Distinguish source motion from force on a recipient

Declare the source and observer

A positive source charge q moves along x with speed v; the observation point is at displacement r = (0,y,1 m) from its instantaneous source position. In the stated low-speed quasistatic approximation, B = mu0 q(v cross r)/(4pi r³). This is not the full retarded relativistic field of arbitrary accelerated motion. The source-to-observer vector points from source to recipient, not the reverse.

Evaluate the cross product before magnitudes

With v = (v,0,0), v cross r = (0,−v times 1 m,vy). Therefore By is negative and Bz positive for positive q and y. Both v dot B and r dot B vanish. At fixed distance and speed, the source field magnitude varies as sin(theta) and is greatest when velocity and displacement are perpendicular; it vanishes along the velocity line in this approximation.

Apply a separate force law

A different test charge with velocity u feels magnetic force qtest(u cross B), not qsource(v cross B) unless those symbols actually refer to that recipient. Reversing only test charge reverses its force but does not change the prescribed source field. For a supplied B = +2 z-hat tesla, a test body with q = +1 C and u = +3 x-hat m/s has F = −6 y-hat N; replacing q by −1 C reverses it. These are mathematical quantities, not real apparatus specifications.

Combine electric and magnetic forces explicitly

The Lorentz force is q(E + u cross B). In a virtual crossed-field packet, E points +y and B points +z. A positive-x-moving body has magnetic force toward −y; cancellation requires u = E/B. With E = 8 N/C and B = 2 T, the cancellation speed is 4 m/s. Reversing charge reverses both forces, so the selected speed is unchanged under these stipulated directions.

Worked original case

For qsource = 1 nC, v = 2×10⁵ m/s and y = 1 m, r = sqrt(2) m. By ≈ −7.07107 picotesla and Bz ≈ +7.07107 picotesla, giving magnitude 10 picotesla. The tiny units are retained rather than formatted as zero.

Investigation

Predict both component signs before evaluating. Vary source speed and observer y separately and retain both perpendicularity checks. Then solve the supplied recipient-force and crossed-field packets while keeping source and test quantities distinct.

Changed-case practice

Try a new practice case

In the crossed-field packet E = 12 N/C and B = 3 T, what positive-x speed balances the forces, in m/s?

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