Solenoids, toroids and parallel currents

A coil that makes a steady field

Wind a long wire in a tight coil, like a spring. The coil is called a solenoid. When it carries a current, each turn makes a field, and the fields of all the turns add up inside. Doorbells, relays and the lifting magnets of a scrap yard all use a solenoid.

The field lines run along the inside of a solenoid, nearly parallel and evenly spaced. Outside, they spread out and the field is weak. So the field is almost uniform inside. This looks like the field of a bar magnet.

The long solenoid

Look at the picture. The solenoid has $n$ turns in each metre of its length. On the left, we draw a rectangle $abcd$ as an Amperian loop. Its side $ab$ is inside the solenoid and has a length $h$. The side $cd$ is outside.

Go round the rectangle.

• On $ab$, the field is along the side, so this side adds $Bh$.
• On $bc$ and $da$, the field is at right angles to the side, or zero. These sides add nothing.
• On $cd$, outside a long solenoid, the field is almost zero. So this side adds almost nothing.

The circulation is $Bh$. The loop meets $nh$ turns, each with a current $I$, so $I_{enc} = nhI$. Ampere's law gives $Bh = \mu_0 nhI$, and

The length $h$ cancelled, and so did the distance from the axis. The field is the same at every point inside the solenoid, and it does not depend on the radius. Here $n = N/L$, where $N$ is the total number of turns and $L$ is the length.

At an end of a long solenoid, the field is half as large, $\tfrac{1}{2}\mu_0 n I$. To see this, join two equal solenoids end to end. At the join the field is $\mu_0 n I$, and each half gives half of it.

A worked example. A tube $60$ cm long carries $900$ turns of wire. The current is $2.4$ A. What is the field inside, and at an end?

1. The turns in each metre. $n = \dfrac{900}{0.60\ \text{m}} = 1500\ \text{m}^{-1}$.
2. The field inside. $B = \mu_0 n I = 4\pi \times 10^{-7}\ \text{T m/A} \times 1500\ \text{m}^{-1} \times 2.4\ \text{A} = 4.5 \times 10^{-3}\ \text{T}$, that is $4.5$ mT.
3. At an end. About half, $2.3$ mT.

Make the solenoid short and wide. Does the field in the middle still equal $\mu_0 n I$? Then make it long and thin.

The toroid

A toroid is a solenoid bent into a ring. Look at the picture on the right. It has $N$ turns in all. Take a circle of radius $r$ inside the ring, running round the ring. By symmetry, the field has the same size all round the circle and points along it. The circulation is $B \times 2\pi r$. The circle goes through $N$ turns, so $I_{enc} = NI$. Then

Now take a circle in the hole, or outside the ring. In the hole, no current goes through the circle. Outside the ring, each turn goes through the circle twice, once in and once out, so the net current is zero. So $B = 0$ in the hole and outside the ring. The field is shut inside the ring.

The field is not quite uniform. It is a little larger near the inner edge than near the outer edge. For a thin ring, $2\pi r$ is the length of the ring and $N/(2\pi r) = n$. Then $B = \mu_0 n I$, the same as in a solenoid.

A worked example. A toroid has $1200$ turns and carries $0.80$ A. Its inner radius is $6.0$ cm and its outer radius is $10$ cm. Find the field at the inner edge, at the outer edge and at the mean radius.

1. The constant. $\dfrac{\mu_0 N I}{2\pi} = 2 \times 10^{-7}\ \text{T m/A} \times 1200 \times 0.80\ \text{A} = 1.92 \times 10^{-4}\ \text{T m}$.
2. The inner edge. $B = \dfrac{1.92 \times 10^{-4}\ \text{T m}}{0.060\ \text{m}} = 3.2 \times 10^{-3}\ \text{T}$, that is $3.2$ mT.
3. The outer edge. $B = \dfrac{1.92 \times 10^{-4}\ \text{T m}}{0.10\ \text{m}} = 1.9\ \text{mT}$.
4. The mean radius, $8.0$ cm. $B = \dfrac{1.92 \times 10^{-4}\ \text{T m}}{0.080\ \text{m}} = 2.4\ \text{mT}$.

Two long parallel wires

Wire 1 carries a current $I_1$. At a distance $d$ its field is $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$. Wire 2 is parallel to it and carries $I_2$. A length $L$ of wire 2 sits in this field, at right angles to it. The force on a current in a field is $F = BIL$. So

Here $F/L$ is the force on each metre. The same size of force acts on the other wire, in the opposite direction, as Newton's third law says.

Which way is the force? Use the right-hand rule for the field of wire 1 at wire 2, and then the rule for the force on a current in a field. The result is simple.

• Currents in the same direction: the wires attract.
• Currents in opposite directions: the wires repel.

Two parallel currents pull together, even though like electric charges push apart.

A worked example. Two long parallel wires are $7.0$ cm apart. They carry $14$ A and $35$ A in the same direction. What is the force on each metre?

1. The field of wire 1 at wire 2. $B_1 = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 14\ \text{A}}{0.070\ \text{m}} = 4.0 \times 10^{-5}\ \text{T}$.
2. The force on each metre. $\dfrac{F}{L} = I_2 B_1 = 35\ \text{A} \times 4.0 \times 10^{-5}\ \text{T} = 1.4 \times 10^{-3}\ \text{N/m}$.
3. In words. That is $1.4$ mN on each metre, and the wires attract.

What if the currents are opposite? The size of the force stays $1.4$ mN on each metre, but the wires repel. Between the wires, the two fields now point the same way, so they add.

Make one current negative. What happens to the sign of the force? Then double the distance. What happens to its size?

The old definition of the ampere

Put $I_1 = I_2 = 1$ A and $d = 1$ m in the formula. The force on each metre is $\dfrac{\mu_0}{2\pi} = 2 \times 10^{-7}\ \text{N/m}$. For many years this was the definition of the ampere. Since 2019, the ampere is fixed by the value of the charge of the electron instead. The force law still gives the same value to very good accuracy, and it still explains why two wires pull on each other.

Practise with new numbers

Try it in Python

The program finds the force on each metre between two long wires of $14$ A and $35$ A, for five spacings. Then it finds the force on each metre for a few equal currents at a spacing of $1$ m.

Common mistakes

• Using the whole length of a solenoid as $n$. The number $n$ is the turns in each metre, $N/L$.
• Putting a $2\pi r$ in the solenoid formula. Inside a long solenoid, $B = \mu_0 n I$ has no distance in it.
• Saying the field of a toroid is uniform. It goes as $1/r$ across the ring.
• Saying the field outside a toroid is $\mu_0 n I$. It is zero.
• Getting the force between wires backwards. Parallel currents attract, and opposite currents repel.
• Forgetting that the force acts on both wires. The two forces are equal and opposite.

Remember these

• Long solenoid: $B = \mu_0 n I$ inside, with $n = N/L$. At an end the field is half as much. Outside it is almost zero.
• Toroid: $B = \dfrac{\mu_0 N I}{2\pi r}$ inside the ring, and $B = 0$ in the hole and outside.
• Two long parallel wires: $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$.
• Same direction attracts, opposite direction repels.
• Equal currents of $1$ A at $1$ m give $2 \times 10^{-7}$ N on each metre.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In a long solenoid, the current is halved and the number of turns in each metre is doubled. What happens to the field inside?

2. A long solenoid has a field $B$ in its middle. A student says that the field at its end is $B/2$. Is this right?

3. Why is the field in the hole of a toroid zero?

4. Two long wires carry the same current in opposite directions. What is the direction of the force on each, and what happens to the size of the force if the distance is made three times larger?