Textbooks show three wires in a row, where the forces cancel. Bend them into an L and the answer changes completely.
Part 1 — The concept, explained simply
Three wires in a straight line is the case every textbook shows, and the forces cancel. Bend the arrangement into an L and they no longer cancel — they combine at right angles.
1. The familiar case — and why it is misleading
Three parallel wires in a row, equally spaced, equal currents in the same direction. The middle wire is pulled left by one neighbour and right by the other. Equal and opposite ⇒ net force zero.
That answer is correct, but it teaches a bad habit: students start assuming the forces always cancel. Move the wires out of a straight line and they do not.
2. The L-shaped arrangement
Put wire B at the corner, wire A directly below it and wire C directly to its right, each a distance d away. All three carry the same current in the same direction.
A pulls B downward (same-direction currents attract).
C pulls B rightward.
These two pulls are at 90° to each other. They cannot cancel — they combine.
Same three wires, same currents, same spacing. Only the geometry changed — and the answer went from zero to √2 times a single force.
3. Combining the two forces
Each neighbour pulls with the standard force per unit length:
f1 = f2 = μ0i² / 2πd
They are perpendicular and equal, so the resultant is √2 times either one:
f = √2 × μ0i²/2πd = μ0i² / (√2 πd)
and it points along the diagonal, at 45° between the two pulls.
Two equal forces at angle θ combine to2f cos(θ/2).
At 0°: 2f (same direction, add). At 90°: √2 f. At 120°: f. At 180°: 0 (cancel).
Only the last case is the one textbooks usually show — hence the bad habit.
4. Three wires at the corners of an equilateral triangle
A common variant. Each wire feels two pulls, each of size μ0i²/2πd, separated by 60°. Using the formula above with θ = 60°:
f = 2f1 cos30° = √3 × μ0i²/2πd
directed towards the centre of the triangle, so the three wires try to collapse inwards.
The step everyone skips. Before combining, draw the two force arrows and check the angle between them. It is 180° only when the three wires are in a straight line. In any other arrangement it is something else, and the answer is not zero.
Part 2 — Formula sheet
Situation
Result
Note
Force per unit length, one pair
f = μ0i1i2/2πd
The building block for every case.
Two equal forces at angle θ
fnet = 2f cos(θ/2)
The master combining formula.
Three wires in a line, equal currents
0
θ = 180°, so they cancel.
Three wires in an L (right angle)
√2 μ0i²/2πd = μ0i²/√2πd
θ = 90°. Resultant at 45°.
Three wires, equilateral triangle
√3 μ0i²/2πd
θ = 60°. Points to the centre.
Perpendicular components generally
f = √(fx² + fy²)
When the two forces are unequal.
Direction of the resultant
tanα = fy/fx
Measured from the fx direction.
Force on a length L
F = f × L
Convert cm to m first.
Attraction or repulsion
parallel → attract; antiparallel → repel
Decides which way each arrow points.
The four-step method
Compute each pairwise force with μ0i1i2/2πd.
Draw each arrow — towards the other wire if the currents are parallel, away if antiparallel.
Measure the angle between the arrows from the geometry.
Combine with 2f cos(θ/2), or by components if the two are unequal.
Part 3 — 20 questions with step-by-step solutions
Attempt each on paper first. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1L arrangement
Three parallel wires perpendicular to the page carry equal currents in the same direction. B is at the corner of a right angle, with A and C each a distance d away along the two arms. The force per unit length on B is:
(a) μ0i²/2πd
(b) 2μ0i²/πd
(c) √2μ0i²/πd
(d) μ0i²/√2πd
Show step-by-step solution
GivenThree wires, equal currents i, same direction; B at a right-angle corner, A and C at distance d
AskedForce per unit length on B
ConceptTwo equal pulls at 90°. Perpendicular forces combine by Pythagoras, not by cancellation.
Formulaf = μ0i²/2πd each; resultant √2 f
SolutionEach neighbour pulls B with f = μ0i²/2πd. The two pulls are along perpendicular directions. Resultant = √2 × μ0i²/2πd = μ0i²/(√2 πd).
Answer: μ0i²/√2πd
Q2In a line
Three parallel wires lie in a straight line, equally spaced by d, carrying equal currents in the same direction. The net force per unit length on the middle wire is:
(a) zero
(b) μ0i²/2πd
(c) 2μ0i²/πd
(d) √2μ0i²/2πd
Show step-by-step solution
GivenThree collinear wires, equal spacing d, equal currents, same direction
AskedNet force on the middle wire
ConceptThe two pulls are equal and exactly opposite — the one case where they do cancel.
Formulaf = 2f1cos(θ/2), θ = 180°
SolutionEach outer wire attracts the middle one with μ0i²/2πd. They pull in exactly opposite directions (θ = 180°). cos90° = 0, so the resultant is zero.
Answer: zero
Q3Combining
Two equal forces f act at 60° to each other. Their resultant is:
(a) f
(b) √2 f
(c) √3 f
(d) 2f
Show step-by-step solution
GivenTwo equal forces f at 60°
AskedResultant
ConceptUse the master formula for two equal forces.
Three long parallel wires at the corners of an equilateral triangle of side d carry equal currents i in the same direction. The force per unit length on any one wire is:
(a) μ0i²/2πd
(b) √2μ0i²/2πd
(c) √3μ0i²/2πd
(d) 2μ0i²/2πd
Show step-by-step solution
GivenThree wires at the corners of an equilateral triangle of side d, equal currents
AskedForce per unit length on one wire
ConceptTwo equal pulls at 60°, both directed towards the other two wires.
Formulafnet = 2f cos30°
SolutionEach pull: f = μ0i²/2πd. The angle between them is 60° (the triangle's interior angle). fnet = 2f cos30° = √3 μ0i²/2πd, directed towards the centre.
Answer: √3μ0i²/2πd
Q5Numerical
Two forces of 3 N/m and 4 N/m act at right angles on a wire. The resultant force per unit length is:
(a) 7 N/m
(b) 5 N/m
(c) 1 N/m
(d) 3.5 N/m
Show step-by-step solution
Givenf1 = 3 N/m, f2 = 4 N/m, perpendicular
AskedResultant
ConceptUnequal perpendicular forces: use Pythagoras.
Formulaf = √(f1²+f2²)
Solutionf = √(9 + 16) = √25 = 5 N/m.
Answer: 5 N/m
Q6Direction
In the L-shaped arrangement of three wires with equal currents in the same direction, the resultant force on the corner wire is directed:
(a) along one arm
(b) at 45° between the two arms, into the corner region
(c) away from both wires
(d) perpendicular to the page
Show step-by-step solution
GivenL arrangement, equal same-direction currents
AskedDirection of the resultant
ConceptTwo equal perpendicular pulls give a resultant along the diagonal between them.
Formulatanα = fy/fx = 1
SolutionThe pulls are equal in size and at 90°. tanα = 1, so α = 45°. The resultant bisects the angle, pointing into the region between the two wires.
Answer: at 45° between the two arms, into the corner region
Q7Mixed directions
In an L arrangement, wire A carries current in the same direction as B but wire C carries it in the opposite direction. The two forces on B are:
(a) both attractive
(b) both repulsive
(c) one attractive and one repulsive, still at 90°
(d) zero
Show step-by-step solution
GivenA parallel to B, C antiparallel to B
AskedNature of the two forces
ConceptEach pair is judged separately: parallel attracts, antiparallel repels. The geometry (90°) is unchanged.
Formulaparallel attract; antiparallel repel
SolutionA and B parallel ⇒ B is attracted towards A. C and B antiparallel ⇒ B is repelled from C. The two arrows still lie along perpendicular directions, so the resultant is still √2 f.
Answer: one attractive and one repulsive, still at 90°
Q8Numerical
Three wires in an L arrangement each carry 10 A, with d = 10 cm. The force per unit length on the corner wire is:
(a) 2 × 10−4 N/m
(b) 2.83 × 10−4 N/m
(c) 4 × 10−4 N/m
(d) 1.41 × 10−4 N/m
Show step-by-step solution
Giveni = 10 A each, d = 0.1 m, right-angle arrangement
AskedForce per unit length
ConceptCompute one pairwise force, then multiply by √2.
Formulaf = √2 × (2×10−7)i²/d
SolutionOne pair: f = 2×10−7 × 100/0.1 = 2 × 10−4 N/m. Two such forces at 90°. Resultant = √2 × 2×10−4 = 2.83 × 10−4 N/m.
Answer: 2.83 × 10−4 N/m
Q9Unequal
Wire B is pulled by A with 6 × 10−5 N/m and by C with 8 × 10−5 N/m, the two pulls being perpendicular. The resultant is:
(a) 14 × 10−5 N/m
(b) 10 × 10−5 N/m
(c) 2 × 10−5 N/m
(d) 7 × 10−5 N/m
Show step-by-step solution
GivenPerpendicular pulls of 6 and 8 (×10−5) N/m
AskedResultant
ConceptUnequal and perpendicular, so use components rather than the equal-force shortcut.
Formulaf = √(f1²+f2²)
Solutionf = √(36 + 64) × 10−5 = 10 × 10−5 N/m.
Answer: 10 × 10−5 N/m
Q10Three wires
Three long parallel wires D, C and G carry 30 A, 10 A and 20 A in the same direction, spaced 3 cm and 2 cm apart in a line (C in the middle). The force on a 25 cm length of C is:
(a) 10−3 N
(b) 2.5 × 10−3 N
(c) zero
(d) 1.5 × 10−3 N
Show step-by-step solution
GivenD 30 A at 3 cm, G 20 A at 2 cm, C 10 A in between; all same direction; L = 0.25 m
AskedForce on C
ConceptCollinear, so the two pulls are opposite — subtract, then multiply by the length.
Formulaf = (2×10−7)i1i2/d
SolutionFrom D: f = 2×10−7 × 30 × 10/0.03 = 2 × 10−3 N/m (towards D). From G: f = 2×10−7 × 20 × 10/0.02 = 2 × 10−3 N/m (towards G). Opposite directions and equal ⇒ they cancel. Net force on any length = zero.
Answer: zero
Q11Concept
Two equal forces cancel completely only when the angle between them is:
(a) 0°
(b) 90°
(c) 120°
(d) 180°
Show step-by-step solution
GivenTwo equal forces at angle θ
AskedAngle for complete cancellation
ConceptUse the combining formula and set the result to zero.
Four long parallel wires at the corners of a square of side a carry equal currents in the same direction. The net force on any one wire is directed:
(a) towards the centre of the square
(b) away from the centre
(c) along one side
(d) zero
Show step-by-step solution
GivenFour wires at the corners of a square, equal same-direction currents
AskedDirection of the net force
ConceptEach wire is attracted by all three others; by symmetry the resultant points inward along the diagonal.
Formulaparallel currents attract
SolutionSame-direction currents attract, so every pull is towards another wire. The two side neighbours pull along the two sides; the diagonal neighbour pulls along the diagonal. By symmetry all three combine to point towards the centre.
Answer: towards the centre of the square
Q13Angle
Two equal forces f give a resultant of magnitude f. The angle between them is:
(a) 60°
(b) 90°
(c) 120°
(d) 150°
Show step-by-step solution
GivenTwo equal forces f with resultant f
AskedThe angle between them
ConceptSet the combining formula equal to f and solve.
Before combining two forces on a wire, the essential step is to:
(a) convert to newtons
(b) determine the angle between them from the geometry
(c) find the magnetic moment
(d) assume they cancel
Show step-by-step solution
GivenMulti-wire force problems
AskedThe essential step
ConceptThe angle decides everything, and it is 180° only for collinear arrangements.
Formulafnet = 2f cos(θ/2)
SolutionCompute each pairwise force first. Then draw the arrows and read the angle from the figure. Only then combine. Assuming cancellation is the commonest error.
Answer: determine the angle between them from the geometry
Q15Numerical
Two wires exert perpendicular forces of equal magnitude 5 × 10−5 N/m on a third wire. The resultant is:
(a) 5 × 10−5 N/m
(b) 7.07 × 10−5 N/m
(c) 10 × 10−5 N/m
(d) zero
Show step-by-step solution
GivenTwo equal perpendicular forces of 5×10−5 N/m
AskedResultant
ConceptEqual and at 90°, so multiply by √2.
Formulafnet = √2 f
Solutionfnet = √2 × 5 × 10−5 = 7.07 × 10−5 N/m.
Answer: 7.07 × 10−5 N/m
Q16Trap
A student sees three parallel wires with equal currents and immediately answers 'zero net force on the middle one'. This is:
(a) always correct
(b) correct only if the three wires are collinear and equally spaced
(c) never correct
(d) correct only for antiparallel currents
Show step-by-step solution
GivenThree wires, equal currents
AskedWhen the zero answer holds
ConceptCancellation needs the two pulls to be exactly opposite and exactly equal.
Formulafnet = 2f cos(θ/2)
SolutionEqual spacing makes the two forces equal in size. Collinearity makes them exactly opposite (θ = 180°). Both conditions are needed. In an L or a triangle the answer is not zero.
Answer: correct only if the three wires are collinear and equally spaced
Q17Components
Forces of 4 N/m along +x and 3 N/m along +y act on a wire. The angle of the resultant with the x-axis is:
(a) 30°
(b) 37°
(c) 45°
(d) 53°
Show step-by-step solution
Givenfx = 4 N/m, fy = 3 N/m
AskedAngle of the resultant
ConceptUse the tangent of the component ratio.
Formulatanα = fy/fx
Solutiontanα = 3/4 = 0.75. α = tan−1(0.75) ≈ 37°. (The 3–4–5 triangle gives the standard 37°/53° pair.)
Answer: 37°
Q18Concept
In the L arrangement, if the middle wire's current is doubled while the others stay the same, the resultant force on it:
(a) stays the same
(b) doubles
(c) becomes four times
(d) halves
Show step-by-step solution
GivenCorner wire's current doubled
AskedEffect on the force on it
ConceptEach pairwise force contains the product of the two currents, so both scale together.
Formulaf ∝ i1i2
SolutionEach pairwise force is proportional to the corner wire's own current. Doubling it doubles both forces. The geometry is unchanged, so the resultant also doubles.
Answer: doubles
Q19Square numeric
Four wires at the corners of a square of side 0.1 m each carry 5 A in the same direction. The force per unit length between two ADJACENT wires is:
(a) 5 × 10−5 N/m
(b) 2.5 × 10−5 N/m
(c) 1 × 10−4 N/m
(d) 2 × 10−4 N/m
Show step-by-step solution
GivenAdjacent wires, i = 5 A each, separation 0.1 m
AskedPairwise force per unit length
ConceptJust the basic two-wire formula for one adjacent pair.
Which statement about forces on a wire from two neighbours is correct?
(a) they always cancel
(b) they always add
(c) they combine as vectors, and cancel only if exactly opposite
(d) they are always perpendicular
Show step-by-step solution
GivenGeneral multi-wire arrangement
AskedThe correct statement
ConceptForce is a vector; cancellation is a special case, not the rule.
Formulafnet = 2f cos(θ/2)
SolutionForces combine as vectors, so the angle matters. Complete cancellation needs equal magnitudes and θ = 180°. Any other geometry gives a non-zero resultant.
Answer: they combine as vectors, and cancel only if exactly opposite