Assign the source field to the correct current recipient

Exclude the recipient’s self-field

The left long wire lies at x = 0 with current +2 A along z. The right wire at x = d carries signed current I2 along z. At the right wire, the left-source magnetic field is +y with magnitude mu0 I1/(2pi d). Force on a wire element is I dl cross Bexternal; its own singular ideal-wire field is not inserted into that recipient-force expression.

Derive force signs and reciprocity

For the right wire, +z cross +y points −x. Thus F2x per length = −mu0 I1I2/(2pi d). Same-sign currents attract, while opposite currents repel. Calculating the left recipient from the right source gives the opposite force per length. A negative component specifies direction in the fixed axes, not a negative magnitude.

Integrate a nonuniform assigned field

For a different straight segment along +x from 0 to L, carrying I, let Bz(x) = B0[1 + alpha(x/L)²]. Then Fy = −I integral Bz dx = −IB0L(1 + alpha/3). With I = 2 A, B0 = 3 mT, L = 0.5 m and alpha = 3, the force is −0.006 N. Using only the center field would give −0.00525 N and misses the quadratic average. This is an original force integral, not a claim that the assigned field is generated by the parallel-wire model.

Keep mechanical and microscopic energy accounts distinct

A magnetic force on an individual charge is perpendicular to its instantaneous velocity. A macroscopic current source can nevertheless exchange energy with a moving conductor through the complete electric/current system. The static two-wire packet computes forces only; it does not establish a free-energy machine or solve source dynamics.

Worked original case

For I1 = I2 = 2 A and d = 1 m, the right-wire force per length is −0.8 micronewton/m and the left is +0.8 micronewton/m. Reversing I2 changes both signs, yielding repulsion while preserving equal-and-opposite pair forces.

Investigation

Predict the right-wire force sign for positive, zero and negative I2. Double separation and retain the inverse-distance ratio. Then independently integrate the quadratic assigned-field packet and compare its exact average with the midpoint shortcut.

Changed-case practice

Try a new practice case

For I1 = 2 A, I2 = −3 A and d = 2 m, what is right-wire force per length in micronewton/m?

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: Magnetic source integration. The supplied cases, derivations and diagrams here are original; the source’s older exact-mu0 convention and classical spin language are not adopted literally.