The magnetic field of a solenoid is the uniform field inside an ideal, tightly wound coil, with magnitude B = μ₀NI/ℓ = μ₀nI, where N is the total number of turns, I is the current, ℓ is the length, and n = N/ℓ is the turn density. It points along the axis and is derived using Ampère's law.
A solenoid is a long coil of wire wound into many closely spaced loops. When current I runs through it, the loops' fields stack up to create a strong, nearly uniform magnetic field inside the coil, pointing along the axis, while the field just outside an ideal solenoid is approximately zero. The magnitude inside is B = μ₀NI/ℓ, often written B = μ₀nI where n = N/ℓ is the number of turns per unit length.
The formula comes straight from Ampère's law (Topic 12.4). You draw a rectangular Amperian loop with one side inside the solenoid and one side outside. Only the inside segment contributes to ∮B·dℓ, and the enclosed current is the current I threading through each of the turns the loop captures. The key insight is that B depends on how densely the turns are packed (n), not on the raw turn count or the radius of the coil. A solenoid is basically a machine for making a uniform B field, the magnetic cousin of the parallel-plate capacitor's uniform E field.
This lives in Topic 12.4, Ampère's Law, in Unit 12 of AP Physics C: E&M. The solenoid is one of the canonical Ampère's law setups, alongside the long straight wire and the toroid, so you're expected to derive B = μ₀nI from scratch, not just memorize it. That means choosing a smart Amperian loop, arguing from symmetry which segments contribute, and counting the enclosed current correctly.
It also matters downstream. The solenoid's uniform internal field is the starting point for computing magnetic flux, which feeds directly into Faraday's law and the self-inductance of an inductor in Unit 13. If you can't get B inside a solenoid quickly, half of electromagnetic induction gets harder.
Keep studying AP® Physics C: E&M Unit 12
Enclosed Current (Unit 12)
In the solenoid derivation, the Amperian loop is pierced by the wire multiple times, so the enclosed current is NI_loop-turns, not just I. Counting how many turns thread your loop is the whole game in this derivation.
Permeability (Unit 12)
The μ₀ in B = μ₀nI is the permeability of free space. Stuff the solenoid with a ferromagnetic core and the effective permeability jumps, which is why real electromagnets have iron inside.
Superposition Principle (Unit 12)
Conceptually, the solenoid's field is just the superposition of many single-loop fields. Inside the coil the loop fields reinforce; outside they nearly cancel, which is why the ideal solenoid has B ≈ 0 outside.
Inductance and Faraday's Law (Unit 13)
The self-inductance of a solenoid, L = μ₀N²A/ℓ, is derived by computing the flux NBA using B = μ₀nI. Every inductor problem in Unit 13 secretly starts with this field formula.
Multiple-choice questions love the stretching trap: a solenoid of length L and N turns is stretched to 2L while N and I stay fixed. Since B depends on turn density n = N/ℓ, halving the density halves the field, so B_new = B₀/2. If you memorized B = μ₀NI and forgot the ℓ, you'll miss this. Other MCQ stems compare solenoids with different n, I, or radius (radius is a distractor; B inside doesn't depend on it for an ideal solenoid).
On the free-response side, expect to derive B = μ₀nI using Ampère's law with a rectangular loop, justifying why only one side contributes, or to use the solenoid field as step one of an induction problem (find B, then flux, then EMF or inductance). Show the Amperian loop and the enclosed-current count explicitly; that's where derivation points live.
B = μ₀nI and B = μ₀NI/ℓ are the same formula, but mixing up n and N is the most common solenoid error. N is the total number of windings; n = N/ℓ is how densely they're packed. The field cares only about density. Stretch the coil and N stays the same, but n drops and so does B. On a problem, check the units of what you're given: 'turns' means N, 'turns per meter' means n.
The field inside an ideal solenoid is B = μ₀NI/ℓ = μ₀nI, uniform, and directed along the axis; outside an ideal solenoid the field is approximately zero.
B depends on turn density n = N/ℓ, so stretching a solenoid to twice its length with the same number of turns cuts the field in half.
The field inside an ideal solenoid does not depend on the coil's radius or on your position inside it, which is what makes it uniform.
You derive the formula with Ampère's law using a rectangular Amperian loop, where only the segment inside the solenoid contributes to ∮B·dℓ and the enclosed current is I times the number of turns the loop captures.
Use the right-hand rule on the current in the loops to find the field direction; your curled fingers follow the current and your thumb points along B inside the coil.
The solenoid field is the foundation for inductance, since L = μ₀N²A/ℓ comes from plugging B = μ₀nI into the flux.
Inside an ideal solenoid, B = μ₀NI/ℓ (equivalently μ₀nI with n = N/ℓ), uniform and pointing along the axis. It's derived using Ampère's law in Topic 12.4 with a rectangular Amperian loop.
Yes. Stretching to length 2L with the same N and I halves the turn density n, so B drops to half its original value. This exact setup is a favorite multiple-choice question.
No. For an ideal solenoid, B = μ₀nI has no radius dependence, and the field is the same magnitude everywhere inside. Radius shows up later only when you compute flux (Φ = BA) for induction problems.
A solenoid is a straight coil with a uniform internal field B = μ₀nI; a toroid is a solenoid bent into a donut, where B = μ₀NI/(2πr) depends on the distance r from the center, so it is not uniform. Both are standard Ampère's law applications in Topic 12.4.
For an ideal (infinitely long, tightly wound) solenoid, yes, the external field is taken as zero, and that assumption is what makes the Ampère's law derivation work. Real solenoids leak a small field near the ends, but AP problems treat them as ideal unless told otherwise.
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