ARCHETYPE: Floating Wires & Magnetic Balances

A copper wire has no business hovering in mid-air. Yet run enough current through two parallel wires the right way round, and the upper one floats — steadily, measurably, adjustably — on nothing but the magnetic push of its partner below. The force per metre is the $\mu_0 I_1 I_2 / 2\pi d$ of Parallel Currents & the Definition of the Ampere; what is new here is the balance sheet: set that magnetic force against a weight and you have the floating-wire archetype, a JEE and NEET fixture because it welds two chapters — magnetism and mechanics — into one three-line calculation. The gym has a small, fixed set of machines. The float itself: which current directions levitate, at what spacing, and whether the float is stable — nudge the wire and watch the ledger respond. The inversion: hang the wire below on attraction — same formula, opposite verdict. The three-wire sandwich: a middle wire balanced between two outer currents, at a point that turns out to be an old acquaintance from Net Fields of Wire Systems & Zero-Field Points. And the instrument the setup grew into: the current balance, which weighed the ampere for a century and, rebuilt by Bryan Kibble, ended up weighing the kilogram itself. Every equation in this article is a balance. Learn to write both sides in newtons per metre and the archetype has no variants left to surprise you.

Key takeaways

A wire floats above a fixed parallel wire when the magnetic force per length repels it upward and matches its weight per length: $\dfrac{\mu_0 I_1 I_2}{2\pi d} = \lambda g$ — repulsion needs opposite currents.
The float is vertically stable: rise and the $1/d$ repulsion weakens (weight wins, wire falls back); sink and it strengthens (pushed back up). The mirror setup — hanging below on attraction (same-direction currents) — is unstable.
Work per length in the balance: everything in N/m. Mass per length $\lambda$ (kg/m) meets force per length $\mu_0 I_1 I_2/2\pi d$ ; the wire's actual length cancels.
Three-wire sandwich: a middle wire balances where the two forces cancel — the same location as the zero-field line of Net Fields of Wire Systems, at distances $d_1/d_2 = I_1/I_2$ from the outer wires.
In gravity-free balances the equilibrium point is a zero-field point; in the floating-wire setup the field at the wire is decidedly non-zero — weight supplies the other side of the ledger.
KEY IDEA: every balancing-wire problem is one line of statics — magnetic force per length equals the other force per length, with $2\times10^{-7}\,I_1I_2/d$ on the magnetic side.

The core insight

A current-carrying wire balances when the magnetic force per unit length from its neighbours, $\mu_0 I_1 I_2/2\pi d$ toward or away from each, cancels the remaining forces per unit length (usually weight $\lambda g$ ) — and stability is decided by re-testing the ledger a small displacement away.

Where this lives today

Walk into any national metrology institute — NIST, NPL, BIPM — and the most carefully engineered object in the building is this archetype grown up: a Kibble balance, a coil carrying just enough current that the magnetic force on it balances a kilogram-scale mass. Since 2019 that balance is how the kilogram is realised. The same weigh-a-current logic runs smaller trades: MEMS accelerometers balance proof masses on the identical nudge-test logic, and short-circuit testing still confronts the violent side of wire-on-wire forces — the busbar-bending $I^2$ blows that Parallel Currents prices. The rails-and-rod version of a balanced conductor — with induced currents paying for the balancing — is coming up in E6 as The Rod on Rails.

The setup

The canonical machine: a long wire clamped to the bench carrying $I_1$ , and above it, parallel, a free wire of mass-per-length $\lambda$ carrying $I_2$ . Gravity pulls the free wire down at $\lambda g$ per metre. The clamped wire, by the chained argument of Parallel Currents & the Definition of the Ampere — my field at your wire, $F = BIL$ from Force on a Current-Carrying Conductor — pushes or pulls at $\mu_0 I_1 I_2/2\pi d$ per metre. For a float we need push: currents in opposite directions, repulsion aimed upward.

Figure

The floating wire — repulsion from below balances weight, and the 1 over d force makes the float self-correcting

Before any algebra, notice what kind of problem this is: statics with exotic ingredients — free-body diagram, forces balanced, stability by displacement test, exactly as in Rigid-Body Equilibrium. The only new entry is a force whose formula contains $\mu_0$ . Treat it as "a statics problem with a magnetic force" and it finishes in three lines.

The balance ledger

The magnetic side of every ledger in this gym is the force per unit length between long parallel wires, quoted from Parallel Currents & the Definition of the Ampere (which built it by chaining $B = \mu_0 I_1/2\pi d$ through $F = BIL$ — we never re-derive it, we spend it):

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2\times10^{-7}\, I_1 I_2}{d}\ \text{N/m} \qquad (1)

where: $d$ — the perpendicular spacing; the $2\times10^{-7}$ shortcut is exact because $\mu_0/2\pi = 2\times10^{-7}$ T·m/A exactly (in the pre-2019 convention carried by every exam). Direction from the same module's rule: parallel currents attract, antiparallel repel. The $1/d$ — not $1/d^2$ — dependence is about to do real mechanical work in the stability argument. The mechanical side, for the standard float, is weight per unit length. If the wire has mass per length $\lambda$ (kg/m), gravity claims $\lambda g$ newtons from every metre. Balance:

\frac{\mu_0 I_1 I_2}{2\pi d} = \lambda g \qquad (2)

where: both sides are per metre — the wire's total length has cancelled out of the problem, which is why no examiner bothers to give it. Solve for whichever symbol the stem withholds: the float height $d = \mu_0 I_1 I_2/2\pi\lambda g$ , a current $I_2 = 2\pi d \lambda g/\mu_0 I_1$ , or the linear density $\lambda$ . One equation, three costumes — that is the entire find-the-current / find-the-distance / find-the-mass ladder.

Derivation AThe floating condition and its stability verdict

We want the equilibrium spacing $d_0$ for a wire of mass per length $\lambda$ carrying $I_2$ above a fixed antiparallel wire carrying $I_1$ , and — the part worth marks at the Advanced level — a proof that the float is vertically stable. The strategy: write the vertical force ledger per unit length as a function of the spacing $d$ , find where it closes, then re-audit the ledger at $d_0 \pm$ a small nudge and read which way the net force points. The $1/d$ shape of the magnetic force does all the work; no calculus is needed, only the direction of an inequality.

Step 1. Ledger at spacing $d$ , per metre of wire, taking up as positive: repulsion supplies $+\dfrac{\mu_0 I_1 I_2}{2\pi d}$ , gravity supplies $-\lambda g$ .

Step 2. Equilibrium closes the ledger at $d = d_0$ :

\frac{\mu_0 I_1 I_2}{2\pi d_0} = \lambda g \;\;\Rightarrow\;\; d_0 = \frac{\mu_0 I_1 I_2}{2\pi \lambda g}

Why opposite currents: only repulsion can point up from a wire below; same-direction currents attract, and no attraction from beneath ever held anything up.

Step 3. Nudge the wire up to $d = 1.1\,d_0$ . The weight is unchanged, but the repulsion has fallen to $\dfrac{1}{1.1} \approx 0.91$ of $\lambda g$ — net force down: the wire is pulled back toward $d_0$ .

Step 4. Nudge it down to $d = 0.9\,d_0$ . The repulsion has grown to $\dfrac{1}{0.9} \approx 1.11$ of $\lambda g$ — net force up: pushed back again. Restoring in both directions: vertically stable. (It even oscillates about $d_0$ if disturbed — a pendulum with a magnetic floor.)

Step 5. Run the same audit on the inverted setup — wire hanging a distance $d_0$ below the fixed wire, held by attraction (currents parallel). Sag to $1.1\,d_0$ : attraction weakens to $0.91\,\lambda g$ , weight wins, the wire falls further. Rise to $0.9\,d_0$ : attraction strengthens, and the wire is reeled in until it hits the upper wire. Escaping or colliding, never returning: unstable. Verified: numbers from Board 1 below — $I_1 = 50$ A, $I_2 = 25$ A, $\lambda = 5.0$ g/m, $g = 10$ m/s²: $d_0 = 2\times10^{-7}\times1250/0.05 = 5.0\times10^{-3}$ m ✓. Dimension check on (2): $\mathrm{(T\,m/A)A^2/m} = \mathrm{N/m}$ against $\mathrm{(kg/m)(m/s^2)} = \mathrm{N/m}$ ✓. The stability asymmetry matches the sign of the slope of $1/d$ : force falling with distance stabilises a push, destabilises a pull.

The three-wire sandwich

Lay the setup flat so gravity leaves the stage: two fixed parallel wires A and C, distance $D$ apart on a horizontal plane, currents the same way; a free middle wire B carrying any current $I_B$ , parallel to both. A attracts (or repels) B; so does C; equilibrium demands the two tugs cancel:

\frac{\mu_0 I_A I_B}{2\pi x} = \frac{\mu_0 I_C I_B}{2\pi (D-x)} \;\;\Rightarrow\;\; \frac{I_A}{x} = \frac{I_C}{D-x} \qquad (3)

where: $x$ — B's distance from wire A. Note what cancelled: $I_B$ itself. The middle wire's own current sets how hard it is held, not where — and reversing $I_B$ flips both forces together, moving nothing. The position rearranges to $x = D\,I_A/(I_A + I_C)$ : closer to the smaller outer current, by the ratio rhyme $d_1/d_2 = I_1/I_2$ . That rhyme is the same equation that located the zero-field point between like currents in Net Fields of Wire Systems & Zero-Field Points — and the coincidence is a theorem, not luck.

Go deeper · Derivation B · 1 minWhy the force balance point is exactly the zero field line

Derivation BWhy the force balance point is exactly the zero field line

We show that for a straight wire parallel to a system of other straight parallel wires, "net force zero" and "net field zero" happen at the same place — the fact that lets the sandwich borrow every result from the zero-field census. The strategy is one structural observation about $F = BIL$ : the force on wire B is proportional to the net field of the others evaluated at B's own location. Watch the geometry requirement — all wires parallel — because that is what makes every contributing field perpendicular-to-B's-current and lets magnitudes add as signed numbers on one axis.

Step 1. At wire B's location, each outer wire contributes a field by $B = \mu_0 I/2\pi d$ , all lying in the plane perpendicular to B (directions by the right-hand grip, as catalogued in Net Fields of Wire Systems & Zero-Field Points). Call their vector sum $\vec B_{\text{net}}$ .

Step 2. The force per length on B, from Force on a Current-Carrying Conductor, is $F/L = B_{\text{net}} I_B$ — built from the net field, because forces superpose exactly as the fields that cause them do.

Step 3. Therefore $F = 0 \iff B_{\text{net}} = 0$ (for any nonzero $I_B$ ). The sandwich equilibrium of equation (3) and the zero-field line of the census are the same locus: between the wires for like outer currents, at $x = D\,I_A/(I_A+I_C)$ . Why the floating wire is different: add gravity and the ledger becomes $B_{\text{net}} I_B = \lambda g \neq 0$ — the wire balances where the field is decidedly non-zero (2 mT in Board 1's setup). Force-balance points and field-zero points coincide only when magnetic force is the ONLY entry in the ledger. *Verified: with $I_A = 8$ A, $I_C = 2$ A, $D = 10$ cm: equation (3) gives $x = 0.10\times8/10 = 8.0$ cm from A — 8/0.08 = 100 = 2/0.02 ✓, and the point sits nearer the 2 A wire, matching the census rule "zero hides nearer the smaller current".*

Figure

The three-wire sandwich — the middle wire balances at the zero-field line, nearer the smaller current

Stability in the sandwich is decided by B's own current direction. Parallel to the outer pair, both outer wires attract it: displaced toward A, the near attraction grows — B runs away. Unstable. Antiparallel, both repel: displaced toward A, the near repulsion pushes it back. Stable along the line. Same location, opposite verdicts — position from the field census, stability from the force ledger.

The variants at a glance

setup	condition	stability verdict
float above fixed wire	opposite currents; $\mu_0 I_1 I_2/2\pi d = \lambda g$	stable vertically (repulsion falls with distance)
hang below fixed wire	same-direction currents; same magnitude condition	unstable — sags away or reels in
middle wire, like outer currents	$I_A/x = I_C/(D-x)$ , independent of $I_B$	$I_B$ parallel: unstable · $I_B$ antiparallel: stable along the line
hanging-wire tension problems	string tension $T$ joins the ledger: $2T = (\lambda g \mp F/L)L$	attraction upward slackens strings; repulsion loads them
current balance (instrument)	unknown $F/L$ weighed against calibrated masses	the ampere's working standard for a century

One honesty note on the float: the stability proof covers vertical nudges only. Slide the wire sideways and the repulsion — always along the line joining the wires — acquires a component pushing it further off; real current balances guide the conductor mechanically. If a stem says "displaced horizontally", the answer flips to unstable.

Common trapA wire floats where the magnetic field is zero

Seductive, symmetric — and wrong in the setup examiners use most. The floating wire of equation (2) balances force against weight: the fixed wire's field at the float is $\mu_0 I_1/2\pi d$ , fully alive ( $2.0$ mT in Board 1 — four hundred times Earth's field), and it must be, because a zero field exerts zero force and zero force lifts nothing. The confusion comes from the gravity-free sandwich, where force balance genuinely does coincide with field zero — because there the magnetic force is the only ledger entry. Sort every problem by its ledger first: magnetic-vs-magnetic → equilibrium at the zero-field line of Net Fields of Wire Systems; magnetic-vs-gravity → equilibrium where the field is exactly strong enough, never zero.

Worked examples

Board level 1 — the standard float (NEET/JEE-Main pattern, direct)

A long horizontal wire fixed on a bench carries $I_1 = 50$ A. A second wire of linear mass density $\lambda = 5.0$ g/m, carrying $I_2 = 25$ A antiparallel to the first, floats freely above it. Find the height of the float. Take $g = 10$ m/s².

Step 1. Check the directions earn a float: antiparallel → repulsion → upward push from below ✓.

Step 2. Balance per metre, equation (2): $\dfrac{2\times10^{-7} I_1 I_2}{d} = \lambda g$ , with $\lambda g = 5.0\times10^{-3} \times 10 = 5.0\times10^{-2}$ N/m.

Step 3. Solve:

d = \frac{2\times10^{-7} \times 50 \times 25}{5.0\times10^{-2}} = \frac{2.5\times10^{-4}}{5.0\times10^{-2}} = 5.0\times10^{-3}\ \text{m} = \boxed{5.0\ \text{mm}}

Step 4. Sanity: millimetres, not metres — magnetic forces between bench-scale currents are gentle, so floats sit close. And the answer is independent of the wire's length: both sides of the ledger were per metre.

Board level 2 — find the mass (CBSE/ISC pattern, 2–3 marks)

Two long parallel wires $4.0$ mm apart carry equal currents of $20$ A in opposite directions. What linear mass density must the upper wire have to float at this spacing? Take $g = 9.8$ m/s².

Step 1. Magnetic force per length available at $4.0$ mm:

\frac{F}{L} = \frac{2\times10^{-7} \times 20 \times 20}{4.0\times10^{-3}} = 2.0\times10^{-2}\ \text{N/m}

Step 2. For the float this must carry the weight per length: $\lambda g = F/L$ , so

\lambda = \frac{2.0\times10^{-2}}{9.8} = 2.04\times10^{-3}\ \text{kg/m} \approx \boxed{2.0\ \text{g/m}}

Step 3. Sanity: a couple of grams per metre is thin copper — a strand of roughly a third of a millimetre. The float is physically buildable, which is exactly how lecture demonstrations of this problem are staged.

Competitive level 1 — the sandwich with unequal currents (JEE-Main pattern)

Wires A and C, fixed $10$ cm apart on a horizontal plane, carry $8.0$ A and $2.0$ A in the same direction. A free wire B, parallel to both, is placed between them. (a) Where is B in equilibrium? (b) Does the answer depend on B's current? (c) Is the equilibrium stable if B's current is antiparallel to A's and C's?

Step 1. Equilibrium demands the two pulls per metre cancel, equation (3): $\dfrac{I_A}{x} = \dfrac{I_C}{D - x}$ with $x$ from A.

Step 2. Solve: $x = D\dfrac{I_A}{I_A + I_C} = 0.10 \times \dfrac{8}{10} = \boxed{8.0\ \text{cm from A}}$ — that is, $2.0$ cm from C, nearer the smaller current, matching the zero-field census of Net Fields of Wire Systems & Zero-Field Points (check: $8/0.08 = 2/0.02 = 100$ ).

Step 3. (b) No: $I_B$ cancelled from equation (3). B's current decides how strongly it is held and which verdict stability takes — never the position.

Step 4. (c) Antiparallel $I_B$ : both outer wires repel B. Nudge it toward A and the near-side repulsion grows, pushing it back — stable along the line joining the wires. Timing note: ninety seconds once you recognise the position is the zero-field point — quote $x = D I_A/(I_A+I_C)$ and spend your thinking only on the stability verdict, which is where the marks hide.

Competitive level 2 — strings, slack and reversal (JEE-Advanced flavour)

A straight wire of mass $10$ g and length $0.50$ m hangs horizontally from two light vertical strings. A fixed parallel wire carrying $I_1 = 100$ A runs $5.0$ mm directly above it. (a) What current $I_2$ , and in which direction, makes the string tensions zero? (b) With that $I_2$ reversed, find the tension in each string. Take $g = 10$ m/s².

Step 1. For zero tension the magnetic force must carry the whole weight, $Mg = 0.010 \times 10 = 0.10$ N, and must point up — toward the fixed wire above. Upward attraction means $I_2$ runs parallel to $I_1$ .

Step 2. Set the total-force ledger (length $L = 0.50$ m is finite, so use forces, not per-length):

\frac{2\times10^{-7} I_1 I_2 L}{d} = Mg \;\;\Rightarrow\;\; \frac{2\times10^{-7} \times 100 \times I_2 \times 0.50}{5.0\times10^{-3}} = 0.10

Step 3. The left side is $2.0\times10^{-3}\,I_2$ , so $I_2 = \dfrac{0.10}{2.0\times10^{-3}} = \boxed{50\ \text{A, parallel to } I_1}$ .

Step 4. Reverse $I_2$ : attraction becomes repulsion, $0.10$ N now pointing down. The strings carry weight plus push:

2T = Mg + F = 0.10 + 0.10 = 0.20\ \text{N} \;\;\Rightarrow\;\; T = \boxed{0.10\ \text{N per string}}

Step 5. Sanity: reversal exactly doubled the load the strings feel compared with the no-current case ( $0.05$ N each) — the magnetic term flipped from cancelling gravity to seconding it. A hidden trap dodged: this equilibrium is maintained by the strings, so no stability analysis is owed; the free-float stability story applies only when magnetism alone holds the wire. Timing note: three minutes. The only decision is direction in (a) — "must attract upward → parallel currents", one sentence. Everything after is arithmetic; write the ledger once and reuse it for (b).

Exam traps

Attract/repel inverted. Parallel currents attract, antiparallel repel — the reverse of the two-charges instinct. Mis-set this and the float becomes a crash; Parallel Currents & the Definition of the Ampere owns the rule, borrow it consciously. A float above needs repulsion (opposite currents); a hang below needs attraction (same currents).

$2\times10^{-7}$ versus $10^{-7}$ . The pair formula carries $\mu_0/2\pi = 2\times10^{-7}$ ; the field formula carries $\mu_0/4\pi = 10^{-7}$ only inside Biot–Savart costumes. Factor-of-two errors here are the archetype's most harvested distractor.

Millimetres. Float spacings arrive in mm ( $5.0$ mm above); leave $d$ unconverted and the answer inflates a thousandfold. Convert on the first line.

Mass versus mass-per-length. $\lambda g$ balances force per length; $Mg$ balances total force with the length $L$ reinstated (Competitive 2). Decide which ledger the stem hands you before writing anything.

Stability asserted, not tested. The verdict needs the nudge test — displace, re-compare, state which wins. And the sideways honesty: the float is stable vertically only.

Field-zero confused with force-balance. Gravity in the ledger ⟹ the field at the wire is not zero (it cannot be). Only gravity-free wire-on-wire balances sit at the zero-field lines of Net Fields of Wire Systems.

Constants. $g = 9.8$ m/s² is the CBSE/NEET default; JEE stems usually grant $10$ (both flagged above). $\mu_0 = 4\pi\times10^{-7}$ T·m/A exactly, so $\mu_0/2\pi = 2\times10^{-7}$ exactly — no rounding lives in that number.

Origins · 1 minBryan Kibble and how this was discovered

Bryan Kibble

1938–2016 · National Physical Laboratory, Teddington

Bryan Kibble spent his career at Britain's National Physical Laboratory doing what this archetype does — balancing weights against magnetic forces on current-carrying conductors — to the most unforgiving precision on Earth. In 1975 he proposed the instrument now bearing his name: a coil whose current-borne force is weighed against a mass, with a second, moving-coil measurement that cancels every hard-to-know geometric factor. The trick converted "weigh a force" into "measure voltage and resistance" — quantities quantum standards could pin to the Planck constant. In 2019 the SI retired the platinum kilogram and redefined mass through $h$ , measured by Kibble balances; he died three years before the redefinition his balance made possible. A wire balanced on a magnetic push, done sufficiently well, outweighed a lump of metal guarded for 130 years.

History

1820

— Ampère measures the forces between current-carrying conductors within weeks of Ørsted's discovery — wires pushing wires becomes quantitative physics.

1884

— Rayleigh's current balance at Cambridge weighs the ampere against gravity: this archetype as a national standard.

1948

— the SI defines the ampere via the force between parallel wires, the story told in Parallel Currents and the Definition of the Ampere.

1975

— Kibble proposes the watt balance: weigh a magnetic force, cancel the geometry, reach the Planck constant.

2019

— the kilogram is redefined through h, realised worldwide by Kibble balances — a balanced conductor now anchors mass itself.

Going further

Texts. NCERT Physics Class 12, Ch. 4 (force between parallel currents; the floating-wire exercise) · HC Verma, Concepts of Physics Vol. 2, Ch. 35 exercises (suspended and balanced wires) · Irodov §3.2 (three-wire equilibria and stability).

Provenance. All numbers recomputed for this article: $d = 5.0$ mm (Board 1); $\lambda = 2.0$ g/m (Board 2); $x = 8.0$ cm with the $100 = 100$ ratio check (Competitive 1); $I_2 = 50$ A and $T = 0.10$ N per string (Competitive 2); FAQ figures $1.0\times10^{-3}$ N/m vs $5.0\times10^{-2}$ N/m and the $2.0$ mT field at the float.

Prerequisites. Parallel Currents & the Definition of the Ampere · Force on a Current-Carrying Conductor. Related. Net Fields of Wire Systems & Zero-Field Points (the census this gym's sandwich borrows) · Rigid-Body Equilibrium (mechanics) · Magnetic Forces Between Currents (UG deep dive). Coming up. The Rod on Rails in E6 — the balanced conductor, with induction paying for the current.

Why is floating above stable but hanging below unstable, when the formula is the same?

Because stability is about how the force changes with distance, not its value at balance. Floating on repulsion: rise → weaker push → pulled back; sink → stronger push → shoved back up. Restoring both ways. Hanging on attraction: sag → weaker pull → sags more; rise → stronger pull → reeled in. Amplifying both ways. Same magnitudes at $d_0$ , opposite verdicts — the response to a nudge is the entire story.

Is the floating wire stable if I push it sideways instead of up or down?

No — and stems that mention horizontal displacement are testing exactly this honesty. The repulsion always points along the line joining the two wires. Displace the float sideways and that line tilts: the push now has a horizontal component aimed away from the balanced position, so the wire slides off. Vertically stable, horizontally unstable. Demonstrations and real current balances add mechanical guides so only the stable axis is in play.

Why does the length of the floating wire never appear in the answer?

Both ledger entries scale identically with length: ten metres weigh ten times as much and receive ten times the push, so the balance divides out. Length re-enters only when something breaks the per-length symmetry — strings at the ends (Competitive 2 works in total forces), or mass and length handed separately, in which case build $\lambda = M/L$ first.

Can the middle wire in the sandwich be in equilibrium if the outer currents are in opposite directions?

Not between them. With antiparallel outer currents, the two forces on any middle wire point the same way between the wires (both toward, or both away from, the same side), so the ledger cannot close there — just as the field census of Net Fields of Wire Systems finds no zero between unlike currents. Equilibrium migrates outside the pair, beyond the weaker current, where the near wire's force is matched by the far wire's. Same location as the census's outside zero; the derivation above explains why the two problems must always agree.

Is this how maglev trains float?

Different mechanism, same ledger. Commercial maglev floats on forces from induced currents (arriving with electromagnetic induction in E6) or actively controlled electromagnets, not on two steady parallel currents. But the balance sheet at cruising height is literally equation (2)'s grammar: magnetic force upward equals weight. The floating wire is maglev's simplest ancestor.

How badly does the Earth's field interfere with a real floating-wire demonstration?

Pleasingly little — and checking is good practice. Earth's field is about $40\ \mu$ T; on Board 1's 25 A float it exerts $BI = 1.0\times10^{-3}$ N/m against a weight of $5.0\times10^{-2}$ N/m — a 2% nuisance, mostly horizontal. The partner wire's field at the float is $2.0$ mT, fifty times Earth's. Bench-scale balances legitimately ignore the planet; the instruments that cannot — compasses and dip needles — get their own module in E5.