IB Physics HLCharges Moving in FieldsPaper 1 & 2F/L = μ₀I₁I₂/2πr~15 min read
Force Between Parallel Wires
Take two wires. Each one makes a magnetic field. Each one therefore sits in the other’s field — and a current in a magnetic field gets pushed. So the wires push on each other, with no contact and nothing between them but empty space. Run the currents the same way and they pull together. Run them opposite and they fly apart. The effect is so clean and so reproducible that for over a century the ampere itself was defined by it.
📘 What you need to know
Every current-carrying wire produces a magnetic field, whose direction comes from the right-hand grip rule
Currents in the same direction → the field lines between the wires cancel → the wires attract
Currents in opposite directions → the field lines between them pile up → the wires repel
F/L = μ0I1I2 / 2πr — the force per unit length
μ0 = 4π × 10−7 N A−2, the permeability of free space. So μ0/2π = 2 × 10−7 exactly
It follows from F = BIL combined with B = μ0I / 2πr for the field round a straight wire
The force goes as 1/r — not an inverse square law
The forces on the two wires are equal and opposite, by Newton’s third law, even if the currents differ
Why they push and pull
Look at what happens in the gap between them. Each wire’s field curls round it, and in the space in between the two fields either oppose each other or reinforce each other.
Both sets of field lines were computed, not sketched. In the left panel the field at the exact midpoint really is zero — the two contributions cancel to the last decimal place.
Here is a memory hook that never fails: think of the field lines as elastic bands under tension, which also repel each other sideways. Where the lines cancel between the wires, there is nothing pushing them apart, and the tension in the surrounding lines drags them together. Where the lines pile up between the wires, they shove the wires apart. Same picture, both cases.
The equation, and where it comes from
You are expected to be able to build this one. It takes two ingredients you already have.
Step 1 — wire 2 sits in wire 1’s fieldF = B1I2L sin θ and here θ = 90°, so F = B1I2Lthe field of a straight wire is a circle, so it is always perpendicular to the other wire
Step 2 — the field of a long straight wireB1 = μ0I1 / 2πr
Step 3 — substitute, and divide by LF = ( μ0I1 / 2πr ) I2LF / L = μ0I1I2 / 2πrμ0 = 4π × 10−7 N A−2 • r = separation of the wires (m)
Current I1
makes a field
B1 = μ0I1/2πr
which pushes on I2
F/L = μ0I1I2/2πr
Two things fall out of that formula for free. First, μ0/2π = 4π × 10−7 / 2π = 2 × 10−7 exactly — so in practice you can write F/L = 2 × 10−7I1I2/r and skip a step. Second, the formula is symmetric in I1 and I2. Swap the wires and you get the same number. That is Newton’s third law appearing out of the algebra, unasked.
It is not an inverse square law
Careful here. Coulomb’s law has an r2. Newton’s law of gravitation has an r2. This one does not. Double the separation and the force per unit length halves.
Two infinitely long wires, one metre apart, each carrying one amp, feel 2 × 10−7 N on every metre of their length. That was the definition of the ampere until 2019, when the SI switched to defining it from the elementary charge instead.
Getting the directions right
Two steps, in this order, every time. Find the field at one wire caused by the other. Then use Fleming’s left-hand rule on that wire.
At X, wire Y’s field curls anticlockwise and points down. Fleming’s left hand (field down, current out of the page) then gives a force to the right. Repeat at Y and everything reverses. Newton’s third law, drawn.
Currents
Field between the wires
Result
Same direction
The two fields cancel. There is a neutral point
Forces towards each other — they attract
Opposite directions
The two fields add. Lines crowd into the gap
Forces away from each other — they repel
🔗 Working a parallel-wires question
Same or opposite? Same → attract. Opposite → repel. Decide before calculating.
Convert.r in metres. mm and cm are where the marks go.
Use the shortcut.μ0/2π = 2 × 10−7, so F/L = 2 × 10−7I1I2 / r.
Force, or force per metre? The formula gives F/L. Multiply by L for the actual force.
Ratios?F/L ∝ I1I2/r. Note the singler.
Directions? Field at one wire from the other (grip rule), then Fleming’s left hand.
WE 1
Two long parallel wires 4.0 cm apart carry currents of 3.0 A and 5.0 A in the same direction. (a) Calculate the force per unit length between them. (b) Calculate the force acting on a 25 cm length of one wire. (c) State whether the wires attract or repel. (μ0 = 4π × 10−7 N A−2)
(a) Step 1 — convert, then substituter = 4.0 cm = 0.040 mF/L = μ₀I₁I₂ / 2πr = (4π × 10⁻⁷)(3.0)(5.0) / (2π × 0.040)Step 2 — the 2π cancels most of the 4πF/L = (2 × 10⁻⁷)(15) / 0.040 = (3.0 × 10⁻⁶) / 0.040F/L = 7.5 × 10⁻⁵ N m⁻¹(b) Step 3 — multiply by the lengthF = (7.5 × 10⁻⁵) × 0.25F = 1.9 × 10⁻⁵ N(c) Step 4 — same directionthey attractNotice how tiny that force is — about the weight of a grain of pollen. Wires only fly apart dramatically when the currents are enormous, which is exactly what happens inside a short-circuited cable.
WE 2
Two parallel wires exert a force per unit length F/L on each other. Determine the new force per unit length, in terms of the original, if: (a) both currents are doubled; (b) the separation is doubled; (c) both currents are doubled and the separation is doubled.
Step 1 — write down what mattersF/L ∝ I₁I₂ / r
Note the single r on the bottom — not r².
(a) both currents ×2factor = 2 × 2 = 44 × the original(b) separation ×2factor = 1/2half the original(c) both togetherfactor = 4 / 22 × the originalPart (b) is the whole point of the question. If you halved it to a quarter, you were thinking of Coulomb’s law. Magnetic force between wires goes as 1/r, because the field of a straight wire itself goes as 1/r.
WE 3
(a) Derive the expression for the force per unit length between two long parallel wires. (b) Two vertical wires carry currents out of the page. Wire P is on the left and wire Q on the right. Determine the direction of the magnetic field at P due to Q, and hence the direction of the force on P. (c) Wire Q carries twice the current of P. Explain why the forces on the two wires are still equal in magnitude.
(a) the derivation
The field of wire 1 at the position of wire 2 is B₁ = μ₀I₁ / 2πr.
That field is perpendicular to wire 2, so sin θ = 1.
F = B₁I₂L = (μ₀I₁ / 2πr) I₂ L
Divide both sides by L:
F/L = μ₀I₁I₂ / 2πr(b) field first, then force
By the right-hand grip rule, Q’s field curls anticlockwise (seen from the front).
At P, which lies to the left of Q, that field points downwards.
Fleming’s left hand: field down, current out of the page.
the force on P is to the right, towards Q — they attract(c) why the forces matchF/L = μ₀I₁I₂ / 2πr is symmetric in I₁ and I₂.
Swapping the wires gives the same number.
equal and opposite, exactly as Newton’s third law demandsPart (c) trips people up because Q makes a bigger field. True — but P carries a smaller current to be pushed. The two effects cancel exactly, and the product I₁I₂ is the same either way round.
💡 Top tips
Same direction → attract. Opposite → repel. Learn it as a pair, and it is the opposite of what charges do.
Use μ0/2π = 2 × 10−7 and save yourself a line of algebra.
The equation gives F/L, a force per metre. Multiply by L if the question wants a force.
1/r, not 1/r2. Double the separation and the force halves.
To find a direction: grip rule for the field, then Fleming’s left hand for the force.
The forces are always equal and opposite, however different the currents.
Be able to derive it. Start from F = BIL and B = μ0I/2πr.
⚠ Common mistakes
Squaring the r. This is not an inverse square law
Getting attract and repel the wrong way round. Like currents attract — unlike charges, which repel
Quoting F/L as the force. It is the force per unit length
Leaving r in cm or mm
Thinking the wire with the bigger current feels the bigger force. Both feel the same
Jumping straight to Fleming’s rule without first finding the field at that wire
Forgetting that θ = 90° here, because a wire’s field lines are circles around it
Quick recap: Each wire sits in the other’s magnetic field, so each feels a force. Currents in the same direction make the field between them cancel, and the wires attract; opposite currents make it add, and they repel. Combining F = BIL with B = μ0I/2πr gives F/L = μ0I1I2/2πr, with μ0 = 4π × 10−7 N A−2. The force goes as 1/r, not 1/r2, and the two forces are always equal and opposite. One amp in each of two wires one metre apart gives 2 × 10−7 N per metre.
Step back for a moment. We started with a force on a whole wire, and then we explained the force between two wires. But a current is nothing more than charges in motion. So the real thing being pushed is not the wire at all — it is each individual electron drifting through it, and the wire only feels the force because the electrons are trapped inside. Take one charge out of the wire entirely, fire it through a field on its own, and it will still be pushed. Next page: Magnetic Force on a Charge.
Attract or repel? 1/r or 1/r2?
Book a free meeting and we’ll settle the direction rules, the derivation and the inverse-square trap for good.
