ARCHETYPE: Net Fields of Wire Systems & Zero-Field Points

Put two current-carrying wires near each other and their magnetic fields do not negotiate — they simply add, vector by vector, at every point in space. Out of that blunt arithmetic comes rich geography: places where the two circular field patterns reinforce, and special places where they cancel exactly — points of pure magnetic silence a few centimetres from amperes of current. Examiners have mined this geography for decades, because it tests three skills at once: recalling the field of a single wire, aiming it with the right hand, and adding vectors like you mean it. The setup family is small and completely learnable. Currents the same way: the zero-field point hides between the wires, nearer the weaker current. Opposite ways: the between-fields gang up instead, and the zero point escapes outside, beyond the weaker wire. Add a third wire, or a loop beside a straight wire, and the same recipe runs — magnitudes from the standard formulas, directions from the right hand, then components. This archetype drills exactly that recipe. The field formulas — $\mu_0I/2\pi d$ for a long wire, $\mu_0I/2R$ at a loop's centre — are owned and derived by Fields of Straight Wires & Circular Loops; the tug-of-war they exert on the wires is Parallel Currents & the Definition of the Ampere. Here the fields are ingredients; the dish is the map — net field at a named point, and the hunt for the point where the map goes blank.

Key takeaways

Magnetic fields superpose: net $\vec B$ = vector sum of each wire's $\dfrac{\mu_0 I}{2\pi d}$ , aimed by the right-hand grip rule.
Like (parallel) currents: the between-fields oppose → a zero-field point sits between the wires at $x = \dfrac{d\,I_1}{I_1 + I_2}$ from wire 1, closer to the smaller current.
Unlike (antiparallel) currents: the between-fields add → the zero point moves outside, beyond the smaller current, at $y = \dfrac{d\,I_2}{I_1 - I_2}$ from wire 2.
Perpendicular contributions combine by Pythagoras; corner-of-a-square and wire-plus-loop setups are component bookkeeping.
A wire feels no force from its own field — only from the others' fields at its location.
Key idea: sketch each wire end-on with ⊙/⊗, draw each field vector at the target point with the right hand, then add components — the whole archetype is one disciplined recipe.

The core insight

The net magnetic field of any wire system at a point is the vector sum of the individual fields $\mu_0I/2\pi d$ (each aimed by the right-hand grip rule), so like currents cancel somewhere between them and unlike currents cancel somewhere outside — at the position where the two magnitudes match.

Where this lives today

Every circuit-board designer fights this archetype daily: layout guidelines insist a signal trace's return path hug it, because closely spaced opposite currents cancel their far fields (our $d \to 0$ limit), keeping one part of a board from whispering noise into another. The same logic twists live and neutral together in mains cable and sums three-phase line currents to zero — the bundle's distant field nearly vanishes. Engineers exploit the loud zones too: busbar current sensors sit where fields reinforce. And deliberately engineered zero-regions — the null points of tokamak divertors, where field lines are peeled apart — turn this exam problem into a billion-euro design constraint.

The setup

Draw the wires end-on. That single habit converts every problem in this family into flat geometry: a current towards you is a dot ⊙ (an arrowhead approaching), away is a cross ⊗ (the fletching leaving), and each wire's field becomes concentric circles you can actually draw — anticlockwise around ⊙, clockwise around ⊗.

Figure

Two parallel wires end-on — like currents cancel between, unlike currents cancel outside

The ingredients arrive from two neighbouring modules: Fields of Straight Wires & Circular Loops supplies the magnitudes and the right-hand rules that aim them; Parallel Currents & the Definition of the Ampere covers what the fields do to the wires. This module asks a different question — standing at a point $P$ , what single vector $\vec B$ does the whole system produce there? The answer is a disciplined march that never gets harder than components:

Step 1. Sketch end-on: ⊙/⊗ each wire, mark $P$ , write the wire-to- $P$ distances.

Step 2. Magnitudes: $B_i = \mu_0I_i/2\pi d_i$ (or the loop-centre formula). Keep $\mu_0/2\pi = 2\times10^{-7}$ loaded — every magnitude becomes " $2\times10^{-7}$ times current over distance" mental arithmetic.

Step 3. Directions: at $P$ , each field is tangent to the circle through $P$ — anticlockwise around ⊙, clockwise around ⊗.

Step 4. Add as vectors: collinear contributions carry signs; perpendicular ones go through $B = \sqrt{B_x^2 + B_y^2}$ . The exam-defining special case is the zero-field point — the magnetic neutral point — where the contributions cancel exactly. If you met Null Points & the Third-Charge Equilibrium in electrostatics, the craft will feel familiar: same hunt, same "closer to the weaker source" logic, one twist. Electric fields point along the source-to-point line; wire fields point around the wires, so cancellation is possible only where geometry aligns them — for two parallel wires, on the line joining them (and, as we will derive, along a whole line parallel to the wires).

Where the field goes silent: the two cancellation regimes

Stand between two wires whose currents both come out of the page (left panel). The left wire's anticlockwise field points up there; the right wire's — you are on its left side — points down. The fields fight, and somewhere the weaker wire matches the stronger at shorter range: the field vanishes. That "somewhere" is pure algebra:

\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi (d - x)} \qquad (1)

where: $x$ — distance of the zero point from wire 1; $d$ — the separation. The $\mu_0/2\pi$ factors cancel before you compute anything — the equation is really just $I_1/x = I_2/(d-x)$ . Solving,

x = \frac{d\,I_1}{I_1 + I_2} \qquad (2)

where: the position splits the gap in the ratio of the currents, $x/(d-x) = I_1/I_2$ — closer to the smaller current, which can only match the strong wire's field at short range. Equal currents put the silence exactly midway. Flip one current and the geography inverts. Between antiparallel wires both fields point the same way (right panel: both up) — the middle is loud, and cancellation is exiled outside. Out there the two fields oppose, but magnitudes can match only beyond the weaker wire (outside the stronger one, its field is bigger and nearer: no contest). With $I_1 > I_2$ and $y$ measured outward from wire 2:

\frac{\mu_0 I_1}{2\pi(d + y)} = \frac{\mu_0 I_2}{2\pi y} \;\;\Rightarrow\;\; y = \frac{d\,I_2}{I_1 - I_2} \qquad (3)

where: $y$ — distance beyond wire 2, away from wire 1. Read the denominator physically: nearly equal currents push the zero far away ( $y \to \infty$ as $I_2 \to I_1$ — an equal-and-opposite pair never cancels at finite distance); a feeble $I_2$ pulls it in close.

Derivation ALocating the zero-field point for both current arrangements

We derive the zero-point positions (2) and (3) for two long parallel wires a distance $d$ apart, carrying $I_1$ and $I_2$ . The strategy is the master recipe frozen at the special point: first use directions to find where cancellation is even possible, then match magnitudes there. The subtlety worth guarding is the direction census — on which segment of the line the two fields oppose. We work on the line through both wires, wire 1 at position $0$ , wire 2 at $d$ .

Step 1. Direction census, like currents (both ⊙). Between the wires: wire 1's anticlockwise circulation points up, wire 2's — seen from its far side — points down. Opposed ⟹ cancellation possible. Outside the pair both give the same sense ⟹ never zero outside.

Step 2. Match magnitudes between the wires:

\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi(d-x)}

Step 3. Cancel $\mu_0/2\pi$ , cross-multiply, solve:

I_1(d - x) = I_2\,x \;\;\Rightarrow\;\; x = \frac{d\,I_1}{I_1 + I_2}

Why closer to the smaller current: the weaker wire's $I/d$ ratio must equal the stronger's — a smaller numerator needs a smaller denominator. The distance from wire 2, $d - x = dI_2/(I_1+I_2)$ , shows the ratio-split symmetrically.

Step 4. Direction census, unlike currents ( $I_1$ ⊙ at $0$ , $I_2$ ⊗ at $d$ ). Between the wires both fields agree ⟹ never zero there. Outside they oppose on both segments — but magnitudes can only match beyond the weaker wire. With $I_1 > I_2$ and a point at distance $y$ beyond wire 2:

\frac{\mu_0 I_1}{2\pi(d+y)} = \frac{\mu_0 I_2}{2\pi y} \;\;\Rightarrow\;\; I_1\,y = I_2\,(d + y) \;\;\Rightarrow\;\; y = \frac{d\,I_2}{I_1 - I_2}

Why beyond the weaker wire only: outside the stronger wire its field is bigger and closer simultaneously — no match possible. Demanding the zero beyond wire 1 instead yields a negative distance: the mathematics refusing the geometry for you.

Step 5. The answer is really a line: every cross-section along the wires sees the same geometry, so the zero locus runs parallel to both wires. Verified numerically: like currents 3 A and 6 A at $d = 10$ cm give $x = 3.33$ cm from the 3 A wire — both fields $1.8\times10^{-5}$ T there ✓. Unlike currents 8 A and 2 A at $d = 15$ cm give $y = 5$ cm beyond the 2 A wire — both fields $8\times10^{-6}$ T ✓.

Two remarks make these results working knowledge. First, at a zero-field point a moving charge feels no magnetic force, and a third wire laid along the zero line feels none either — how examiners disguise this computation as equilibrium, and a bridge to Floating Wires & Magnetic Balances later in E4. Second, a wire does not feel its own field: its force comes only from the other wires' fields at its location — the tug-of-war of Parallel Currents & the Definition of the Ampere.

Beyond two wires: components, corners and loops

Figure

Three into-page wires on a square's corners — rotate each wire-to-P line 90 degrees clockwise, then add components

Nothing new happens with more wires — only more rows in the vector table, as the figure's corner drill shows. One configuration deserves its own paragraph because it fuses the two field formulas: the wire-plus-loop cancellation. A loop's centre field $\mu_0I_L/2R$ points along its axis. Park a long straight wire in the loop's plane at distance $D$ from the centre: the wire's field there, $\mu_0I_w/2\pi D$ , is also perpendicular to that plane — antiparallel to the loop's field if the current runs the right way. Antiparallel and equal is a design condition:

\frac{\mu_0 I_L}{2R} = \frac{\mu_0 I_w}{2\pi D} \;\;\Rightarrow\;\; I_w = \frac{\pi D}{R}\,I_L \qquad (4)

where: $I_w$ — the straight-wire current that silences the loop's centre. The $\pi$ survives because the loop formula carries no $\pi$ downstairs while the wire formula does — a factor lost so often that (4) is worth engraving. With $D = 2R$ the wire needs $2\pi \approx 6.3$ times the loop's current: circles concentrate field; straight wires waste it on the space around them.

configuration	net field / zero point	meaning
like currents, between	zero at $x = \dfrac{dI_1}{I_1+I_2}$ from wire 1	gap split in current ratio; nearer the weaker wire
like currents, midpoint	fields oppose: $B \propto \lvert I_2 - I_1\rvert$	SUBTRACT; zero only if currents equal
unlike currents, between	fields add: $B \propto I_1 + I_2$	loud zone — no zero between, ever
unlike currents, outside	zero at $y = \dfrac{dI_2}{I_1 - I_2}$ beyond weaker wire	nearly equal currents exile the zero to infinity
wire in loop's plane	centre silent when $I_w = \pi D I_L/R$	the stray $\pi$ — loop formula has none, wire formula does
any zero point	zero force on charges and third wires there	cancellation ⟹ equilibrium disguises

Limiting cases

case	result	meaning
$I_1 = I_2$ , like currents	zero exactly midway	symmetry decides before algebra does
$I_1 = I_2$ , unlike currents	no finite zero point	equal-opposite pair: fields never cancel, midfield maximally loud
$I_2 \to 0$	$x \to d$ : zero point rides onto wire 2	switch a wire off and "cancellation" collapses to the lone-wire picture
$d \to 0$ , unlike currents	fields cancel far away, add nearby	the cable limit: a cord's go-and-return currents make distant fields nearly vanish
$P$ very far from all wires	$B \to \mu_0 I_{\text{net}}/2\pi r$	from far away, a wire bundle looks like one wire carrying the algebraic sum

That last row is quietly deep: from across a room, a lamp cord carrying $5$ A out and $5$ A back is magnetically almost invisible — which is why mains wiring doesn't swing every compass in the house. The circulation-based version of "net enclosed current decides" belongs to Ampère Setups — Cables, Sheets & Cavities, coming later in E4.

Common trap“Parallel currents attract, so their fields must add up between the wires”

Force language and field language run in opposite directions here, and swapping them is this archetype's signature error. Like currents attract each other — yet their fields between the wires point in opposite senses and partially cancel: that is exactly why the zero point lives there. Unlike currents repel — yet between them their fields reinforce. Attraction/repulsion is about the force one wire's field exerts on the other wire; the between-the-wires map is about what both fields do at an empty point. If you catch yourself writing "fields add because the wires attract", stop and redo the right-hand census — the two stories connect only through $\vec F = I\vec L \times \vec B$ , never through wishful symmetry.

Worked examples

Board level 1 — midpoint field and the zero point (CBSE pattern, 3 marks)

Two long parallel wires A and B, $10$ cm apart, carry $3.0$ A and $6.0$ A in the same direction. Find (a) the field at the midpoint of the line joining them, and (b) the point on that line where the net field is zero.

Step 1. End-on sketch: both ⊙, midpoint $5.0$ cm from each wire. Census: between like currents the fields oppose.

Step 2. Magnitudes at the midpoint, with $\mu_0/2\pi = 2\times10^{-7}$ :

B_A = \frac{2\times10^{-7}\times 3.0}{0.050} = 1.2\times10^{-5}\ \mathrm{T}, \qquad B_B = \frac{2\times10^{-7}\times 6.0}{0.050} = 2.4\times10^{-5}\ \mathrm{T}

Step 3. Opposing directions ⟹ subtract: $B = 2.4\times10^{-5} - 1.2\times10^{-5} = \boxed{1.2\times10^{-5}\ \mathrm{T}}$ , directed the way the stronger wire (B) dictates.

Step 4. Zero point — between, nearer the weaker current: $x = \dfrac{d\,I_A}{I_A + I_B} = \dfrac{10 \times 3}{9} = \boxed{3.3\ \mathrm{cm\ from\ A}}$ . Sanity: fields match at $1.8\times10^{-5}$ T each ✓.

Board level 2 — antiparallel pair: loud middle, quiet outside (ISC pattern)

Two long parallel wires $15$ cm apart carry $8.0$ A and $2.0$ A in opposite directions. Find (a) the net field midway between them, and (b) the zero-field point.

Step 1. Sketch: $8$ A ⊙, $2$ A ⊗. Census: between unlike currents both fields point the same way — they add.

Step 2. Midpoint magnitudes ( $7.5$ cm each):

B_1 = \frac{2\times10^{-7}\times 8.0}{0.075} = 2.13\times10^{-5}\ \mathrm{T}, \qquad B_2 = \frac{2\times10^{-7}\times 2.0}{0.075} = 0.53\times10^{-5}\ \mathrm{T}

Step 3. Same direction ⟹ add: $B = \boxed{2.7\times10^{-5}\ \mathrm{T}}$ .

Step 4. Zero point sits outside, beyond the weaker wire: $y = \dfrac{d\,I_2}{I_1 - I_2} = \dfrac{15 \times 2}{6} = \boxed{5.0\ \mathrm{cm\ beyond\ the\ 2\,A\ wire}}$ . Check: $2\times10^{-7}(8)/0.20 = 2\times10^{-7}(2)/0.050 = 8.0\times10^{-6}$ T — matched, opposed ✓.

Competitive level 1 — silencing a loop's centre (JEE-Main pattern)

A circular loop of radius $5.0$ cm carries $2.0$ A. A long straight wire in the plane of the loop, $10$ cm from its centre, is to cancel the field at the centre exactly. Find the required current and comment on its direction.

Step 1. Loop's centre field (formula from Fields of Straight Wires & Circular Loops):

B_L = \frac{\mu_0 I_L}{2R} = \frac{(4\pi\times10^{-7})(2.0)}{2(0.050)} = 2.51\times10^{-5}\ \mathrm{T}

along the loop's axis (right-hand curl).

Step 2. The wire lies in the loop's plane, so its field at the centre is perpendicular to that plane — antiparallel to $B_L$ for the right current direction. Cancellation:

\frac{\mu_0 I_w}{2\pi D} = \frac{\mu_0 I_L}{2R} \;\;\Rightarrow\;\; I_w = \frac{\pi D}{R} I_L = \frac{\pi (0.10)}{0.050}\,(2.0) = 4\pi \approx \boxed{12.6\ \mathrm{A}}

Step 3. Direction: the wire's grip-rule circles at the centre must oppose the loop's axial field — state it concretely from your figure in an exam.

Step 4. Sanity: $I_w$ is $6.3$ times $I_L$ — the wire splays its field around a full circumference and needs a heavy current to match what the loop focuses into its centre. Timing note: the only hazard is the stray $\pi$ . Write both formulas symbolically, cancel $\mu_0$ , and the answer is one line: $I_w = \pi D I_L / R$ . Ninety seconds.

Competitive level 2 — three wires on a square's corners (JEE-Advanced flavour)

Three long wires pass perpendicular to the page through corners A $(0,0)$ , B $(a,0)$ and C $(a,a)$ of a square of side $a = 20$ cm, each carrying $5.0$ A into the page. Find the net field at the fourth corner D $(0,a)$ .

Step 1. Distances to D: wires A and C at $a$ (directly below and directly right), wire B across the diagonal at $a\sqrt 2$ . Near-wire magnitude:

B_0 = \frac{2\times10^{-7} \times 5.0}{0.20} = 5.0\times10^{-6}\ \mathrm{T}, \qquad B_{\text{diag}} = \frac{B_0}{\sqrt 2} = 3.5\times10^{-6}\ \mathrm{T}

Step 2. Directions (into-page currents circulate clockwise: rotate each wire-to-D line $90°$ clockwise): wire A (below D) → $+x$ ; wire C (right of D) → $+y$ ; wire B (diagonally down-right) → along the $+x{+}y$ diagonal, components $\left(\tfrac{B_0}{\sqrt2}\right)\tfrac{1}{\sqrt 2} = \tfrac{B_0}{2}$ each.

Step 3. Component ledger:

B_x = B_0 + \tfrac{B_0}{2} = 7.5\times10^{-6}\ \mathrm{T}, \qquad B_y = B_0 + \tfrac{B_0}{2} = 7.5\times10^{-6}\ \mathrm{T}

Step 4. Magnitude and direction:

B = \sqrt{B_x^2 + B_y^2} = 7.5\sqrt2 \times10^{-6} \approx \boxed{1.1\times10^{-5}\ \mathrm{T}}

at $45^\circ$ , along the diagonal from A to C. The closed form $B = \tfrac{3}{2}\sqrt{2}\,B_0$ is worth noticing: symmetric setups usually hide one. Timing note: ~4 minutes, almost all on Step 2. Draw D, draw the three wire-to-D lines, rotate each 90° clockwise — arrows first, arithmetic second. Reaching for cosines before fixing directions is how students drown here; the component table is three rows once the arrows exist.

Exam traps

The add/subtract flip. Like currents ⟹ fields oppose between (zero point between); unlike ⟹ fields add between (zero point outside, beyond the weaker). Half the lost marks in this family are these rules interchanged — usually via the force-language shortcut dismantled above. The right-hand census costs ten seconds; do it every time.

Zero point placed near the stronger current. The silence always shelters beside the weaker wire — it must, to let the weak field match the strong one at shorter range. If your $x$ lands nearer the big current, you inverted the ratio.

Using $\mu_0 I/2\pi d$ for a loop's centre. The loop formula is $\mu_0I/2R$ — no $\pi$ downstairs, radius not distance. Mixed setups (Competitive 1) punish the confusion with a clean factor of $\pi$ .

Perpendicular distances. The $d$ is the perpendicular distance from the wire (a line!) to the point — on corner setups sometimes a diagonal, never "the nearest labelled length". Write each distance on the sketch first.

Forgetting the field is a vector at the last step. Three correct magnitudes added as plain numbers is the classic three-wire fumble. Non-collinear contributions demand components — Pythagoras is not optional.

Constants and craft. All boards use $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$ ; the working shortcut is $\mu_0/2\pi = 2\times10^{-7}$ exactly. Convert cm → m on the sketch line; typical answers run $10^{-6}$ – $10^{-4}$ T — millitesla from a few amperes at centimetres means recheck.

Self-force fantasies. A wire's own field wraps around it and exerts no net force on it; only the other wires' fields at its location count.

Origins · 1 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · Royal Institution, London

Faraday could not follow the mathematics of Ampère's treatise — and out of that limitation invented something better: a picture. Scattering iron filings around magnets and wires, he watched them chain into curves and took the curves seriously as physics, christening them lines of force; his 1851–52 papers argued the heresy that the lines were not bookkeeping but the actual seat of magnetic action. Every diagram in this article — nested circles around each wire, loud zones where lines crowd, neutral points where opposing patterns thin into silence — is drawn in Faraday's language. Maxwell said Faraday had been a mathematician of the highest order without knowing it. (His grandest experiment, the 1831 induction ring, opens a later chapter of this course.)

History

1820

— Oersted's compass twitches beside a wire; within months Ampère maps the forces between conductors.

1820

— Biot and Savart measure the

1/d

fall-off of a wire's magnetic action — the law this archetype deploys.

1851

— Faraday's On the Physical Character of Lines of Magnetic Force elevates field-line maps — crowded, sparse, neutral — into physics.

1873

— Maxwell's Treatise turns Faraday's superposed line-pictures into vector mathematics: fields add component by component.

1900

s — neutral-point surveying becomes laboratory craft: finding the silent spot where an apparatus field cancels Earth's.

Going further

Texts. NCERT Physics Class 12, Ch. 4 (field due to straight conductors) · HC Verma, Concepts of Physics Vol. 2, Ch. 35 (wire-pair exercises) · SL Arora XII (neutral-point problems).

Provenance. Standard exam lore; a verifier should recompute: Board 1's $1.2\times10^{-5}$ T midpoint and $3.3$ cm zero (fields match at $1.8\times10^{-5}$ T); Board 2's $2.7\times10^{-5}$ T midpoint and $5.0$ cm outside zero (match at $8.0\times10^{-6}$ T); Competitive 1's $I_w = 4\pi \approx 12.6$ A; Competitive 2's $B_x = B_y = 7.5\times10^{-6}$ T, net $1.06\times10^{-5}$ T at $45°$ (cross-checked by full vector summation).

Prerequisites. Fields of Straight Wires & Circular Loops · Parallel Currents & the Definition of the Ampere. Leads to. Floating Wires & Magnetic Balances (coming up in E4) · Ampère Setups — Cables, Sheets & Cavities (coming up in E4) · Magnetic Field of Steady Currents (UG deep dive).

Why does the zero point sit closer to the smaller current and not the bigger one?

Because each wire's field weakens as $I/d$ , and the point must be where the two ratios match: a small numerator needs a small denominator — a short distance — to keep up with the strong wire's field arriving from farther away. Same logic as the electrostatic null point between unequal charges (Null Points & the Third-Charge Equilibrium), with $I/d$ replacing $q/r^2$ as the currency.

Is the zero-field location a point or a line?

A line. Long straight wires make the picture identical in every perpendicular plane, so the "zero point" on your end-on sketch is the footprint of a zero-field line parallel to the wires. Exam answers in centimetres-from-wire-A are fine — but say "line" if the question asks for the locus.

Between like currents the fields cancel somewhere — but the wires attract. Doesn't zero field mean zero force?

Zero field at the empty point between the wires, not at the wires themselves. Each wire sits in the other wire's decidedly nonzero field and feels $F = BIL$ from it; the cancellation point is just a spot where a third object would feel nothing. Keeping "field map at a point" and "force on a wire" as separate ledgers is the discipline this archetype trains.

Why doesn't a wire feel a force from its own magnetic field?

Its own field circles around it with perfect symmetry about its axis — every filament pushed equally in opposing senses, summing to nothing. More fundamentally, a body cannot accelerate itself with its own internal interactions. Only external fields — the other wires' — produce net force, which is why every formula in Parallel Currents & the Definition of the Ampere pairs wire 1's field with wire 2's current, never a wire with itself.

Do I need to worry about the Earth's magnetic field in these problems?

In the idealized exam setup, no — "find the field due to the wires" silently excludes Earth's $\sim 4\times10^{-5}$ T. But notice: that is the same order as our answers at centimetre distances. A real neutral-point measurement absolutely includes Earth's field — locating where a coil's field cancels it is a classic practical. If a question gives a horizontal external field, add it as one more vector in the ledger; the recipe doesn't change.

How do I keep the right-hand rule fast under exam pressure?

Work end-on and memorize two words: ⊙ anticlockwise, ⊗ clockwise. At any target point the field is tangent to the circle centred on the wire — draw the wire-to-point line and rotate it $90°$ in the memorized sense. Never re-grip an imaginary wire in 3D mid-exam; the flat tangent-rotation is the same rule with the thinking pre-done. Ten seconds per wire, and Competitive 2's difficulty evaporates.