One wire in the field of another
A long straight wire carrying a current \(I\) sets up a magnetic field around it. The field lines are circles round the wire. At a distance \(d\) the field has the size \(B = \mu_0 I/2\pi d\), where \(\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A}\) is the permeability of free space. We quote this result here; the chapter Sources of the Magnetic Field derives it.
The direction of the circles follows a right-hand grip rule. Grip the wire with the right hand, thumb along the current. The fingers curl the way the field lines run. A current out of the page has field lines running anticlockwise round it.
Now lay a second long wire parallel to the first, a distance \(d\) away. Wire 2 sits in the field of wire 1, which is perpendicular to it. So every metre of wire 2 feels a force \(B_1 I_2 \times 1 = \mu_0 I_1 I_2/2\pi d\). Wire 1 sits in the field of wire 2 and feels the same force per metre.
Two long parallel wires carry currents in the same direction. What do the wires do to each other?
The figure above shows why. Wire 1 carries its current out of the page, so its field at wire 2, on its right, points up. Wire 2 also carries its current out of the page, and \(\hat k \times \hat j = -\hat i\). So wire 2 is pushed to the left, towards wire 1. The same argument pushes wire 1 to the right: like currents attract. Reverse one current and both forces reverse: unlike currents repel.
The force on wire 1 and the force on wire 2 have the same formula, \(\mu_0 I_1 I_2/2\pi d\). It does not matter that one current is larger. The larger current makes a stronger field, but it acts on the smaller current. The two effects balance exactly, as Newton's third law requires.
(a) Two long parallel wires \(0.1\,\text{m}\) apart carry \(10\,\text{A}\) and \(20\,\text{A}\) in the same direction. Find the force on each metre of each wire. (b) Two wires \(0.1\,\text{m}\) apart carry \(100\,\text{A}\) each in opposite directions. Find the force per metre and say whether they attract or repel.
(a) \(f = 2 \times 10^{-7} \times 10 \times 20/0.1 = 4 \times 10^{-4}\,\text{N/m}\) on each wire. The currents are alike, so the wires attract.
(b) \(f = 2 \times 10^{-7} \times 100 \times 100/0.1 = 0.02\,\text{N/m}\). The currents are opposite, so the wires repel.
Values used: \(\mu_0/2\pi = 2 \times 10^{-7}\,\text{T m/A}\).
Two long parallel wires carry currents in opposite directions. What happens?
A wire carrying \(10\,\text{A}\) runs parallel to a wire carrying \(20\,\text{A}\). How does the force on the \(10\,\text{A}\) wire compare with the force on the \(20\,\text{A}\) wire?
Two long parallel wires \(4\,\text{cm}\) apart carry \(5\,\text{A}\) and \(8\,\text{A}\). Find the force on each metre of either wire, in units of \(10^{-4}\,\text{N/m}\).
Both currents in a pair of long parallel wires are doubled, and the distance between the wires is also doubled. The force per metre becomes
The flex of a room heater has two wires about \(3\,\text{mm}\) apart, carrying \(5\,\text{A}\) in opposite directions. Estimate the force per metre that pushes the two wires apart.
A force that defines a unit
Put \(I_1 = I_2 = 1\,\text{A}\) and \(d = 1\,\text{m}\) into the formula. The force is \(2 \times 10^{-7} \times 1 \times 1/1 = 2 \times 10^{-7}\,\text{N}\) on each metre. For many years this was the definition of the ampere. A force and a length can be measured with great care, so a current could be set by measuring them.
Let the same steady current flow in two infinitely long, thin, straight parallel wires placed \(1\,\text{m}\) apart in vacuum. The current is one ampere when the force between them is \(2 \times 10^{-7}\,\text{N}\) on each metre of length.
In 2019 the SI changed its footing. It now fixes the charge of the electron, \(e = 1.602176634 \times 10^{-19}\,\text{C}\), exactly. One ampere is then a flow of a fixed number of electron charges each second. The force between parallel wires still comes out as \(2 \times 10^{-7}\,\text{N/m}\), to about one part in a billion. So the old definition remains a true statement about nature.
Two long parallel wires \(1\,\text{m}\) apart each carry \(1\,\text{A}\). What is the force on a \(3\,\text{m}\) length of one wire, in units of \(10^{-7}\,\text{N}\)?
In September 1820 news reached Paris that Hans Christian Ørsted had deflected a compass with a current. André-Marie Ampère heard it at the Academy of Sciences on 11 September. Within about two weeks he showed the Academy that two wires carrying currents push and pull each other, with no magnet anywhere. Like currents attracted and unlike currents repelled. The unit of current carries his name.
A loop beside a long wire
Place a rectangular loop beside a long straight wire, with two of its sides parallel to the wire. The near side carries current in the same direction as the wire, so it is attracted. The far side carries current the other way, so it is repelled. But the far side is farther away, in a weaker field. The attraction wins.
The two short sides run straight away from the wire. They carry opposite currents through the same pattern of field, so their forces are equal and opposite. They cancel. In Fig 3.1 the near side feels \(2 \times 10^{-7} \times 20 \times 10 \times 0.2/0.05\) \(= 1.6 \times 10^{-4}\,\text{N}\). The far side feels \(8 \times 10^{-5}\,\text{N}\). The net force is \(8 \times 10^{-5}\,\text{N}\) towards the wire.
Three long parallel wires lie in one plane, \(0.1\,\text{m}\) apart, each carrying \(10\,\text{A}\) in the same direction. Find the force per metre on the middle wire and on an outer wire.
Every pair of like currents attracts. The middle wire has one neighbour on each side at \(0.1\,\text{m}\). An outer wire has neighbours at \(0.1\,\text{m}\) and \(0.2\,\text{m}\), both on the same side.
Each pull is \(f = 2 \times 10^{-7} \times I^2/d\).
Middle wire: \(2 \times 10^{-4}\,\text{N/m}\) to each side, so the net is zero. Outer wire: \(2 \times 10^{-7} \times 100/0.1 = 2 \times 10^{-4}\) plus \(2 \times 10^{-7} \times 100/0.2 = 10^{-4}\).
The middle wire feels nothing; each outer wire is pulled inward with \(3 \times 10^{-4}\,\text{N/m}\).
A long wire carries \(10\,\text{A}\). A rectangular loop carrying \(5\,\text{A}\) has its \(0.1\,\text{m}\) sides parallel to the wire, at \(0.02\,\text{m}\) and \(0.04\,\text{m}\) from it. The near side carries current the same way as the wire. Find the net force on the loop, in units of \(10^{-5}\,\text{N}\).
A substation carries power on thick copper or aluminium bars, called busbars, mounted side by side. If a fault shorts the line, the current can jump to tens of thousands of amperes in a moment. Take \(20\,\text{kA}\) in two bars \(0.2\,\text{m}\) apart. Each metre is then pushed with \(2 \times 10^{-7} \times (2 \times 10^4)^2/0.2 = 400\,\text{N}\), the weight of a \(40\,\text{kg}\) load. Engineers brace the bars with strong insulators so that a fault cannot bend them.
- Taking like currents to repel, as like charges do.Like currents attract and unlike currents repel; the force is \(I\vec L \times \vec B\), not Coulomb's law.
- Giving the larger current the larger force.Both wires feel \(\mu_0 I_1 I_2/2\pi d\); the forces form a third-law pair.
- Using \(1/d^2\) for parallel wires.The field of a long wire falls as \(1/d\), so the force per metre does too.
- Forgetting the far side of a loop beside a wire.Add the pull on the near side and the push on the far side; only the short sides cancel.