Choose the correct Amperian path for slabs and solenoids

Fix sheet current before changing thickness

An infinite slab occupies −a ≤ x ≤ a with current density J along +z. Fixing total sheet current K means J = K/(2a). Reflection and translation symmetries give B along y, odd across the center. A rectangular Amperian path has two nonzero side contributions, not one.

Integrate the enclosed slab current

For |x| ≤ a, the symmetric rectangle encloses current 2J|x| times its side length, giving By(x) = mu0 Jx. Outside, By = sign(x)mu0 K/2. Doubling thickness at fixed K halves the interior slope while preserving exterior magnitude. Reversing current reverses the field; the right-hand orientation gives +By at positive x for +Jz.

Transfer the loop argument to a long solenoid

A tightly wound very long solenoid with n turns per length and current I is modeled with uniform axial interior field and negligible exterior field away from ends. A rectangular loop with one long side inside gives B ell = mu0(n ell)I, hence B = mu0 nI in vacuum. For n = 500 per metre and I = 0.2 A, B ≈ 125.664 microtesla. Doubling n at the same current doubles this ideal result. Finite ends and nonlinear material cores require new assumptions.

Use superposition with orientation

Two long coaxial solenoids with opposing winding-current directions and products n1I1 = 100 A/m and n2I2 = 60 A/m give net axial B = mu0(40 A/m) ≈ 50.2655 microtesla in their shared ideal interior. Adding magnitudes would ignore orientation. A zero net field at one location need not imply zero source currents.

Name the dynamic boundary without overextending the law

Ampère’s current-only circulation law applies here to steady symmetric sources. Maxwell’s extension accounts also for magnetic fields generated by changing electric fields. Understanding that mechanism is core context, but quantitative changing-electric-field calculations are outside the stated Unit 12 requirement and are not silently applied in these static models.

Worked original case

For K = 2 A/m and a = 1 m, J = 1 A/m². At x = 0.5 m, By ≈ 0.628319 microtesla; the positive exterior limit is 1.25664 microtesla. At x = −0.5 m the field has equal magnitude and opposite sign.

Investigation

Compare half-thicknesses at fixed K and retain signed interior/exterior predictions. Draw the two contributing loop sides. Then derive the solenoid expression, solve the opposing-solenoid packet and state why a broken symmetry or changing source requires a different argument.

Changed-case practice

Try a new practice case

For a long vacuum solenoid with n = 200 per metre and I = 0.5 A, what is B in microtesla?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and model boundary

Original teaching checked 31 October 2026 against the current official framework, Unit 12. The displayed magnetic calculations use the stipulated introductory approximation mu0 = 4pi×10⁻⁷ SI, not a claim that this is exact in modern SI. All apparatus and records are virtual, supplied or generated. No real magnet handling, electrical current setup, particle apparatus or material operation is requested. Public feedback and retained reports are practice evidence, not independent mastery or performed laboratory time.

Supplemental primary teaching reference: Ampère’s law. The supplied cases, derivations and diagrams here are original; the source’s older exact-mu0 convention and classical spin language are not adopted literally.