Force Between Parallel Conductors
Until 20 May 2019, the SI ampere was defined mechanically: the constant current that, if maintained in two infinitely long parallel wires 1 m apart in vacuum, produces a force of exactly $2 \times 10^{-7}\ \text{N}$ per metre of length. $F/L = \mu_0 I_1 I_2 / (2\pi d)$. The 2019 redefinition replaced this with a fixed value of the elementary charge, but the underlying physics, and the formula, has not changed.
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Two long straight wires are held parallel, 5.0 cm apart, each carrying 3.0 A.
Predict 1: If the currents flow in the same direction, do the wires attract or repel?
Predict 2: If the separation is doubled to 10 cm (currents unchanged), does the force per unit length increase, decrease, or stay the same?
Two parallel wires carrying current in the same direction will:
Know
- Force per unit length between parallel wires: $F/L = \mu_0 I_1 I_2 / (2\pi d)$
- Same-direction currents attract; opposite-direction currents repel
- The ampere is defined using this mechanical force
Understand
- Why $F/L \propto 1/d$ (not $1/d^2$)
- Why this interaction gives a practical way to define and measure current
- Why short-circuit fault currents create dangerous mechanical forces on transmission lines
Can Do
- Calculate $F/L$ for two parallel current-carrying wires
- Determine whether two parallel wires attract or repel
- Predict how $F/L$ changes when current or separation is changed
Core Content
Run two wires side by side, both carrying current in the same direction, and something unexpected happens: they pull toward each other. Reverse the current in one wire and they push apart instead. This is the opposite of what happens with static charges, where like repels and unlike attracts. Here, each wire creates its own circular magnetic field (L13), and that field exerts a motor-effect force (L15) on the other wire's current. Because each wire's field wraps around it in a specific rotational sense, the force this produces on same-direction currents is always attractive, and on opposite-direction currents is always repulsive.
$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$
$F/L$: force per unit length (N m⁻¹); $I_1$, $I_2$: the two currents (A); $d$: separation between the wires (m); $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$
$F/L \propto I_1$ and $F/L \propto I_2$: doubling either current doubles the force per unit length.
$F/L \propto 1/d$, not $1/d^2$: doubling the separation halves the force per unit length.
Two wires each carry 2.0 A in opposite directions, separated by 10 cm. Calculate the force per unit length.
- $F/L = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{(4\pi \times 10^{-7})(2.0)(2.0)}{2\pi \times 0.10}$
- $= \dfrac{(2 \times 10^{-7})(4.0)}{0.10} = \dfrac{8 \times 10^{-7}}{0.10}$
- $= 8.0 \times 10^{-6}\ \text{N m}^{-1}$, repulsive (opposite directions)
Two wires each carry 5.0 A in the same direction, separated by 20 cm. Find the total force on a 3.0 m length of one wire.
- $F/L = \dfrac{(4\pi \times 10^{-7})(5.0)(5.0)}{2\pi \times 0.20} = 2.5 \times 10^{-5}\ \text{N m}^{-1}$
- $F = (F/L) \times L = (2.5 \times 10^{-5})(3.0) = 7.5 \times 10^{-5}\ \text{N}$, attractive (same direction)
Two parallel current-carrying wires exert a mutual force on each other: $F/L = \mu_0 I_1 I_2/(2\pi d)$ (N m⁻¹). Same-direction currents attract; opposite-direction currents repel. $F/L \propto I_1 I_2$ and $F/L \propto 1/d$ (inverse first power, not inverse square).
Pause, copy the highlighted formula and attraction/repulsion rule into your book before the check below.
Parallel wires carrying current in the same direction attract each other.
The force per unit length between two parallel wires is inversely proportional to the square of their separation.
How do you build a device that measures current directly, without first assuming you already know what current is? For most of the 20th century, the answer was to measure a force. If two long parallel wires, 1 m apart in vacuum, each carry exactly 1 A, they attract with a force of exactly $2 \times 10^{-7}\ \text{N}$ per metre of length. That mechanical force, measurable with a sensitive balance, was the ampere's legal definition until 20 May 2019, when the SI system was redefined around fixed values of fundamental constants, including the elementary charge $e$, instead of a physical apparatus.
1 A in each of two infinite parallel wires, 1 m apart in vacuum, produces $F/L = 2 \times 10^{-7}\ \text{N m}^{-1}$.
Doubling the separation $d$ (currents unchanged) halves $F/L$.
Doubling both currents (same $d$, so $I_1 = I_2 = I$) makes $F/L \propto I^2$, so the new force per unit length is four times the original.
Real transmission lines rarely fail this way in normal operation, but during a short circuit, fault currents can spike from tens of amps to thousands of amps almost instantly. Since $F/L \propto I_1 I_2$, this sudden surge produces enormous mechanical forces between conductors, forces large enough to snap insulators, bend support structures, or fling cables apart. This is one reason circuit breakers must trip within milliseconds of a fault.
The ampere was historically defined via the force between two current-carrying wires: 1 A in each of two wires, 1 m apart in vacuum, gives $F/L = 2 \times 10^{-7}\ \text{N m}^{-1}$. Since 2019, the ampere is instead defined using a fixed value of the elementary charge, but $F/L = \mu_0 I_1 I_2/(2\pi d)$ still describes the same physical force, including the dangerous forces produced by short-circuit fault currents on transmission lines.
Add the ampere's definition and the short-circuit hazard to your notes before the check below.
If the current in each of two parallel wires is doubled while their separation stays the same, the force per unit length becomes:
Two long parallel wires each carry 3.0 A in opposite directions, separated by 8.0 cm.
- Calculate the force per unit length between the wires.
- State whether the wires attract or repel, and explain why.
- If the separation were increased to 16 cm (currents unchanged), recalculate the force per unit length.
Two parallel wires carrying 4.0 A each in opposite directions, separated by 20 cm, have a force per unit length of approximately:
- Using the historical definition of the ampere ($2 \times 10^{-7}\ \text{N m}^{-1}$ at 1 A, 1 A, 1 m apart), calculate the force per unit length between two wires each carrying 50 A, separated by 1.0 m.
- Explain, using $F/L = \mu_0 I_1 I_2/(2\pi d)$, why a short-circuit fault current (which can reach thousands of amps) creates dangerous mechanical forces on transmission line conductors and their supports.
- Two transmission wires carry 800 A each in the same direction, 2.0 m apart. Calculate the force on a 100 m length of one wire, and state whether it is attractive or repulsive.
Three of these are correct statements about parallel current-carrying conductors. Pick the odd one out.
Two wires, 4.0 A each, in the same direction, 5.0 cm apart, experience a force per unit length of _____ × 10⁻⁵ N m⁻¹. (Enter the coefficient only, 2 sig figs.)
ApplyBand 3(3 marks) 3. Two long parallel wires, each carrying 4.0 A in opposite directions, are separated by 8.0 cm. Calculate (a) the force per unit length, (b) whether the wires attract or repel, and (c) the force per unit length if the separation is reduced to 4.0 cm.
AnalyseBand 4(3 marks) 4. Wire A carries 10 A and wire B carries 30 A, parallel to each other in the same direction, 15 cm apart. (a) Calculate the force per unit length between them. (b) Using Newton's third law, explain why the force wire A exerts on wire B has the same magnitude as the force wire B exerts on wire A, despite the different currents. (c) State how the current in wire B would need to change for the wires to repel instead.
EvaluateBand 6(4 marks) 5. Before 2019, the ampere was defined as the constant current that, if maintained in two straight parallel conductors of infinite length placed 1 m apart in vacuum, produces a force of exactly $2 \times 10^{-7}\ \text{N}$ per metre of length. Evaluate why this mechanical definition was replaced in the 2019 SI redefinition, and explain the advantage the new definition (based on a fixed value of the elementary charge) provides over the old one.
Show all answers
Activity 1, Model Answers
- $F/L = \mu_0 I_1 I_2/(2\pi d) = (4\pi \times 10^{-7})(3.0)(3.0)/(2\pi \times 0.08) = (2 \times 10^{-7})(9.0)/0.08 = 2.25 \times 10^{-5}\ \text{N m}^{-1}$
- The currents are in opposite directions, so the wires repel each other, opposite-direction currents always repel.
- Doubling $d$ halves $F/L$: new $F/L = 1.125 \times 10^{-5} \approx 1.1 \times 10^{-5}\ \text{N m}^{-1}$.
Activity 2, Model Answers
- At the reference separation of 1.0 m, $F/L \propto I_1 I_2$: scaling from $2 \times 10^{-7}\ \text{N m}^{-1}$ at 1 A each, 50 A each gives $F/L = 2 \times 10^{-7} \times (50 \times 50) = 5.0 \times 10^{-4}\ \text{N m}^{-1}$.
- Since $F/L \propto I_1 I_2$, a fault current that jumps from tens of amps to thousands of amps increases the force per unit length by a factor of thousands squared, producing forces large enough to damage insulators, bend supports, or separate conductors.
- $F/L = (4\pi \times 10^{-7})(800)(800)/(2\pi \times 2.0) = (2 \times 10^{-7})(640{,}000)/2.0 = 0.064\ \text{N m}^{-1}$. Over 100 m: $F = 0.064 \times 100 = 6.4\ \text{N}$, attractive (same direction).
Short Answer, Model Answers
Q3 (3 marks): (a) $F/L = (4\pi \times 10^{-7})(4.0)(4.0)/(2\pi \times 0.08) = (2 \times 10^{-7})(16)/0.08 = 4.0 \times 10^{-5}\ \text{N m}^{-1}$. (b) Opposite directions, so the wires repel. (c) Halving $d$ doubles $F/L$: new $F/L = 8.0 \times 10^{-5}\ \text{N m}^{-1}$.
Q4 (3 marks): (a) $F/L = (4\pi \times 10^{-7})(10)(30)/(2\pi \times 0.15) = (2 \times 10^{-7})(300)/0.15 = 4.0 \times 10^{-4}\ \text{N m}^{-1}$. (b) By Newton's third law, the force wire A exerts on wire B and the force wire B exerts on wire A form an action-reaction pair, equal in magnitude and opposite in direction, regardless of the size of each individual current, because each wire's field acts on the other's current symmetrically through $F/L = \mu_0 I_1 I_2/(2\pi d)$, which is symmetric in $I_1$ and $I_2$. (c) The current in wire B would need to reverse direction (flow opposite to wire A) for the wires to repel instead of attract.
Q5 (4 marks): The pre-2019 definition of the ampere relied on an idealised physical setup, two infinitely long, perfectly straight parallel wires in a vacuum, which cannot be built exactly in a real laboratory. In practice, national standards laboratories approximated this using current balances, but every real measurement carried some experimental uncertainty from finite wire length, imperfect vacuum, and mechanical imprecision in measuring such a small force. The 2019 SI redefinition instead fixed the numerical value of the elementary charge $e = 1.602176634 \times 10^{-19}\ \text{C}$ exactly, defining the ampere as a flow of a fixed number of elementary charges per second. This ties the ampere to an unchanging property of nature rather than to a specific experimental apparatus, giving a definition that is, in principle, exactly reproducible anywhere in the universe with sufficiently precise equipment, and one that will never need to be revised as measurement technology improves.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Until 20 May 2019, the ampere was defined by exactly this lesson's formula: two infinite parallel wires, 1 A each, 1 m apart in vacuum, attract with $F/L = 2 \times 10^{-7}\ \text{N m}^{-1}$. Every ammeter on Earth was, in principle, calibrated back to that mechanical force. The 2019 redefinition replaced the apparatus with a fixed value of the elementary charge, more precise, but the physics of $F/L = \mu_0 I_1 I_2/(2\pi d)$ that Ampère first quantified has not changed at all.
Now check your Think First answers: two wires 5.0 cm apart, 3.0 A each in the same direction, attract each other. Doubling the separation to 10 cm halves the force per unit length, it does not stay the same, because $F/L \propto 1/d$.