Interpret magnetic flux and coil orientation

Field strength alone does not describe a coil

Two coils with the same area can face the same field differently. One presents its broad face to the field; the other is edge-on. Predict which has greater magnetic flux before moving the angle control. Flux combines field, area and orientation; it is not a force or a current.

The browser has 20 identical turns, each enclosing area 0.020 m². All turns share a uniform external field. The normal is an arrow perpendicular to the coil plane. Its angle θ is measured from reference +z. The side-view diagram shows the coil plane as a dark line and the normal in red. This is a schematic edge view, not a picture of 20 separate wires or a scaled physical coil.

Set ramp rate to zero and compare angles 0°, 60°, 90° and 180°. The applied field stays at +1 T throughout the two-second interval. Predict the signed flux and linkage before reading the ledger.

Use the normal, not the surface angle

For a uniform field and planar surface, magnetic flux per turn is Φ = Bz A cos θ in this geometry. Φ (phi) is flux in webers (Wb), Bz is the signed field along z in teslas, A is area in square metres, and θ is the normal’s angle from +z. One weber is one tesla square metre. The cosine selects the field component along the normal. The formula uses the normal-to-reference angle, not the angle between the field and the coil plane.

At 0°, the normal and positive field are aligned and cos θ = 1. With Bz = +1 T and A = 0.020 m², Φ = +0.020 Wb. At 60°, cos θ = 0.5, giving +0.010 Wb. At 90°, the field lies in the plane and the flux is zero, even though the magnetic field is not zero. At 180°, the chosen normal points against +z, giving −0.020 Wb. A negative flux is a direction relative to a reference, not a negative field strength or negative surface area.

The graph uses signed Bz. If the ramp later reverses the field, θ remains the normal’s angle from reference +z, and the negative Bz carries the reversal. Do not change both θ and the field sign merely because the field points the other way.

Separate flux per turn from linkage

For N turns with the same flux through each turn, flux linkage = NΦ. With N = 20 and θ = 0°, the initial linkage is 20 × 0.020 = 0.400 Wb turns. The word “turns” makes the counting explicit; turns are dimensionless. The field through one turn has not become 20 times stronger. Each turn contributes to the circuit’s induced emf when linkage changes.

If turns had different areas or sampled different fields, add their individual fluxes instead of multiplying one value by N. Nonuniform fields require subdividing the surface, calculating each patch’s normal field times area, then adding. The browser assumes uniform field and equal turns, so no patch integration is required here.

A reference sign is not a physical effect

At 0° and 180°, the coil has the same unoriented plane but opposite normal references. Signed flux changes sign. The positive circulation reference also changes: it is counterclockwise viewed from the normal tip toward the coil. A sign change caused only by relabeling the normal must not be described as a physical reversal when viewed from the same fixed side. Keep the viewing direction explicit whenever reporting clockwise or counterclockwise current.

During each plotted run, the angle stays fixed. Moving the angle slider chooses a new static orientation for another run; it does not simulate rotating a coil during the interval. A physically rotating coil can induce emf even in a constant field, but that is outside this investigation.

Contrast value with change

With ramp rate zero and θ = 0°, linkage stays +0.400 Wb turns and induced emf is zero. A nonzero linkage does not imply an induced current. At θ = 90°, both linkage and its change are zero for every available ramp rate, because the changing field remains in the plane. The next lesson uses the slope of linkage to determine voltage.

Transfer to another coil

A 50-turn coil has 30 cm² area per turn in a uniform +0.20 T field. Its normal makes 60° with the positive field direction. Calculate flux per turn and linkage. What if the stated 60° were instead the angle between the field and the coil plane?

Check the flux transfer

30 cm² = 30 × 10⁻⁴ m² = 0.0030 m². Flux per turn is 0.20 × 0.0030 × 0.5 = 0.00030 Wb. Linkage is 50 times that, 0.015 Wb turns. If 60° were measured from the plane, the normal angle would be 30°; flux would instead be approximately 0.000520 Wb and linkage 0.0260 Wb turns. State the angle convention before calculating.

Use captures to explain a zero-flux case with nonzero field and a nonzero-flux case with zero induced emf. Distinguish a field property from the flux through a chosen oriented surface.

Reference: OpenStax: magnetic flux and Faraday’s law. The orientation comparisons and transfer problem are original.