9478 / 2027
Electromagnetic Forces overview

Lesson 6 of 9

Forces between parallel currents

  • I can explain the forces between current-carrying conductors and predict their directions.Syllabus

    Syllabus 9478, 17(i). Explain the forces between current-carrying conductors and predict their directions.

Make a guess. You are not marked.

Two long parallel wires carry currents in the same direction. What happens to them?

  1. They repel, because like currents push apart just as like charges do.
  2. They attract, because each sits in the other's field.
  3. Nothing, because the wires are not made of a magnetic material.
Show the answer

They attract, because each sits in the other's field.

Each current makes a magnetic field at the other wire. Fleming's left-hand rule gives a force on each wire towards the other, so currents in the same direction attract.

Each current-carrying wire produces a magnetic field at the other wire. Use that other-wire field to find each force; the source field and the responding conductor must be named separately.

Place two long parallel wires vertically in the page, wire 1 on the left and wire 2 on the right. Let both conventional currents initially point up. At wire 2, wire 1 produces a field into the page. An upward current in that field experiences a force left, towards wire 1.

At wire 1, wire 2's field is out of the page, so its upward current experiences a force right. Both forces point inward: parallel currents attract. These are paired forces on different wires, not two forces to add on one selected wire.

Parallel currents: the wires attract

Parallel currents: the wires attractTwo portions of long vertical wires have current one of four amperes upward at the left and current two of three amperes upward at the right. At wire two, the field from wire one remains into the page and has magnitude forty microtesla. At wire one, the field from wire two is out of the page and has magnitude thirty microtesla. The left wire force is right and the right wire force left, so they attract. The equal-length purple arrows show equal force magnitudes, sixty micro-newtons, on the two considered half-metre portions. Each force uses the other wire's field.Wire 1: 4.00 AWire 2: 3.00 AFrom wire 2From wire 130 µT out40 µT ind = 0.0200 m

Teal dot/cross symbols show the field made by the other wire at that location. Brown arrows show current. The two forces are equal and opposite despite the unequal currents and local fields; the considered length is 0.500 m.

Reverse one current: the wires repel

Reverse one current: the wires repelTwo portions of long vertical wires have current one of four amperes upward at the left and current two of three amperes downward at the right. At wire two, the field from wire one remains into the page and has magnitude forty microtesla. At wire one, the field from wire two is into the page and has magnitude thirty microtesla. The left wire force is left and the right wire force right, so they repel. The equal-length purple arrows show equal force magnitudes, sixty micro-newtons, on the two considered half-metre portions. Each force uses the other wire's field.Wire 1: 4.00 AWire 2: 3.00 AFrom wire 2From wire 130 µT in40 µT ind = 0.0200 m

Teal dot/cross symbols show the field made by the other wire at that location. Brown arrows show current. The two forces are equal and opposite despite the unequal currents and local fields; the considered length is 0.500 m.

Each local field label names the other wire as its source. Reversing only the right-hand current reverses that wire's response and the field it produces at the left-hand wire; both forces become outward.

After reversing wire 2's current, wire 1's field at wire 2 remains into the page, but the reversed current now feels force right. Wire 2's field at wire 1 changes to into the page, giving the unchanged upward current a force left. Thus antiparallel currents repel. This follows from the field and force rules, not from treating currents as north or south poles.

Different currents, equal force magnitudes

Use I1 = 4.00 A, I2 = 3.00 A, separation d = 0.0200 m and a considered common length l = 0.500 m, away from the wire ends. With μ0 = 4π × 10-7 H/m:

B1 at 2 = μ0I1/(2πd)
= 4.00 × 10-5 T
Fon 2 = B1 at 2I2l
= (4.00 × 10-5)(3.00)(0.500)
= 6.00 × 10-5 N

For parallel upward currents this force is left. Calculate the partner using the other source:

B2 at 1 = μ0I2/(2πd)
= 3.00 × 10-5 T
Fon 1 = B2 at 1I1l
= (3.00 × 10-5)(4.00)(0.500)
= 6.00 × 10-5 N

The local source fields are unequal, yet the forces are equal and opposite. Combining the two relationships gives a useful result for these long parallel wires:

F/l = μ0I1I2/(2πd)

The symmetric product I1I2 explains the equal magnitudes. Reversing one current changes attraction to repulsion while leaving the magnitudes unchanged if both current magnitudes and the geometry stay fixed.

Check your understanding

Two long parallel wires carry unequal upward currents and attract. Only the right-hand current is reversed, with both current magnitudes unchanged. What happens to the interaction forces?

  1. They repel, but the larger-current wire feels the larger force.
  2. They repel with equal and opposite forces of the same magnitudes as before.
  3. They still attract, because the size of each current is unchanged by reversing one.
Show the answer

They repel with equal and opposite forces of the same magnitudes as before.

The other wire supplies each local field. Reversing one current makes both forces outward; mu_0 I_1 I_2 l/(2 pi d) retains the same magnitude when the current magnitudes and geometry stay fixed.