Ampère’s law relates the line integral of B around a closed Amperian loop to the total current enclosed: ∮ B · dℓ = μ0 Ienc. In practice you pick a loop that matches the symmetry of the problem so B is constant along parts of the path or zero elsewhere. For a long straight wire this gives B = μ0I/(2πr); for a long ideal solenoid (many closely spaced turns, negligible outside field) B = μ0 n I inside and ≈0 outside. Use the right-hand rule to get B’s direction around currents and superpose fields from multiple conductors when needed. Don’t try Ampère’s law for nonsymmetric setups—the AP C: E&M exam expects it only for symmetrical cases (long wires, long solenoids, cylinders). Maxwell’s addition (displacement current) exists but you won’t need the full form on the AP exam. For a focused review and worked examples, see the Ampère’s law study guide (Fiveable) here: https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX—and practice problems at (https://library.fiveable.me/practice/ap-physics-c-e-m).

You use an Amperian loop because Ampère’s law turns a hard vector field problem into a simple scalar integral when the situation has symmetry. Instead of trying to measure B at every point, you choose a closed path where B is either constant or perpendicular/parallel to each segment so the line integral ∮B·dℓ = μ0 Ienc gives B directly (e.g., long straight wire → B = μ0I/2πr, long solenoid → B = μ0 n I). Directly computing B from the Biot–Savart law requires integrating vector contributions from every current element and is much harder except for simple geometries. Remember AP C only expects quantitative Ampère’s-law use for highly symmetric cases (long wires, long solenoids, cylinders), so pick Amperian loops that exploit that symmetry. For a quick refresher and worked examples, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and plenty of practice problems at (https://library.fiveable.me/practice/ap-physics-c-e-m).

Think of a long solenoid as lots of closely spaced current loops stacked along an axis. Using Ampère’s law with a rectangular Amperian loop that runs partly inside and partly outside the coil, the line integral around the loop picks up the current enclosed only from the turns inside the loop. For an ideal long solenoid the result is B_inside = μ0 n I (uniform, along the axis) and the contribution from the outside path is essentially zero, so B_outside ≈ 0. That’s why AP problems (and the CED) say long solenoids have uniform internal fields and negligible external fields (12.4.A.1.ii and 12.4.A.1.iii). Real, finite solenoids leak field at the ends and produce nonzero B outside (you’d use Biot–Savart to get that detail), but for AP-C you should apply the “long solenoid” idealization unless told otherwise. For a focused review, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and practice problems at Fiveable (https://library.fiveable.me/practice/ap-physics-c-e-m).

Use the shape and symmetry of the current to pick the formula. For a single long, straight wire use B_wire = μ0/(2π) · I / r—that comes from Ampère’s law with a circular Amperian loop centered on the wire (cylindrical symmetry). For an ideal long solenoid use B_sol = μ0 n I—that comes from an Amperian rectangle that encloses turns, assuming the field is uniform inside and negligible outside. Quick checks: is the conductor essentially one long straight current (use the wire formula)? Is it a tightly wound long solenoid with many turns per length and you want the inside field (use the solenoid formula)? The AP C exam expects only these symmetric cases (long straight wires and long solenoids)—use Ampère’s law with an Amperian loop aligned to the symmetry. For more examples and practice, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and more problems at (https://library.fiveable.me/practice/ap-physics-c-e-m).

Short answer: use symmetry. For a long straight wire the field is azimuthal and falls off with distance; Ampère’s law with a circular Amperian loop centered on the wire gives B_wire = μ0 I / (2π r). For an ideal (very) long solenoid the field inside is nearly uniform and parallel to the axis and essentially zero outside; choose an Amperian rectangle that runs partly inside and partly outside to get B_sol = μ0 n I (n = turns per length). Key differences to remember for the AP exam: the wire has cylindrical symmetry so B ∝ 1/r; the solenoid has translational symmetry along its axis so B inside is constant and independent of r. Pick your Amperian loop to match symmetry (circle for a wire, rectangle coaxial with solenoid loops). Use the right-hand rule for direction and superposition if multiple sources are present. The CED limits quantitative Ampère’s-law work to these symmetric cases (see Topic 12.4 in the CED and the Fiveable study guide) (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX). For more review and practice, check the unit overview (https://library.fiveable.me/ap-physics-c-e-m/unit-12) and 1000+ practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

In Bsol = μ0 n I, each symbol is literal: μ0 is the permeability of free space (μ0 = 4π × 10^−7 T·m/A), n is the number of turns per unit length of the solenoid (turns/m), and I is the current through each turn (A). Physically: each loop of wire creates a small magnetic field along the axis; having n loops per meter means n·I is the total “current-turns” per meter that add up. μ0 converts that current-turns density into magnetic field strength (in tesla). Derivation quick-check (Ampère’s law, AP-allowed): choose a rectangular Amperian loop that runs inside along the axis and back outside where B ≈ 0. ∮B·dl = Bℓ = μ0 (nℓ) I ⇒ B = μ0 n I. AP CED assumes a very long solenoid (uniform inside, negligible outside), so this formula is exactly the one you’ll use on the exam. For extra practice, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) or try problems at (https://library.fiveable.me/practice/ap-physics-c-e-m).

For an ideal, very long solenoid the magnetic field outside is (for AP problems) essentially zero because of symmetry and Ampère’s law. Pick a rectangular Amperian loop that goes partly inside the solenoid and partly outside. Ampère’s law says ∮B·dl = μ0 Ienc. For that loop Ienc equals μ0 times the net current piercing the loop; if the loop runs outside the winding region it encloses no net current (the many turns’ currents cancel for a very long, tightly wound solenoid), so the line integral around the outside portion is zero. By symmetry the field inside is uniform (B = μ0 n I) and the outside field must be negligible to make the integral consistent—so B_out ≈ 0. Remember this is the AP “ideal long solenoid” result (CED 12.4.A.1.ii and 12.4.A.1.iii). For more worked explanations and examples using Amperian loops, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and unit review (https://library.fiveable.me/ap-physics-c-e-m/unit-12).

Pick the Amperian loop to match the symmetry so B·dℓ is easy and |B| is constant along parts of the path. Quick rules you can use on AP C (Topic 12.4): - Identify symmetry: cylindrical (long straight wire, solid cylinder) → choose concentric circular loops about the wire axis so B is tangential and constant (gives B = μ0I/2πr). - Solenoid (very long, uniform inside): use a rectangular loop with one side inside (where B ≈ μ0nI) and one outside (B ≈ 0) so only the inside segment contributes. - Cylindrical current distributions: use circular Amperian loops coaxial with the cylinder; inside use enclosed current fraction (current density). - Always check that for chosen segments B is either parallel, antiparallel, or perpendicular to dℓ so the integral simplifies. Use right-hand rule to get direction and the sign of Ienc. AP exams only expect symmetric cases listed in the CED, so focus practice there. For a short guided review and examples, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

Use superposition: the net B at a point is the vector sum of the fields from each wire. For each long straight wire you can use the CED formula B = μ0 I / (2πr) (direction from the right-hand rule); then add those B vectors (account for sign/direction). Ampère’s law (∮B·dℓ = μ0 Ienc) is great when the geometry has high symmetry (long straight wire, long solenoid, concentric cylinders) so you can get B directly for that source, but it doesn’t give a simple shortcut for arbitrary multiple-wire arrangements. For non-symmetric sets use the Biot–Savart law or compute each wire’s B (magnitude + direction) and sum vectorially. On the AP exam you’ll only be asked to use Ampère’s law in symmetric cases—otherwise rely on superposition and the single-wire result (and the right-hand rule). For a quick refresher, check the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

Maxwell’s addition to Ampère’s law adds the “displacement current” term, μ0ε0 dΦE/dt, so the full law is ∮B·dℓ = μ0Ienc + μ0ε0 dΦE/dt. That means a changing electric flux (a changing electric field) produces a magnetic field just like a real current does. Practically this fixes the capacitor paradox: between capacitor plates there’s no conduction current, but the changing E-field gives a displacement current that produces the same B-field around the gap, keeping Ampère’s law and charge conservation consistent. For AP C: E&M you don’t need to solve many problems using the displacement-current term, but you must understand conceptually that changing E creates B (CED 12.4.A.4). Review the Topic 12.4 study guide for a clear summary (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX). For broader review and lots of practice problems, check the unit page (https://library.fiveable.me/ap-physics-c-e-m/unit-12) and the practice set (https://library.fiveable.me/practice/ap-physics-c-e-m).

Because magnetic fields are vectors and produced linearly by currents, the principle of superposition lets you find the net B at a point by vector-adding the contributions from each conductor. That’s crucial because Ampère’s law (CED 12.4.A.1) is only directly solvable for highly symmetric situations (long straight wire, long solenoid, cylinders). In most real setups with multiple wires or loops you can: - Use Biot–Savart or the known B for each simple element (e.g., B_wire = μ0I/(2πr)) - Treat each element’s field as a vector and add them componentwise to get the net field (CED 12.4.A.3) Superposition also tells you how directions combine—fields from currents in the same sense add, opposite senses subtract—so you can predict cancellations, reinforcement, or null points. For AP problems, use symmetry + Ampère where allowed; otherwise compute individual contributions and superpose. For a focused review on Ampère’s law and examples, see the topic study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX). For more practice, check the AP problem bank (https://library.fiveable.me/practice/ap-physics-c-e-m).

"Enclosed current" in Ampère’s law means the net electric current that actually passes through the surface spanned by your chosen closed Amperian loop. Mathematically Ienc is the algebraic sum of currents piercing that loop (currents going one way count positive, the opposite way negative, based on your chosen orientation). Ampère’s law uses that Ienc in the line integral ∮B·dℓ = μ0 Ienc to relate the circulation of B around the loop to the currents threading the loop. In practice pick an Amperian loop that matches the symmetry (e.g., a circle around a long straight wire or a rectangle through a solenoid) so it’s easy to tell which currents pass through the loop. For AP purposes you won’t need the displacement-current term except as conceptual context (Maxwell’s addition)—focus on counting physical currents through the loop. See the Topic 12.4 study guide for worked examples (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and more practice (https://library.fiveable.me/practice/ap-physics-c-e-m).

Use symmetry and the Amperian loop. If the current source is a long straight wire (cylindrical symmetry) choose a circular Amperian loop centered on the wire—B is tangent and depends on radius r, so Ampère’s law gives B = μ0 I/(2πr) (the 1/r falloff). For a long solenoid you have translational symmetry along its axis and many closely spaced turns: pick a rectangular Amperian loop that goes inside and outside. Inside the ideal (very long) solenoid B is uniform and parallel to the axis, and outside is negligible, so Ampère’s law gives B = μ0 n I (constant inside). Quick memory trick: “single straight conductor → radial dependence (1/r). many turns packed into a long tube → inside uniform (constant).” For more AP-aligned practice and worked examples, see the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and Unit 12 review (https://library.fiveable.me/ap-physics-c-e-m/unit-12).

Short answer: around a long straight wire the magnetic field circles the wire and falls off as 1/r (B = μ0 I / 2πr), while inside a long solenoid the field is nearly uniform and along the axis (B = μ0 n I) with almost zero field outside. Why that difference? Ampère’s law uses symmetry and an Amperian loop: for a straight wire you pick a circular loop centered on the wire and get the 1/r result. For an ideal long solenoid you choose a rectangular Amperian loop that encloses many turns; symmetry gives a constant B inside and negligible B outside (AP tests only require these symmetrical cases per the CED). Use the right-hand rule for field direction and superposition if you combine sources. Want practice applying Ampère’s law and solenoid/wire problems? See the Topic 12.4 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and thousands of practice Qs (https://library.fiveable.me/practice/ap-physics-c-e-m).

You assume solenoids are “very long” so you can use symmetry and ignore edge effects when applying Ampère’s law. For a long solenoid the turns are effectively translationally symmetric along the axis, so the magnetic field inside is uniform and parallel to the axis (B = μ0 n I) and the field outside is negligible. That symmetry makes a simple Amperian loop (rectangular around the axis) give a clean ∮B·dℓ = μ0 Ienc result. Real, finite solenoids have fringing fields near the ends, so the inside field isn’t perfectly uniform and there’s nonzero outside field—but AP CED limits quantitative Ampère applications to “long solenoids,” so you’re allowed to ignore those complications on the exam. For the official topic study guide and practice problems on Ampère’s law, see the Fiveable resource (https://library.fiveable.me/ap-physics-c-e-m/unit-5/4-amperes-law/study-guide/RURo66Hv1aueyDWX) and the unit page (https://library.fiveable.me/ap-physics-c-e-m/unit-12).