Forces Between Parallel Conductors
In 1820, André-Marie Ampère at the École Polytechnique, Paris, demonstrated that two parallel wires carrying current in the same direction attract each other, and measured the force per unit length as F/L = μ₀I₁I₂/2πd. This result became so precise that it defined the SI ampere for nearly 200 years — 1 A was officially the current in two parallel wires 1 m apart producing 2 × 10⁻⁷ N/m of force — until the 2019 redefinition fixed the elementary charge e = 1.602 176 634 × 10⁻¹⁹ C exactly.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Two straight wires hang vertically side by side, a few centimetres apart. When current flows through both wires in the same direction, do you think they will attract, repel, or remain unaffected?
What do you think happens if the currents flow in opposite directions? Write your predictions with a brief reason before working through the lesson.
Warm-up — the force on a current-carrying wire in a magnetic field is given by F = BIl sin θ. The force is MAXIMUM when the angle between the current and the field is…
Know — Force Law
- The force per unit length between two parallel current-carrying conductors is $F/l = \mu_0 I_1 I_2 / (2\pi d)$
- Same-direction currents attract; opposite-direction currents repel
- The ampere was historically defined using this force relationship
Understand — Why Attraction and Repulsion?
- Each wire produces a magnetic field that acts on the other wire via the motor effect
- The direction of force is predicted using the right-hand grip rule and $F = BIl$
- This interaction underpins the SI definition of electric current
Can Do — Calculate and Predict
- Calculate the force per unit length given currents and separation
- Predict attraction or repulsion from current directions
- Explain how this principle is used to define the ampere
Core Content
Revisiting the right-hand grip rule before we add a second wire
Stretch two long wires parallel to each other on a bench, connect each to its own battery, and switch both on. Watch: the wires twitch toward each other if the currents flow the same way, or push apart if the currents oppose. This is Ampère's 1820 observation — two current-carrying wires exert a force on each other — and understanding why requires first knowing what magnetic field a single wire creates around itself. The right-hand grip rule gives the answer: wrap your right hand around the wire with your thumb pointing in the direction of conventional current, and your curled fingers show the field lines.
Figure 1 — Concentric magnetic field circles around a current-carrying wire (current out of page)
The magnetic field strength at a distance $r$ from a long straight wire carrying current $I$ is:
$B = \dfrac{\mu_0 I}{2\pi r}$
The field decreases with distance — strongest near the wire, falling off as $1/r$. This equation is the key step in deriving the force between two parallel wires.
Right-hand grip rule: thumb = current direction, fingers curl = $\vec{B}$ direction (concentric circles). $B = \mu_0 I / (2\pi r)$ (T) — strongest close to the wire, falls as $1/r$. $\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$.
Pause — copy the highlighted grip rule and field formula into your book before moving on.
A long straight wire carries current vertically upward. At a point horizontally to the east of the wire, the magnetic field direction is…
Each wire sits in the magnetic field produced by the other
We just saw that a current-carrying wire creates a circular magnetic field $B = \mu_0 I/2\pi r$ around itself. That raises a question: if a second current-carrying wire sits in that field, does it feel a force? This card answers it → yes, via the motor effect; same-direction currents attract, opposite-direction repel.
Consider two long parallel wires separated by distance $d$, carrying currents $I_1$ and $I_2$. Wire 1 produces a magnetic field at the location of Wire 2. Because Wire 2 carries current, it experiences a motor-effect force in that field. By Newton's third law, Wire 1 experiences an equal and opposite force.
$\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$
F/l — force per unit length (N/m) · μ₀ = 4π × 10⁻⁷ T m A⁻¹ · d — perpendicular separation (m)
Same-direction currents → wires attract each other
Opposite-direction currents → wires repel each other
Figure 2 — Two wires with currents in the same direction attract each other
Two long parallel wires are separated by 8.0 cm. Wire A carries 12 A upward and Wire B carries 8.0 A downward.
(a) Calculate the magnitude of the force per unit length. (b) Determine whether the force is attractive or repulsive. (c) If both currents are reversed, what changes?
- Part (a) — magnitude. Use $F/l = \mu_0 I_1 I_2 / (2\pi d)$ with $I_1 = 12$ A, $I_2 = 8.0$ A, $d = 0.080$ m.
$\dfrac{F}{l} = \dfrac{(4\pi \times 10^{-7})(12)(8.0)}{2\pi(0.080)} = 2.4 \times 10^{-4}$ N/m - Part (b) — direction. Currents are in opposite directions → force is repulsive.
- Part (c) — reversing both. If both currents are reversed, they are still in opposite directions relative to each other. The force remains repulsive with the same magnitude.
Force per unit length: $F/l = \mu_0 I_1 I_2 / (2\pi d)$ (N/m). Same-direction currents → attract; opposite-direction → repel. $F/l \propto I_1 I_2$ and $\propto 1/d$. Reversing both currents leaves magnitude and direction unchanged.
Add the highlighted parallel-wire force formula and direction rule to your notes before the check below.
Two parallel wires 20 cm apart each carry 10 A in the same direction. The force per unit length between them is…
The SI standard for electric current rests on this force
We just saw how to calculate the force per unit length between two parallel wires. That raises a question: if this force is so fundamental, could it be used to define the unit of current itself? This card answers it → yes, the historical SI definition of the ampere is exactly 1 A = the current producing $2 \times 10^{-7}$ N/m between two wires 1 m apart.
The ampere is one of the seven SI base units. Its historical definition (and the basis for μ₀) relies directly on the force between parallel conductors:
"One ampere is the constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to 2 × 10⁻⁷ newtons per metre of length."
This definition fixes the value of the permeability of free space:
$\mu_0 = 4\pi \times 10^{-7}$ N A⁻² (exactly)
Notice: set $I_1 = I_2 = 1$ A and $d = 1$ m in the force law, and you get exactly $F/l = 2 \times 10^{-7}$ N/m — confirming the definition.
In HSC problems you may be asked to "show that the ampere is consistent with μ₀ = 4π × 10⁻⁷." Substitute $I_1 = I_2 = 1$ A and $d = 1$ m into $F/l = \mu_0 I_1 I_2 / (2\pi d)$ and confirm you get $2 \times 10^{-7}$ N/m.
1 ampere ≡ the constant current producing $2 \times 10^{-7}$ N/m between two infinite parallel wires 1 m apart in vacuum. This fixes $\mu_0 = 4\pi \times 10^{-7}$ N A$^{-2}$ exactly. Verification: sub $I_1 = I_2 = 1$ A, $d = 1$ m → $F/l = 2 \times 10^{-7}$ N/m.
Pause — write the highlighted ampere definition and verification into your book before moving on.
Two parallel wires each carrying 1.0 A in the same direction, 1.0 m apart, exert an attractive force of 2 × 10⁻⁷ N/m on each other.
Doubling the separation between two parallel current-carrying wires doubles the force per unit length between them.
The force per unit length on wire 1 due to wire 2 equals the force per unit length on wire 2 due to wire 1 (Newton's third law).
Wire X and Wire Y are parallel. The force per unit length is halved while the currents remain unchanged. Which change is consistent with this?
Practise using the force per unit length formula
- Set I₁ = 5.0 A, I₂ = 5.0 A, d = 10.0 cm. Calculate F/l and state the direction (both currents same direction).
- Keep the same settings but switch to opposite directions. What happens to the magnitude? What changes?
- Return to same direction. Double both currents to 10.0 A (keep d = 10 cm). By what factor does the force change? Explain why.
- Return I₁ = I₂ = 5.0 A. Halve the separation to 5.0 cm. By what factor does the force change?
Show that the force law is consistent with the definition of the ampere
Two parallel wires 1.0 m apart in vacuum each carry 2.0 A in the same direction.
- Calculate the force per unit length between the wires.
- Explain why this is four times larger than the force in the ampere definition (which uses 1.0 A).
- A student claims: "Reversing both currents would change the force to repulsive." Is this correct? Justify your answer.
Three separate ideas lock together in this lesson:
- A current-carrying wire produces a magnetic field (right-hand grip rule).
- A second current-carrying wire in that field experiences a motor-effect force ($F = BIl$).
- Combining these gives the force law $F/l = \mu_0 I_1 I_2 / (2\pi d)$, which is so precise it defines the ampere.
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(3 marks) 1. Two long parallel wires are 5.0 cm apart. Wire 1 carries 15 A upward and Wire 2 carries 10 A upward. (a) Calculate the force per unit length on Wire 2 due to Wire 1. (b) Explain why the force on Wire 1 due to Wire 2 has the same magnitude. (c) State whether the force is attractive or repulsive.
1 mark: correct calculation with units · 1 mark: Newton's third law explanation · 1 mark: direction correct
AnalyseBand 5(4 marks) 2. A third wire carrying 5.0 A downward is placed midway between Wire 1 (15 A upward, left) and Wire 2 (10 A upward, right), so all three wires are 2.5 cm apart from their neighbours. (a) Determine the force per unit length on the third wire due to Wire 1. (b) Determine the force per unit length on the third wire due to Wire 2. (c) State the direction of each force. (d) Calculate the magnitude of the net force per unit length on the third wire and state its direction.
1 mark each: correct magnitudes for forces from Wire 1 and Wire 2 · 1 mark: correct directions · 1 mark: correct net force magnitude and direction
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Multiple Choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (3 marks): (a) $F/l = \mu_0 I_1 I_2 / (2\pi d) = (4\pi \times 10^{-7} \times 15 \times 10) / (2\pi \times 0.050) = 6.0 \times 10^{-4}$ N/m. (b) By Newton's third law, the force exerted by Wire 2 on Wire 1 is equal in magnitude and opposite in direction to the force exerted by Wire 1 on Wire 2. (c) Both currents are in the same direction (upward) → the force is attractive.
Q2 (4 marks): Wire 3 (downward) is 2.5 cm from Wire 1 (upward) and 2.5 cm from Wire 2 (upward). (a) Force from Wire 1: $F/l = (4\pi \times 10^{-7} \times 15 \times 5.0) / (2\pi \times 0.025) = 6.0 \times 10^{-4}$ N/m. (b) Force from Wire 2: $F/l = (4\pi \times 10^{-7} \times 10 \times 5.0) / (2\pi \times 0.025) = 4.0 \times 10^{-4}$ N/m. (c) Wire 3 is opposite to Wire 1 → repulsion (Wire 3 pushed right). Wire 3 is opposite to Wire 2 → repulsion (Wire 3 pushed left). (d) Net force = $6.0 \times 10^{-4} - 4.0 \times 10^{-4} = 2.0 \times 10^{-4}$ N/m directed away from Wire 1 (toward Wire 2 / to the right).
Five timed questions on forces between parallel conductors. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaAt the start you were asked about Ampère's 1820 experiment at the École Polytechnique: two 1 A wires 1 m apart — what force per metre, and what if the currents were reversed?
The answer: $F/l = \mu_0 I_1 I_2 / 2\pi d = (4\pi \times 10^{-7} \times 1 \times 1)/(2\pi \times 1) = 2 \times 10^{-7}$ N/m — attractive, since currents are in the same direction. Reversing one current reverses the force to repulsion. This exact value (2 × 10⁻⁷ N/m per ampere in each wire) defined the SI ampere for nearly 200 years until the 2019 redefinition.
Extend: Two parallel wires carry 5.0 A each in the same direction, 10 cm apart. A student claims the force would double if you moved one wire to 5.0 cm away without changing the currents. Is this correct? Calculate both forces to verify.