Ampere's circuital law

A loop of string round a wire

Hold a long straight wire upright. Tie a loop of string round it. Now imagine that you walk round the string, and at every step you write down the part of the magnetic field that points along your step. Then you add all the numbers.

The total does not depend on the shape of the string. It does not depend on where the wire is inside the loop. It depends only on the current that goes through the loop. This is Ampere's circuital law.

The law

Divide a closed path into short steps $d\vec{l}$. At each step, take the part of $\vec{B}$ along the step, which is $\vec{B}\cdot d\vec{l}$. Add them all round the path. The sum is written $\oint \vec{B}\cdot d\vec{l}$ and is called the circulation of $\vec{B}$. Ampere's law says

Here $I_{enc}$ is the net current that passes through the loop. Use the right-hand rule to find the sign. Curl your right fingers the way you walk round the path. Your thumb shows the direction of positive current.

Three points matter.

• Only the current through the loop counts. A wire outside the loop adds to $\vec{B}$ at every point of the path. But its contribution to the circulation is zero.
• Currents have signs. A current into the page cancels part of a current out of the page.
• The law is always true. It helps us find $B$ only when the symmetry makes the sum easy: $\vec{B}$ must have the same size at every step, and point along the step. Then the sum is just $B$ times the length of the path.

A long straight wire again

Take a circle of radius $r$ with the wire at its centre. By symmetry, $B$ has the same size all round the circle, and it points along the circle. So the circulation is $B \times 2\pi r$. The enclosed current is $I$. Then

This is the result that we found from the Biot-Savart law, with much less work.

A thick wire

A real wire has a radius $a$. Let a current $I$ be spread evenly over its section. Then the symmetry still holds, and we use circles about the axis.

Outside, at $r > a$. All of the current goes through the circle. So $B = \dfrac{\mu_0 I}{2\pi r}$, as if the whole current were along the axis.

Inside, at $r < a$. The circle meets only part of the section. The part of the current that goes through it is in proportion to the area, $I_{enc} = I\dfrac{\pi r^2}{\pi a^2} = I\dfrac{r^2}{a^2}$. So

The field is zero on the axis. It grows in a straight line up to the surface, where $B = \dfrac{\mu_0 I}{2\pi a}$, and then falls as $1/r$.

A worked example. A round wire of radius $3.0$ mm carries $45$ A, spread evenly. Find the field at $1.5$ mm from the axis, at the surface, and at $12$ mm from the axis.

1. At the surface. $B = \dfrac{\mu_0 I}{2\pi a} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 45\ \text{A}}{3.0 \times 10^{-3}\ \text{m}} = 3.0 \times 10^{-3}\ \text{T}$, that is $3.0$ mT.
2. Inside, at half the radius. The field grows in a straight line, so at $r = a/2$ it is half of the surface value, $1.5$ mT. Check: $I_{enc} = 45\ \text{A} \times \dfrac{1}{4} = 11.25$ A, and $B = \dfrac{2 \times 10^{-7} \times 11.25}{1.5 \times 10^{-3}} = 1.5$ mT.
3. Outside, at $12$ mm. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 45\ \text{A}}{12 \times 10^{-3}\ \text{m}} = 7.5 \times 10^{-4}\ \text{T}$, that is $0.75$ mT.

A hollow pipe. A long pipe carries a current along its wall. Take a circle inside the hollow part. No current goes through it, so $I_{enc} = 0$ and $B = 0$ there. Outside the pipe, the pipe acts like a thin wire on the axis.

Counting the current in a loop

The law is useful for several wires too. A loop surrounds some wires and leaves the others outside. Add the currents of those inside, with their signs.

A worked example. A loop surrounds two long wires. One carries $12$ A out of the page, and the other carries $5$ A into the page. A third wire carries $9$ A and is outside the loop. What is the circulation of $\vec{B}$ round the loop, taking out of the page as positive?

1. The net current. $I_{enc} = 12\ \text{A} - 5\ \text{A} = 7\ \text{A}$ out of the page. The third wire does not count.
2. The circulation. $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc} = 4\pi \times 10^{-7}\ \text{T m/A} \times 7\ \text{A} = 8.8 \times 10^{-6}\ \text{T m}$.

Set the three currents so that their sum is zero. What is the circulation round the largest loop? Does this mean that the field is zero everywhere on the loop?

Make the wire thinner. Is the field at the surface larger or smaller? Is the field far outside the wire changed?

Practise with new numbers

Try it in Python

The program adds up $\vec{B}\cdot d\vec{l}$ for $20000$ short steps round a circle. A long wire carries $8.0$ A. It is first at the centre of the circle, then off centre but still inside, and last outside.

Common mistakes

• Using Ampere's law with no symmetry. The law is always true, but the shortcut $B \times \text{length}$ needs a path on which $B$ is constant and along the path.
• Counting a wire outside the loop. It makes a field, but it adds nothing to the circulation.
• Forgetting the signs. Currents in opposite directions through the loop subtract.
• Using all of the current inside a thick wire. At $r < a$, only the part $I\dfrac{r^2}{a^2}$ goes through the circle.
• Saying that zero circulation means zero field. Zero circulation says only that the net current through the loop is zero.

Remember these

• $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$, with the sign from the right-hand rule.
• Only the current through the loop counts.
• Long wire, or a point outside a round wire: $B = \dfrac{\mu_0 I}{2\pi r}$.
• Inside a round wire with an even current: $B = \dfrac{\mu_0 I\,r}{2\pi a^2}$. It is zero on the axis and largest at the surface.
• Inside a hollow pipe, $B = 0$.

Check yourself

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

1. A wire of radius $a$ carries a steady current spread evenly over its section. How does the field at $r = a/2$ compare with the field at $r = 2a$?

2. A loop surrounds a wire with $7$ A out of the page. Another wire with $7$ A into the page also goes through the loop. What is the circulation of $\vec{B}$, and is the field zero on the loop?

3. Why can we not use $B \times 2\pi r = \mu_0 I$ to find the field at a point near the end of a short wire?

4. A long pipe carries $40$ A along its wall. What is the field inside the hollow part, and what is it at a distance of $10$ cm from the axis, outside the pipe?