Force between two parallel currents

Run a modest 5 A through a wire in your ceiling and a modest 8 A through a wire twenty centimetres beneath it, and the two wires quietly pull toward each other. The force is tiny — tens of micronewtons, far less than the weight of a mosquito — yet this feeble tug is one of the most trusted numbers in all of physics. It is trusted because it needs nothing exotic to predict: only a field you already derived (a straight wire's $B=\mu_0I/2\pi r$ ) and a force law you already know ( $F=BIl$ ), chained together. The chain is short but the payoff is enormous. Two current-carrying wires, laid parallel, each sit inside the magnetic field the other one makes. Wire 1 makes a field at the location of wire 2; that field pushes on the current flowing in wire 2. By Newton's third law wire 2 pushes back on wire 1 exactly as hard. Whether that push is a pull-together or a push-apart depends on nothing more subtle than whether the two currents point the same way or opposite ways — a rule that, delightfully, runs backwards compared to the like-charges-repel rule you learned in electrostatics. This article rebuilds that chain from the two results you already have, keeps every worked number from the original derivation, and adds the layer of exam-craft that turns "I understand the idea" into "I can defend six marks under time pressure." Along the way you will meet the reason this particular force law is not just useful but foundational: it is the operational definition of the ampere itself, the base SI unit that every other electrical quantity in your syllabus is built on top of.

Key takeaways

Two long parallel wires each sit inside the other's magnetic field, so each experiences a force — this is Ampere's force law, and it needs no new physics beyond $B=\mu_0I/2\pi r$ and $F=BIl$ .
The force per unit length between them is $\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}$ , where $d$ is the wire separation.
Same-direction currents attract; opposite-direction currents repel — the reverse of the rule for electric charges.
By Newton's third law the force on wire 1 due to wire 2 equals the force on wire 2 due to wire 1 in magnitude, even when $I_1\neq I_2$ .
This law defines the ampere: 1 A is the current that, in two infinite parallel wires 1 m apart in vacuum, produces exactly $2\times10^{-7}$ N of force per metre of length.
Board answers must show the two-step field-then-force logic explicitly, not just quote the final formula; competitive problems often disguise the geometry (unequal currents, asked-for total force, three-wire systems).
Key idea: each parallel current-carrying wire sits in the magnetic field of the other, so combining $B=\mu_0I/2\pi d$ with $F=BIl$ gives a force per unit length $F/l=\mu_0I_1I_2/2\pi d$ that attracts for same-direction currents and repels for opposite-direction currents, and this exact law is what defines the ampere.

The core insight

Because each of two parallel currents sits inside the magnetic field generated by the other, $B_1=\mu_0I_1/2\pi d$ combined with $F=B_1I_2l$ gives a mutual force per unit length $F/l=\mu_0I_1I_2/2\pi d$ that is attractive for like-direction currents and repulsive for opposite-direction currents, and it is precisely this law that fixes the SI definition of the ampere.

Where it fits in the syllabus

In the CBSE Class 12 unit Moving Charges and Magnetism, force between two parallel current-carrying wires is one of the explicitly listed prescribed results, quoted together with "definition of the ampere" ( $F/L=\mu_0I_1I_2/2\pi d$ ). It is deliberately placed after both the straight-wire field result (from Ampere's circuital law) and the force-on-a-conductor result ( $\vec F=I\vec l\times\vec B$ ) have already been built, because it is not an independent law — it is those two results used one after the other. Nothing here goes beyond the syllabus: no toroid, no Hall effect, no cyclotron numerics are needed, and none appear below.

The physical picture

Picture two long, straight wires running parallel to each other, a fixed perpendicular distance $d$ apart, carrying steady currents $I_1$ and $I_2$ . Because each wire is a current, it produces its own circular magnetic field around itself, exactly as you already know for a single straight wire — and that field does not stop at the wire's own surface, it fills all the space around it, including the location of the neighbouring wire.

B_1=\dfrac{\mu_0I_1}{2\pi d}

where: $B_1$ is the magnetic field produced by wire 1, measured at the site of wire 2, a perpendicular distance $d$ away; $I_1$ is the current in wire 1; and $\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space. Physically, this field circles wire 1 like concentric rings, and at the straight line where wire 2 lies, the field is uniform in magnitude and points perpendicular to both wires — exactly the direction needed to exert a sideways force on wire 2's current. A length $l$ of wire 2, immersed in this field $B_1$ and carrying current $I_2$ perpendicular to $B_1$ , feels a magnetic force by the same rule you already used for any current-carrying conductor in a field:

F=B_1I_2l=\dfrac{\mu_0I_1I_2l}{2\pi d}

where: $F$ is the total magnetic force on the chosen length $l$ of wire 2. Dividing through by $l$ gives the more useful, length-independent quantity, the force per unit length, which both wires experience equally by Newton's third law:

\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}

where: $F/l$ has SI units of newtons per metre; $I_1,I_2$ are the two currents in amperes; $d$ is the separation in metres. Read this equation physically: the mutual force grows in direct proportion to each current and shrinks as the wires are moved apart, exactly the same $1/d$ falloff as the field of a single straight wire, because that field is the only distance-dependent ingredient in the whole chain.

Figure

Two vertical parallel wires separated by a gap d, wire one on the left carrying current I one upward and wire two on the right carrying current I two upward, with circular field lines from wire one shown reaching wire two and an arrow labelled force per length pulling the wires together since both currents point the same way

The rule for the direction, worth memorising as a single sentence: currents in the same direction attract, currents in opposite directions repel. This is the mirror image of electrostatics, where like charges repel — a contrast worth stating explicitly in an exam answer because examiners like to see the comparison drawn on purpose, not stumbled into.

Derivation AForce per unit length between two long parallel currents

The goal is to derive $F/l=\mu_0I_1I_2/2\pi d$ from first principles available at this stage of the chapter — nothing beyond the straight-wire field formula (itself a consequence of Ampere's circuital law) and the force-on-a-current-carrying-conductor formula (itself a consequence of $\vec F=q\vec v\times\vec B$ summed over the charge carriers in a wire). The derivation is really an exercise in applying two already-proved results in sequence, and the only new content is recognising which field acts on which current.

Step 1. Set up two long, straight, parallel wires a perpendicular distance $d$ apart, carrying steady currents $I_1$ and $I_2$ . Because both wires are long compared with any distance we care about near their midsection, each behaves, locally, exactly like the idealised infinite straight wire used to derive $B=\mu_0I/2\pi r$ from Ampere's circuital law.

Step 2. Write the magnetic field produced by wire 1 alone, evaluated at the position of wire 2, a distance $d$ away: $B_1=\mu_0I_1/(2\pi d)$ . This field is uniform in magnitude along the entire length of wire 2 (since $d$ does not change as you move along either wire), and its direction, by the right-hand rule, is perpendicular to the plane containing both wires.

Step 3. Because $B_1$ is perpendicular to wire 2's current direction, the force on a length $l$ of wire 2 is the full, un-reduced form of the conductor-force law, $F=B_1I_2l\sin\theta$ with $\theta=90°$ , so $\sin\theta=1$ and $F=B_1I_2l$ . Substituting the expression for $B_1$ from Step 2 gives $F=\dfrac{\mu_0I_1I_2l}{2\pi d}$ .

Step 4. Divide both sides by $l$ to obtain the force per unit length, which is the physically meaningful, length-independent quantity examiners ask for: $\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}$ .

Step 5. Invoke Newton's third law. The force wire 2 exerts back on wire 1, computed by exactly the same two-step logic with the roles of 1 and 2 swapped, gives the identical magnitude $\mu_0I_1I_2/2\pi d$ per unit length, directed oppositely. So even though $I_1$ and $I_2$ may differ (each wire produces a different field), the mutual force per unit length experienced by each wire is exactly the same number — this is why the single formula $F/l=\mu_0I_1I_2/2\pi d$ correctly describes the force on either wire, not just one of them.

Step 6. Read off the direction rule from the geometry: if the two currents point the same way, the two forces (found by applying $\vec F=I\vec l\times\vec B$ with the appropriate field directions) point toward each other — attraction. If the currents point opposite ways, the forces point apart — repulsion. This can be checked quickly with the right-hand rule on a diagram; it does not require a new formula, only care with directions.

Worked examples

Board level - force per unit length between two given currents

Two long parallel wires carry currents of $I_1=5\ \text{A}$ and $I_2=8\ \text{A}$ in the same direction, separated by a distance $d=0.2\ \text{m}$ . Find the force per unit length between them and state whether it is attractive or repulsive.

Step 1. Write the governing formula: $\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}$ , and note the standard shortcut value $\mu_0/2\pi=2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ , worth memorising since it recurs in every problem of this type.

Step 2. Substitute the given values: $\dfrac{F}{l}=2\times10^{-7}\times\dfrac{5\times8}{0.2}$ .

Step 3. Compute the numerator and divide: $\dfrac{F}{l}=2\times10^{-7}\times\dfrac{40}{0.2}=2\times10^{-7}\times200=4\times10^{-5}\ \text{N/m}$ .

Step 4. Fix the direction using the rule from the physical picture: the currents flow in the same direction, so the force is attractive — the wires are pulled toward each other. The force per unit length is $\boxed{4\times10^{-5}\ \text{N/m}}$ , attractive.

Competitive level - total force with unequal currents and a length constraint

Two straight parallel wires, each of length $2\ \text{m}$ , carry currents of $10\ \text{A}$ each but in opposite directions, separated by $d=0.05\ \text{m}$ . Find the total force between them, and then find the separation at which this same pair of currents would instead produce a force per unit length of exactly $2\times10^{-7}\ \text{N/m}$ (the ampere-defining value).

Step 1. Find the force per unit length at the given separation: $\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}=2\times10^{-7}\times\dfrac{10\times10}{0.05}=2\times10^{-7}\times2000=4\times10^{-4}\ \text{N/m}$ .

Step 2. Multiply by the actual wire length to get the total force, since the question explicitly asks for total force, not force per unit length: $F=\left(\dfrac{F}{l}\right)\times l=4\times10^{-4}\times2=8\times10^{-4}\ \text{N}$ .

Step 3. Fix the direction: the currents are in opposite directions, so the force is repulsive — the wires are pushed apart.

Step 4. Now solve the inverse question. We want the separation $d'$ at which $10\ \text{A}$ and $10\ \text{A}$ give $F/l=2\times10^{-7}\ \text{N/m}$ . Rearranging $F/l=\mu_0I_1I_2/2\pi d'$ for $d'$ : $d'=\dfrac{\mu_0I_1I_2}{2\pi(F/l)}=\dfrac{2\times10^{-7}\times(10\times10)}{2\times10^{-7}}=100\ \text{m}$ .

Step 5. Sanity-check by comparison with the ampere's defining case: two $1\ \text{A}$ currents $1\ \text{m}$ apart give exactly $2\times10^{-7}\ \text{N/m}$ . Here both currents are $10\times$ larger, so to keep the same force per unit length the separation must scale up by $10\times10=100$ , matching $d'=100\ \text{m}$ found directly — a useful proportional-reasoning check that avoids redoing the algebra from scratch. So the total force is $\boxed{8\times10^{-4}\ \text{N}}$ , repulsive, and the two $10\ \text{A}$ currents would need to be $\boxed{100\ \text{m}}$ apart to reproduce the ampere-defining force per unit length. Timing note: a two-part numerical like this should take about 3-4 minutes in a competitive setting — the first part is a direct plug-in (roughly 90 seconds), and the second part is faster if you spot the proportional-scaling shortcut in Step 5 instead of re-deriving $d'$ from scratch; always keep the shortcut as a check, not a replacement for showing the rearranged formula, since exams award marks for the algebraic step.

Common traps

Mixing up the attraction/repulsion rule — the correct statement is "same direction attracts, opposite direction repels," which is the reverse of the electrostatic like-charges-repel rule; students who transfer the electrostatics intuition directly get this backwards.
Forgetting to multiply force-per-unit-length by the actual wire length $l$ when a question asks for total force rather than force per metre — a very common one-mark loss in board exams.
Applying the formula $F/l=\mu_0I_1I_2/2\pi d$ to wires that are short, non-parallel, or bent — the derivation strictly assumes long, straight, parallel wires so that the field of one wire is essentially uniform in magnitude ( $B=\mu_0I/2\pi d$ ) over the entire length of the other; for short or angled segments this shortcut is not valid.
Assuming the force on wire 1 differs from the force on wire 2 whenever $I_1\neq I_2$ — by Newton's third law the mutual force is always equal and opposite for both wires, even though the individual fields each wire produces are different.
Confusing the direction convention with the right-hand rule for a single wire's field — students sometimes try to read off attraction/repulsion straight from $B=\mu_0I/2\pi d$ without actually working out the cross product $\vec F=I\vec l\times\vec B$ for each wire, which can give the wrong sign on a quickly-sketched diagram.

You should be able to…

Derive $F/l=\mu_0I_1I_2/2\pi d$ by chaining the straight-wire field formula with the force-on-a-conductor formula, stating each step explicitly.
State and correctly apply the direction rule: same-direction currents attract, opposite-direction currents repel.
Compute either the force per unit length or the total force between two parallel current-carrying wires, and convert cleanly between the two.
Explain why the mutual force is equal and opposite on both wires even when $I_1\neq I_2$ , citing Newton's third law.
State the SI definition of the ampere in terms of this force law and reproduce the numerical check ( $1\ \text{A}$ , $1\ \text{A}$ , $1\ \text{m}$ apart gives $2\times10^{-7}\ \text{N/m}$ ).
Recognise when the formula does not apply — short, finite, or non-parallel wires — and say why.

Ask the tutor: If you want to see exactly how the right-hand rule produces the attract/repel directions step by step, or want extra practice turning force-per-length problems into total-force and separation-solving variants, ask the AI tutor to walk through a few more worked cases.

Why does the same-direction rule for currents come out opposite to the rule for electric charges?

It is not a coincidence you need to memorise blindly, but it also is not something you are expected to re-derive from scratch every time — for CBSE purposes, treat it as a distinct fact about magnetism. The physical root is that magnetic force depends on the cross product of velocity and field, not on a simple radial attraction or repulsion like the electric force between charges, so the geometry works out differently. If you trace through the right-hand rule carefully for two same-direction currents, you will see the field of one wire curls around to push the other wire inward at the point where it sits — that is the honest way to convince yourself, rather than trying to force an analogy with charges.

Do I need to remember the direction of the force, or just the magnitude formula, for board exams?

Both. CBSE board papers regularly ask you to state whether the wires attract or repel given the current directions, sometimes as a full mark on its own, separate from the numerical force calculation. The safest habit is to always write one line stating the rule ("same direction attracts, opposite direction repels") right after computing the magnitude, even if the question does not explicitly ask for direction — it shows the examiner you understand the physics, not just the formula.

Is this the same idea as the force on a single current-carrying wire in an external magnetic field, or something new?

It is the very same idea, applied twice. There is no new force law here — $F/l=\mu_0I_1I_2/2\pi d$ is just $F=BIl$ with $B$ replaced by the specific field a straight wire produces, $B=\mu_0I/2\pi d$ . If you already trust both of those individually, you already know this result; the only genuinely new content is the observation that each wire's own current also acts as the source of a field for the other wire, so the two effects have to be chained together and Newton's third law applied to check consistency.

Why is this particular formula used to define the ampere instead of some other electromagnetic quantity?

Historically it was chosen because it can, in principle, be measured with a mechanical balance — you can literally weigh the force between two calibrated parallel wires and back out the current, without needing to count individual charges. That made it a practical, reproducible laboratory standard for well over a century (modern metrology since 2019 actually fixes the ampere by defining the elementary charge exactly, but the parallel-wire force law remains the conceptual definition taught at this level and gives numerically the same result).