Normalize a radial current profile before using circulation

Normalize the inventory

An infinite cylindrical conductor of radius R carries total current I with Jz(r) = J0(r/R)^n. Integrating annular areas gives I = 2pi J0 R²/(n + 2), so J0 = I(n + 2)/(2pi R²). Changing n at fixed J0 would change total current; the simulator instead holds I = 2 A fixed.

Use symmetry to extract local field

Inside, Ienc = I(r/R)^(n + 2). Cylindrical symmetry makes B azimuthal and constant on a circular path, so Ampère’s static-current law gives B(2pi r) = mu0 Ienc. Thus the interior field scales as r^(n + 1), while exterior B = mu0 I/(2pi r). At r = 0 use the continuous zero-field limit rather than divide by zero.

Distinguish circulation from a pointwise determination

An Amperian path is an imaginary closed integration path, not a physical wire. The law constrains integral B dot dl. Without the stated symmetry, it does not justify replacing the integral by B times circumference or infer a unique local vector from one enclosed current. Different profiles here agree outside while disagreeing inside; an independent interior record can distinguish them.

Add a coaxial return inventory

In an original changed packet, a uniform core of radius 1 m carries +3 A and a shell from 1 to 2 m carries −3 A uniformly in cross-sectional area. At r = 1.5 m, enclosed current is 3 − 3(1.5² − 1)/(2² − 1) = 1.75 A. Therefore B = mu0(1.75 A)/(2pi×1.5 m) ≈ 0.233333 microtesla. Outside r = 2 m, total enclosed current is zero and so is B under this coaxial symmetry, even though fields inside are nonzero.

Worked original case

For n = 1, R = 1 m, I = 2 A and r = 0.5 m, enclosed current is 0.25 A. The field is 0.1 microtesla and circulation is approximately 0.314159 microtesla m. A uniform n = 0 profile at the same radius has different enclosed current despite the same exterior field.

Investigation

Normalize n = 0, 1 and 2 at fixed total I. Predict interior power laws, surface matching and common exterior values. Retain a proposed interior observation that distinguishes two profiles, then solve the written coaxial return packet.

Changed-case practice

Try a new practice case

For n = 2 and r/R = 0.5, what fraction of total current is enclosed?

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.