The Biot-Savart law

A needle that turns

Put a small compass next to a straight wire. Switch on a large current, and the needle turns away from north. Switch it off, and the needle goes back.

Hans Christian Oersted saw this in 1820. A current makes a magnetic field. Now we ask two questions. How strong is the field? Which way does it point?

Add up the field of small pieces

Biot and Savart measured the field of currents. We write their result for a very short piece of wire. Call its length $dl$, and draw the vector $d\vec{l}$ along the current $I$. Take a point $P$, and let $\vec{r}$ run from the piece to $P$. The piece makes a field

The size of it is

where $\theta$ is the angle between $d\vec{l}$ and $\vec{r}$. The constant $\mu_0$ is the permeability of free space. Its value is $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$, so $\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m/A}$.

Four things to notice.

• The field falls as $1/r^2$, like the electric field of a point charge. But the source is a piece of current, $I\,dl$, and not a charge.
• The field is strong for a large current and for a long piece.
• On the line of the piece, $\theta = 0$ and $\sin\theta = 0$. A piece makes no field straight ahead of it or straight behind it.
• The field is at right angles to both $d\vec{l}$ and $\vec{r}$. So the field lines of a current go round the current in circles.

The field at $P$ from a whole wire is the sum of the fields of all its pieces. We add them as vectors.

Which way does it point?

Use the right-hand rule. Point the thumb of your right hand along the current. The fingers curl the way the field goes. Look at the picture on the right: the current comes out of the page, and the field runs anticlockwise.

For a loop, turn the rule round. Curl the fingers of your right hand the way the current goes round the loop. The thumb points the way the field goes through the loop.

A long straight wire

Take a straight wire and a point $P$ at a distance $d$ from it, measured at right angles. Every piece of the wire makes a field at $P$ in the same direction, along the circle through $P$. So we may add the sizes. The sum, over a wire of any length, is

Here $\varphi_1$ and $\varphi_2$ are the angles that the lines from $P$ to the two ends of the wire make with the perpendicular from $P$ to the wire.

For a very long wire, both angles go to $90^\circ$ and both sines go to $1$. Then

The field falls as $1/d$, not as $1/d^2$. A long wire is made of pieces at all distances, and the nearer pieces matter most.

Now take a wire that starts at the point nearest to $P$ and runs far away in one direction. One angle is $0^\circ$ and the other is $90^\circ$. So $B = \dfrac{\mu_0 I}{4\pi d}$, half of the field of a very long wire.

A worked example. A long straight cable carries $36$ A. What is the field at a distance of $12$ cm?

1. The formula. $B = \dfrac{\mu_0 I}{2\pi d}$.
2. The numbers. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 36\ \text{A}}{0.12\ \text{m}} = 6.0 \times 10^{-5}\ \text{T}$.
3. In words. That is $60\ \mu\text{T}$, about the size of the magnetic field of the Earth.

A short wire. A straight wire is $40$ cm long and carries $15$ A. What is the field at a point $15$ cm from its middle, on the line that cuts it at right angles?

1. The angles. Each end is $20$ cm from the middle, along the wire. The line from $P$ to an end is $\sqrt{15^2 + 20^2} = 25$ cm long. So $\sin\varphi = \dfrac{20}{25} = 0.80$ for each end.
2. The field. $B = \dfrac{10^{-7}\ \text{T m/A} \times 15\ \text{A}}{0.15\ \text{m}} \times (0.80 + 0.80) = 1.6 \times 10^{-5}\ \text{T}$, that is $16\ \mu\text{T}$.
3. Compare. A very long wire at the same place gives $20\ \mu\text{T}$. The short wire gives $0.80$ of that.

Make the wire very short, and then very long. When the wire is much longer than the distance, does the field stop changing?

An arc and a circular loop

Now bend the wire into a circle of radius $R$. Take the centre as the point $P$. Every piece is at the same distance $R$ from it, and each piece is at right angles to $\vec{r}$, so $\sin\theta = 1$. All the fields point the same way, along the axis. For an arc that turns through an angle $\theta$ (in radians), the length of the arc is $R\theta$. So

A full circle has $\theta = 2\pi$. At the centre of a loop,

A coil of $N$ turns pressed together gives $N$ times as much, $B = \dfrac{\mu_0 N I}{2R}$. A straight piece of wire that points at the centre makes no field there, because $\sin\theta = 0$.

A worked example. A flat coil has $25$ turns and a radius of $8.0$ cm. It carries $3.2$ A. What is the field at its centre?

1. The formula. $B = \dfrac{\mu_0 N I}{2R}$.
2. The numbers. $B = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 25 \times 3.2\ \text{A}}{2 \times 0.080\ \text{m}}$. This gives $B = 6.3 \times 10^{-4}\ \text{T}$.
3. In words. That is $0.63$ mT, more than ten times the field of the Earth.

On the axis of a loop

Now take a point on the axis at a distance $x$ from the centre. Every piece of the loop is at the same distance $\sqrt{R^2 + x^2}$ from the point. Each field $dB$ is at right angles to that line. It has a part along the axis and a sideways part. For every piece, the piece on the opposite side cancels the sideways part. Only the part along the axis stays, and it is $dB \times \dfrac{R}{\sqrt{R^2 + x^2}}$. Adding all the pieces, whose lengths sum to $2\pi R$, gives

At $x = 0$ this is $\dfrac{\mu_0 I}{2R}$, as before. Far from the loop, $x \gg R$, so $B \approx \dfrac{\mu_0 I R^2}{2x^3}$. The field falls as $1/x^3$. A loop far away looks like a tiny magnet, and you meet that idea again with the magnetic moment of a loop.

A worked example. A single loop of radius $6.0$ cm carries $5.0$ A. Find the field on its axis at $8.0$ cm from the centre, and compare it with the field at the centre.

1. The distance. $R^2 + x^2 = (0.060\ \text{m})^2 + (0.080\ \text{m})^2 = (0.10\ \text{m})^2$. So $(R^2 + x^2)^{3/2} = 1.0 \times 10^{-3}\ \text{m}^3$.
2. The field. The top of $B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ is $4\pi \times 10^{-7}\ \text{T m/A} \times 5.0\ \text{A} \times 0.0036\ \text{m}^2 = 2.3 \times 10^{-8}\ \text{T m}^3$. The bottom is $2 \times 1.0 \times 10^{-3}\ \text{m}^3$. So $B = 1.1 \times 10^{-5}\ \text{T}$, that is $11\ \mu\text{T}$.
3. At the centre. $B_0 = \dfrac{\mu_0 I}{2R} = 52\ \mu\text{T}$. The ratio is $\left(\dfrac{6}{10}\right)^3 = 0.216$.

Practise with new numbers

Try it in Python

The program builds a loop out of $720$ short pieces. It adds the Biot-Savart field of every piece, one by one, at five points on the axis. Then it compares the sum with the formula.

Common mistakes

• Drawing the field along the wire. The field goes round the wire, at right angles to it.
• Leaving out the 2. For a long wire, $B = \dfrac{\mu_0 I}{2\pi d}$, not $\dfrac{\mu_0 I}{\pi d}$ and not $\dfrac{\mu_0 I}{4\pi d}$.
• Mixing cm and m. Put every length in metres before you use $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$.
• Using the long-wire formula for a short wire. Near a short wire, use the two angles.
• Using $\dfrac{\mu_0 I}{2R}$ off the centre. That formula is for the centre only. On the axis, use the longer formula.
• Forgetting the turns. A coil of $N$ turns makes $N$ times the field of one turn.

Remember these

• $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$, with $\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m/A}$. The field is along $d\vec{l}\times\vec{r}$.
• Right-hand rule: thumb along the current, fingers along the field.
• Wire of any length: $B = \dfrac{\mu_0 I}{4\pi d}(\sin\varphi_1 + \sin\varphi_2)$. Very long wire: $B = \dfrac{\mu_0 I}{2\pi d}$.
• Arc at its centre: $B = \dfrac{\mu_0 I\theta}{4\pi R}$. Full loop: $B = \dfrac{\mu_0 N I}{2R}$.
• Axis of a loop: $B = \dfrac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$.

Check yourself

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

1. A very long straight wire carries $24$ A. What is the field at a distance of $6.0$ cm?

2. A short piece of wire carries a current to the east. What is the field that it makes at a point to the east of it, on the same line?

3. Two coils have the same radius and carry the same current. One has $1$ turn and the other has $2$ turns. Then the radius of the second is doubled. How do the fields at the centres compare?

4. A current goes up a vertical wire. A compass is placed to the east of the wire. Which way does the field of the wire point there? Ignore the field of the Earth.

5. On the axis of a loop, at a distance $x = R$ from the centre, what is the field compared with the field at the centre?