Predict the magnetic field produced by a current

A current can be the source of a magnetic field

The previous investigation prescribed a magnetic field without explaining how to make it. Here a steady conventional current in a long straight wire produces the field. Wire 1 stays at x = 0 and carries +2 A toward you. Wire 2 is to its right. The current symbols show direction along the wires; the blue arrows show the other wire’s magnetic field in the plane of the page.

Start with 2 cm separation and +2 A in wire 2. Predict the field from wire 1 at wire 2. Then set wire 2’s current to zero. Does removing that receiving current remove the field produced by wire 1? Capture both cases before comparing their forces.

Follow the circulation, not a radial arrow

Point your right thumb along the conventional current. Your curled fingers give the circulation of the magnetic field around a straight wire. For current toward you (+z, shown by a dot), the field circulates counterclockwise in this view. For current away (−z, shown by a cross), it circulates clockwise. Conventional current is the direction positive charge would move; electron drift in a metal is opposite.

At a point to the right of wire 1, its counterclockwise field points up (+y). At a point to the left of a positive-current wire 2, that wire’s field points down (−y). This is why the two field arrows can point oppositely even with currents in the same direction. Field lines are tangent to circles around their source, not arrows radiating toward or away from it. They describe a field direction, not a particle trajectory.

The displayed arrows show direction only. Read their magnitudes below the positions or in the ledger. The field-versus-separation curve uses fixed numerical axes for comparing strengths. The circles representing wires are symbols, not wire diameters.

Predict strength using distance and current

For an ideal infinitely long thin wire in vacuum, the field magnitude a perpendicular distance r from its axis is B = μ₀ |I|/(2πr). B is field in teslas, I is conventional current in amperes, r is distance in metres, and μ₀ is the vacuum magnetic permeability. The browser uses the rounded coefficient μ₀/(2π) ≈ 2 × 10⁻⁷ T m/A. This is a teaching approximation, not an exact definition of a unit.

At r = 2 cm = 0.020 m with I = 2 A, B = (2 × 10⁻⁷) × 2 / 0.020 = 2 × 10⁻⁵ T = 20 µT. A microtesla (µT) is 10⁻⁶ T. At 4 cm, the field is 10 µT. Doubling the distance halves this long-wire field; it does not quarter it. The inverse-distance rule differs from the inverse-square field of an isolated point charge because the source geometry differs.

Capture separations 1, 2 and 4 cm with wire 2 current fixed. Wire 1’s field at wire 2 is 40, 20 and 10 µT, respectively. Multiply each field by its separation in metres: the product stays approximately 4 × 10⁻⁷ T m. This gives a numerical check of the inverse-distance relationship. The curve is a sweep over different static configurations, not a record of moving wires.

Identify which source is changing

Keep separation 2 cm and compare wire 2 currents +1, +4, −4 and 0 A. Wire 1’s field at wire 2 remains +20 µT because its source current and the sampling distance have not changed. Wire 2’s field at wire 1 instead becomes −10, −40, +40 and 0 µT. Reversing current reverses its own field; doubling current magnitude doubles that field at the same position.

The ledger reports the other wire’s field at each wire, excluding the field of a wire at its own ideal line. The thin-line formula is singular at its source and is not an inside-wire model. Do not add the two ledger entries: they refer to different positions. To find a total field at a third point, evaluate both contributions at that same point and add vectors. Earth’s field and other nearby conductors would contribute in an actual measurement, but are omitted here.

Check the long-wire assumption

The equations assume straight wires extending far beyond the region being considered, with diameter much smaller than their separation. Real circuits have return paths and finite lengths; those can change both strength and direction. No source wiring, heating or motion is simulated. Do not treat the current settings as instructions to connect a bare wire directly across a supply.

For a centered straight segment of half-length L, its own field contribution at a perpendicular distance r from the midpoint is the infinite-wire value multiplied by L / √(L² + r²). For L = 0.10 m and r = 0.020 m, this factor is about 0.981: the segment’s contribution is about 1.9% below the infinite-wire approximation. This does not include the return path and therefore is not an error bound for a complete physical circuit. The browser continues to use the infinite-wire model.

Transfer to a different source

A single long wire carries 3 A toward you. Predict its field 6 cm to its right, then 12 cm to its right. What changes if the current is reversed? What extra information would be needed to predict a real compass reading there?

Check the current-field transfer

At 0.060 m, B ≈ (2 × 10⁻⁷) × 3 / 0.060 = 10 µT along +y. At 0.120 m, it is 5 µT in the same direction. Reversing current reverses both fields. A compass responds to the local total field, so background fields, return-path geometry and the compass orientation must also be considered. These distances and source current are outside the browser controls.

Explain one capture where nonzero field coexists with zero magnetic force on wire 2. Keep source current, sampling position and receiving current distinct. State which approximation you would test before comparing this prediction with an apparatus.

Reference for the long-wire field and finite-segment relation: OpenStax: magnetic field of a straight wire. The activities and numerical comparisons are original.