Enclosed current (I_enc) is the net electric current passing through the surface bounded by an Amperian loop. In Ampère's law, ∮B⃗·dl⃗ = μ₀I_enc, so only the current threading the loop determines the line integral of the magnetic field, not the total current in the conductor.
Enclosed current is the net current that actually pokes through the imaginary surface bounded by your Amperian loop. Ampère's law says ∮B⃗·dl⃗ = μ₀I_enc, and the right side only counts current inside the loop. Current outside the loop still creates a magnetic field everywhere, but its contribution to the line integral around the loop cancels out perfectly. That's the subtle (and very testable) part.
Finding I_enc is usually the hardest step of an Ampère's law problem. If the loop encloses the whole wire, I_enc is just I. If you're inside a conductor, you only count the fraction of current inside radius r. For uniform current density that's I_enc = I(r²/R²), and for non-uniform density like J = kr, you have to integrate, I_enc = ∫J dA = ∫₀ʳ (kr')(2πr' dr'). Currents in opposite directions partially cancel, so 'net' really means net. Two equal and opposite currents through the same loop give I_enc = 0.
Enclosed current lives in Topic 12.4 (Ampère's Law) in AP Physics C: E&M, and it's the variable that makes or breaks every Ampère's law calculation. The symmetry argument tells you ∮B⃗·dl⃗ = B(2πr) for a circular loop, but B itself depends entirely on what you decide I_enc is. Coaxial cables, hollow cylinders, slabs of current, and solenoids all reduce to the same question. Draw a loop, figure out what current threads it, solve for B. If you can set up I_enc correctly, the rest is algebra. It's also conceptually deep, since it explains why the field inside a hollow current-carrying cylinder is zero (no enclosed current means the integral vanishes) even though there's plenty of current nearby.
Keep studying AP® Physics C: E&M Unit 12
Ampère's Law (Unit 12)
Enclosed current is the right-hand side of Ampère's law, the way enclosed charge is the right-hand side of Gauss's law. The whole strategy of Topic 12.4 is choosing a loop where symmetry makes the integral easy, then carefully counting I_enc.
Magnetic field of a solenoid (Unit 12)
The famous B = μ₀nI result comes straight from counting enclosed current. A rectangular Amperian loop that threads N turns of the solenoid encloses I_enc = NI, and that's where the turns-per-length factor n comes from.
Gauss's Law and enclosed charge (Unit 8)
Enclosed current is the magnetic cousin of enclosed charge. In Gauss's law, only charge inside the surface sets the flux; in Ampère's law, only current through the loop sets the line integral. If you mastered q_enc with non-uniform charge density, the integral for I_enc with J = kr is the exact same move.
Permeability (Unit 12)
The constant μ₀ is what converts enclosed current into the line integral of B. It plays the same role for magnetism that ε₀ (sitting in the denominator) plays for Gauss's law in electrostatics.
Ampère's law problems almost always hinge on computing I_enc, and the exam loves making that step nontrivial. Multiple-choice stems give you a loop around a straight wire and ask for ∮B⃗·dl⃗ (answer: μ₀I, regardless of the loop's shape, as long as the wire passes through it). Harder questions hand you a non-uniform current density like J = kr and ask for B at r = R/2, which forces you to integrate J over the enclosed area instead of grabbing the total I. Hollow cylinders test whether you know I_enc = 0 for r < a, so B = 0 there. Current-carrying slabs ask how B varies with distance from the central plane, where I_enc grows linearly with z. On FRQs, expect to draw or describe your Amperian loop, justify the symmetry, write Ampère's law, and explicitly set up I_enc before solving for B. Skipping the I_enc setup is where points die.
The total current I is everything flowing in the conductor; the enclosed current I_enc is only the part passing through your Amperian loop. They're equal when your loop wraps around the entire conductor, but inside a wire of radius R at distance r < R, only a fraction is enclosed. For uniform current density, I_enc = I(r²/R²). Plugging total I into Ampère's law inside a conductor is one of the most common wrong answers on these problems.
Enclosed current is the net current passing through the surface bounded by an Amperian loop, and it equals ∮B⃗·dl⃗ divided by μ₀.
Only current threading the loop counts; currents outside the loop affect B at individual points but contribute zero to the line integral around the loop.
Inside a conductor with uniform current density, I_enc = I(r²/R²), so the field grows linearly with r inside and falls off as 1/r outside.
For non-uniform current density like J = kr, find I_enc by integrating J over the enclosed area using rings: I_enc = ∫₀ʳ J(r')(2πr')dr'.
Inside the hollow region of a cylindrical conductor (r < inner radius), I_enc = 0, so the magnetic field there is zero.
Currents in opposite directions through the same loop subtract, which is exactly why the field outside an ideal coaxial cable is zero.
It's the net electric current passing through the surface bounded by an Amperian loop. Ampère's law states ∮B⃗·dl⃗ = μ₀I_enc, so this enclosed current is what determines the line integral of the magnetic field around the loop.
Yes and no, and this trips people up. Outside current does contribute to B at every point on the loop, but its contributions to the line integral ∮B⃗·dl⃗ cancel out exactly, so it adds nothing to μ₀I_enc. That's why Ampère's law only works cleanly when symmetry lets you pull B out of the integral.
Total current is everything the conductor carries; enclosed current is only what passes through your loop. Inside a uniform wire at r < R, I_enc = I(r²/R²), not I. They only match when the loop encloses the whole conductor.
Integrate the current density over the enclosed area. For J = kr in a cylindrical wire, slice the cross-section into rings of area 2πr'dr', so I_enc = ∫₀ʳ kr'(2πr')dr' = 2πkr³/3. This is the same technique as finding enclosed charge for a non-uniform charge density in Gauss's law.
Yes, for points inside the hollow region (r less than the inner radius). An Amperian loop there encloses zero current, and by symmetry B(2πr) = μ₀(0), so B = 0 even though current flows in the surrounding shell.
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