 \times 10^3 \ \mathrm{N/C}$ , directed radially outward.
Check the shape: $E$ grows linearly with $r$ inside the material, and at $r = 0$ it is zero, which is right because there is no enclosed charge at the axis.

$E = \rho r/(2\varepsilon_0) = 2.3 \times 10^3 \ \mathrm{N/C}$ , pointing radially away from the axis. The field rises linearly with distance from the axis while you are still inside the charge.

Ampere on a current-carrying cylinder: the same algebra, a different picture

A long solid cylindrical conductor of radius $R = 0.040 \ \mathrm{m}$ carries a total current $I = 8.0 \ \mathrm{A}$ distributed uniformly over its cross-section. Find the magnitude of the magnetic field at $r = 0.020 \ \mathrm{m}$ from the axis. Use $\mu_0 = 4\pi \times 10^{-7} \ \mathrm{(T \cdot m)/A}$ .

Symmetry: the current distribution is unchanged by rotating about the axis or sliding along it, so $B$ has the same magnitude at every point of a circle centred on the axis and runs along that circle. Choose that circle as the Amperian loop.
Because $\vec{B}$ is parallel to $d\vec{\ell}$ everywhere on the loop and constant in magnitude, $\oint \vec{B} \cdot d\vec{\ell} = B(2\pi r)$ .
The enclosed current is the fraction of the cross-section inside radius $r$ : $I_{\mathrm{enc}} = I\dfrac{\pi r^2}{\pi R^2} = I\dfrac{r^2}{R^2}$ .
Ampere's law: $B(2\pi r) = \mu_0 I r^2/R^2$ , so $B = \dfrac{\mu_0 I r}{2\pi R^2}$ , linear in $r$ , exactly as the electric case was.
Numerically, use $\mu_0/(2\pi) = 2.0 \times 10^{-7}$ to keep the arithmetic clean: $B = (2.0 \times 10^{-7})\dfrac{I r}{R^2}$ .
$\dfrac{I r}{R^2} = \dfrac{(8.0)(0.020)}{(0.040)^2} = \dfrac{0.16}{0.0016} = 100$ .
$B = (2.0 \times 10^{-7})(100) = 2.0 \times 10^{-5} \ \mathrm{T}$ .
Direction: point the right thumb along the current and the fingers curl the way $\vec{B}$ circles the wire. Unlike the electric case, the field never points away from the axis at all.

$B = \mu_0 I r/(2\pi R^2) = 2.0 \times 10^{-5} \ \mathrm{T}$ , circling the axis. Same enclosed-fraction argument as the charged cylinder, same linear rise with $r$ , and a field that wraps instead of radiating.

Zero circulation with a large field: two antiparallel wires

Two long parallel wires are $0.10 \ \mathrm{m}$ apart. Each carries $5.0 \ \mathrm{A}$ , in opposite directions. An Amperian loop encircles both wires. (a) Find $\oint \vec{B} \cdot d\vec{\ell}$ around that loop. (b) Find the magnitude of $\vec{B}$ at the midpoint between the wires. (c) Say what the pair of answers means.

(a) Walk around the loop in some chosen sense and apply the right-hand rule: one wire's current counts as $+5.0 \ \mathrm{A}$ and the other, running the opposite way, counts as $-5.0 \ \mathrm{A}$ .
$I_{\mathrm{enc}} = +5.0 - 5.0 = 0$ , so $\oint \vec{B} \cdot d\vec{\ell} = \mu_0(0) = 0$ exactly, for every loop that encircles both wires.
(b) Now the field at the midpoint, from the long-wire result. Each wire is $0.050 \ \mathrm{m}$ away, so each contributes $B = \dfrac{\mu_0 I}{2\pi r} = (2.0 \times 10^{-7})\dfrac{5.0}{0.050} = (2.0 \times 10^{-7})(100) = 2.0 \times 10^{-5} \ \mathrm{T}$ .
Directions: for currents running opposite ways, the two circulating fields point the same way at the midpoint between them, so the contributions add rather than cancel.
$B_{\mathrm{mid}} = 2.0 \times 10^{-5} + 2.0 \times 10^{-5} = 4.0 \times 10^{-5} \ \mathrm{T}$ .
(c) The circulation around the loop is zero while the field at a point inside it is twice what one wire alone would give. Ampere's law was applied correctly and told you nothing about $\vec{B}$ , because no loop here has a constant $B$ along it.

$\oint \vec{B} \cdot d\vec{\ell} = 0$ around any loop enclosing both wires, while $B = 4.0 \times 10^{-5} \ \mathrm{T}$ at the midpoint. Zero circulation is not zero field, exactly as zero flux is not zero field on the electric side.

Frequently asked questions

What is the difference between Gauss's law and Ampere's law?

Gauss's law integrates the electric field over a closed surface and sets the result equal to the enclosed charge divided by epsilon zero. Ampere's law integrates the magnetic field around a closed loop and sets the result equal to mu zero times the current enclosed by that loop. So one uses a two-dimensional closed surface and counts charge inside a volume, the other uses a one-dimensional closed curve and counts current threading the area it bounds. Both are exactly true in general and both only yield a field when the symmetry lets you pull the field out of the integral.

Is Ampere's law the magnetic version of Gauss's law?

No. The magnetic version of Gauss's law is the closed surface integral of B dot dA equals zero, which is printed on the AP Physics C: Electricity and Magnetism equation sheet a few lines above Ampere's law. It says the magnetic flux through any closed surface is zero, because there is no magnetic charge to enclose. Ampere's law is the magnetic counterpart of Faraday's law in structure, since both are loop integrals of a field. Gauss's law and Ampere's law resemble each other only in that both close an integral and count an enclosed source.

Why does Gauss's law divide by epsilon zero while Ampere's law multiplies by mu zero?

It follows from how the two constants are defined. Epsilon zero, the vacuum permittivity, appears in the denominator of Coulomb's law through k equals one over four pi epsilon zero, so a larger permittivity means a weaker electric field for the same charge. Mu zero, the vacuum permeability, appears in the numerator of the Biot-Savart law, so a larger permeability means a stronger magnetic field for the same current. The AP sheet prints epsilon zero as 8.85 times ten to the minus twelve and mu zero as four pi times ten to the minus seven.

Does an Amperian loop have to be a circle?

No, but it has to be a path on which the calculation actually simplifies. For a long straight wire or a cylindrical conductor the right choice is a circle centred on the axis, because the field magnitude is constant along it and points along the path. For a long solenoid the right choice is a rectangle with one side inside the coil and one side outside, because the field is uniform inside and negligible outside, so only one side of the rectangle contributes. Any closed path is legal; only some make the integral collapse.

Are Gauss's law and Ampere's law on the AP Physics 2 equation sheet?

Neither one is. The AP Physics 2 CED contains no Gauss's law, no Ampere's law, no electric flux and no Gaussian surfaces. What the Physics 2 sheet prints instead is the finished result B equals mu zero I over two pi r for the field of a long straight wire, which is what Ampere's law produces for that case. Both closed-integral laws are AP Physics C: Electricity and Magnetism content, appearing as Topic 8.6 and Topic 12.4 respectively.

What does it mean when the circulation of B around a loop is zero?

It means the net current threading the loop is zero, with currents counted positive or negative according to the right-hand rule applied to the direction you traverse the loop. It does not mean the magnetic field is zero anywhere. Two parallel wires carrying equal and opposite currents give zero circulation around any loop enclosing both, while the field between them is twice what a single wire would produce. This mirrors the electrostatic case where a Gaussian surface around a dipole has zero net flux and a large field on it.

Which Maxwell's equations are Gauss's law and Ampere's law?

The AP Physics C: Electricity and Magnetism CED names both. Essential knowledge 8.6.A.6 states that Gauss's law is Maxwell's first equation. Essential knowledge 12.4.A.4 states that Maxwell's fourth equation is Ampere's law with Maxwell's addition, which adds that a changing electric field creates a magnetic field alongside the contribution from current. A boundary statement on that topic says the course does not expect students to use the fourth equation with a changing electric field, only to understand that a changing electric field generates a magnetic field.