Magnetic-Field Superposition & Mapping

Linearity is what licenses the whole method: doubling every current doubles every field, and switching on a second wire does not alter what the first one produces. So compute each source's contribution as though it were alone, then add. The adding is where students lose the thread. Contributions are vectors, and the angle between them settles the answer: two wires that each give 60 μT at a point sum to 120 μT when their directions coincide, to 85 μT when they meet at 90°, to 60 μT when they meet at 120°, and to nothing at all when they are opposed. One pair of magnitudes therefore covers every value from zero up to twice either of them. So fix each direction first, with the right-hand rule applied separately to each source — thumb along that current, fingers curling the way its own field circles — then resolve onto one pair of axes and add the components. Never add magnitudes and attach a direction afterwards.

A map is the same short calculation repeated. For a long wire through (x₀, y₀) carrying I in the +z direction, write its contribution at (x, y) straight into components: Bₓ = −(μ₀I/2π)(y − y₀)/r² and B_y = (μ₀I/2π)(x − x₀)/r², with r² = (x − x₀)² + (y − y₀)². The signs already carry the right-hand rule, so nothing is reasoned out twice. Test it on one point: a wire at the origin carrying 10.0 A out of the page gives, at (4.0 cm, 3.0 cm), r = 5.0 cm and magnitude μ₀I/2πr = 40 μT, split as Bₓ = −24 μT and B_y = +32 μT — perpendicular to the radius, as it must be. Now sample a grid, sum the kernel over every wire at each node, and draw an arrow. Arrows at true scale are useless, because the magnitude runs as 1/r and the near ones swamp everything else; draw each arrow along the direction of B with a length going as √|B|, or as ln(|B|/B_ref) against a stated reference field, and say on the figure which you used.

Zero field means every component is zero, so a null in a plane is two equations in two unknowns and is generically a point, not a region. On the line joining two parallel wires it collapses to one equation. With both currents in the same direction the contributions oppose only between the wires, so the null lies there, where μ₀I₁/2πd₁ = μ₀I₂/2πd₂, that is d₁/d₂ = I₁/I₂. With d = 0.20 m, I₁ = 12 A and I₂ = 4.0 A the ratio is 3.0 and d₁ = 0.15 m — three-quarters of the way across, close to the weak wire, where its field has grown enough to match. Reverse one current and the geometry inverts: between the wires the contributions now reinforce, and the null moves outside the pair, beyond the weaker wire, at distance d I₁/(I₁ − I₂) from the stronger one. Make the currents equal as well and that denominator vanishes: there is then no null on the line at all. Add a third source — Earth's field is always present — and the null shifts again, which is why a measured null never sits exactly where a two-wire calculation puts it.

Superposition also lets you design. Take two identical coaxial coils of radius R, N turns each, a distance s apart, carrying the same current the same way round. On the axis the sum is B(z) = (μ₀NIR²/2)[(R² + (z − s/2)²)^(−3/2) + (R² + (z + s/2)²)^(−3/2)]. The arrangement is symmetric about the midpoint, so every odd derivative of B already vanishes at z = 0 and the first thing that can spoil uniformity is the second derivative. Demanding d²B/dz² = 0 at the centre gives exactly s = R, which is what the name Helmholtz records. The leading surviving term is fourth order, B(z) ≈ B₀[1 − (144/125)(z/R)⁴], so the field is flat to better than 0.5% out to a quarter of a radius either side of centre. The centre value is B₀ = 8μ₀NI/(5√5 R) ≈ 0.7155 μ₀NI/R; with N = 130, I = 1.50 A and R = 0.150 m that is 1.17 mT. Pull the coils closer and the field peaks at the centre; push them past s = R and it sags into a double hump.

A Hall probe does not report B; it reports a voltage. Convert with B = (V − V₀)/S, where S is the sensitivity in V T⁻¹ and V₀ the reading in zero field — a few millivolts that drifts with temperature and, at S = 50 V T⁻¹, stands for tens of microtesla. Two habits make a map trustworthy. First, reverse the source current and take half the difference: forward, the probe sees Bₛ + B_bg; reversed, it sees −Bₛ + B_bg, so (V₊ − V₋)/2S returns Bₛ with the offset and every steady background — Earth's field, the magnetised steel bench, the transformer across the room — removed in one step. That matters, because Earth's field is 25 to 65 μT and a 100 μT bench coil cannot ignore it. Second, watch the orientation. The probe senses only B ⋅ n̂, so a misalignment θ reads |B| cos θ: 10° costs 1.5% and 25° costs 9.4%. Mapping a direction, not just a magnitude, means turning the probe through three orthogonal settings at the same point.

Uncertainties travel through superposition the way the fields do — in components, in quadrature, provided the sources are independent: u(Bₓ)² = u₁(Bₓ)² + u₂(Bₓ)² + … . The absolute uncertainty therefore survives cancellation untouched. Two contributions of 200 μT each known to 1% carry 2.0 μT apiece, so their difference near a null is 0 ± 2.8 μT: the value is small, its uncertainty is not, and the relative uncertainty is unbounded. Quoting a null to a percentage is meaningless. What is worth reporting is the position of the null, which is well conditioned. For two parallel wires d₁ = d/(1 + I₂/I₁), so a fractional mismatch δ in the current ratio moves the null by dδ/4 when the currents are nominally equal — 1.0 mm for a 2% mismatch across a 0.20 m gap. That displacement is measurable, and it is how a null earns its keep: as a sensitive comparator of two currents, not as a place where the field is known to be zero.