Biot-Savart Law

A field with no source to point at

In magnetic force you took the field as given. Someone handed you a $\vec{B}$, you crossed it with $q\vec{v}$ or with $I\vec{L}$, and out came a force. Now turn the question around. Where did that field come from?

Hold a compass next to a wire carrying a big current and the needle swings. Not towards the wire and not away from it: it lines up around it, tangent to a circle whose centre is the wire. That is already strange. An electric charge makes a field that points away from itself, so the field has an obvious source sitting at the middle of it. A current makes a field that closes on itself with nothing in the middle to point at. Whatever rule produces $\vec{B}$ from a current has to build that sideways-ness in from the start.

The rule is the Biot-Savart law, and it is the magnetic twin of Coulomb's law: chop the source into tiny pieces, work out what one piece sends to the point you care about, add the pieces up as vectors. Everything below is that one sentence, applied to wires bent into different shapes.

What one piece of current sends out

Take a wire carrying current $I$ and mentally cut out a short length of it. Call that stub $d\vec{\ell}$: a vector whose length is the length of the stub in metres and whose direction is the direction the current flows there. The product $I\,d\vec{\ell}$ is the source: a current element, measured in ampere-metres.

Pick a field point $P$. Let $\vec{r}$ run from the element to $P$, let $r$ be its length, and let $\hat{r}$ be the unit vector along it ($\hat{r} = \vec{r}/r$, length one, direction only). Three things have to be true of the field $d\vec{B}$ this element makes at $P$. It should grow with the current and with the size of the stub. It should fall off with distance the way every point-source field does, as $1/r^2$. And it must come out sideways, because the compass says so. A cross product is the only clean way to write "sideways to both". Put those together:

The constant $\mu_0 = 4\pi\times10^{-7}$ T·m/A is the permeability of free space; the $4\pi$ is packed into it so that it cancels in the tidiest answers. The combination $\mu_0/4\pi = 1.0\times10^{-7}$ in SI units is the number you will actually punch into a calculator, so learn it in that form.

Read the cross product carefully, because it does two separate jobs. Its magnitude is $d\ell\sin\theta$, where $\theta$ is the angle between the element and the line to $P$. So an element sends nothing at all to a point straight ahead of itself or straight behind it, and sends the most to a point off to its side. Its direction is given by the right-hand rule: point your fingers along the current, curl them towards $\hat{r}$, and your thumb gives $d\vec{B}$. For a point beside a wire that means the field circles the wire.

The straight wire

Now add elements up. Put a straight wire along a line and a field point $P$ a perpendicular distance $d$ from it. Every element of the wire lies in the plane of the page, and $P$ is in that plane too, so every $d\vec{\ell}\times\hat{r}$ points the same way: straight out of the page. That is the piece of luck that makes this problem doable. The vector sum collapses into an ordinary sum of magnitudes, and the only question left is how much each element gives.

Measure position along the wire by the angle $\theta$ from the perpendicular through $P$, positive on one side and negative on the other. Doing the integral (it is a standard one, and the substitution $x = d\tan\theta$ is what makes it go) gives, for a segment whose ends sit at angles $\theta_1$ and $\theta_2$,

Memorise this one. Nearly every straight-wire problem in the bank is this formula used once per segment. Two special cases fall straight out. Let the wire run forever both ways: $\theta_2 \to +90^\circ$, $\theta_1 \to -90^\circ$, the bracket becomes $1-(-1) = 2$, and

Let it run in from infinity and stop level with $P$ (a semi-infinite wire): the bracket is $1 - 0 = 1$, and you get exactly half of that, $\mu_0 I/4\pi d$.

Worked example 1. A long straight wire carries 5.0 A. Find $B$ at 10 cm.

Straight substitution into the infinite-wire result:

The $\pi$ cancels against the $\pi$ upstairs, leaving $B = (2\times10^{-7})(5.0)/(0.10) = 1.0\times10^{-5}$ T. Ten microtesla, about a fifth of the Earth's field. That grouping, $B = 2\times10^{-7}\,I/d$, is worth keeping: it saves you from mishandling the $\pi$'s under time pressure.

Now the same formula twice. Two long wires 0.30 m apart carry 8.0 A and 2.0 A in the same direction. Where between them does the field vanish? Between the wires the two fields point opposite ways, so they can cancel; outside, they point the same way and cannot. Let $x$ be the distance from the 8.0 A wire. Setting magnitudes equal,

Everything common cancels, so $8.0(0.30-x) = 2.0x$, giving $2.4 = 10.0x$ and $x = 0.24$ m. The null point sits closer to the weaker wire, four times further from the 8 A wire than from the 2 A one, because the ratio of currents is four. Check that against your intuition before you move on.

Arcs, loops and polygons

Bend the wire into a circular arc of radius $R$ and ask for the field at the centre of curvature. This is the easiest integral in the subject, because every element is the same distance $R$ away and every element is perpendicular to its own $\hat{r}$, so $\sin\theta = 1$ throughout and every contribution points the same way. The magnitudes just add, and the total length of an arc subtending angle $\phi$ (in radians) is $R\phi$:

A full circle has $\phi = 2\pi$ and gives $B = \mu_0 I/2R$. So an arc gives the fraction $\phi/2\pi$ of a full loop: a semicircle gives half, a quarter arc gives a quarter. And straight leads that run radially in or out from the centre contribute nothing at all, since they point along their own $\hat{r}$. That is why so many problems are drawn with radial leads: it is a way of handing you the arc alone.

Worked example 2. A wire carries 10 A in from far away, runs parallel to the $x$ axis, curls round a semicircle of radius 5.0 cm, and heads back out parallel to the $x$ axis on the other side: a hairpin. Find $B$ at the centre.

Three pieces. The semicircle: $\phi = \pi$, so

Each straight lead is semi-infinite and lies a perpendicular distance $R = 0.050$ m from the centre, so each gives

All three point the same way (check with the right hand: the current sweeps the same sense past the centre in all three pieces), so they add: $6.28\times10^{-5} + 2(2.0\times10^{-5}) = 1.03\times10^{-4}$ T. Had the leads run radially instead of parallel, the answer would have been the arc alone.

A polygon is the same trick with the finite-segment formula. For a regular $n$-gon of side $a$, the centre is a distance $d = a/(2\tan(\pi/n))$ from each side, and each side subtends angles $\pm\pi/n$, so each contributes $\frac{\mu_0 I}{4\pi d}\cdot 2\sin(\pi/n)$, and there are $n$ of them.

Worked example 3. A square loop of side 10 cm carries 6.0 A. Find $B$ at one corner.

The trap here is doing the centre calculation by mistake. Sit at the corner and look at the four sides. Two of them start at your corner and run away from you along straight lines that pass through you: $\theta = 0$ or $180^\circ$ for every element, so those two sides give nothing. Only the far two matter, and each is a distance $a = 0.10$ m away. For one of them, the near end is level with you ($\theta_1 = 0$) and the far end is at the opposite corner, at $45^\circ$. So

The two far sides carry current in senses that give fields the same way out of the page, so double it: $B = 8.49\times10^{-6}$ T.

Off the centre: the loop axis

Step off the centre of a circular loop and onto its axis, a distance $z$ away. Now the elements are not all equivalent in direction. Each one is still a distance $\sqrt{R^2+z^2}$ from the field point and still perpendicular to its $\hat{r}$, so each $|d\vec{B}|$ is the same, but each is tilted: it has a component along the axis and a component perpendicular to it. Pair up any element with the one diametrically opposite and the perpendicular components cancel exactly, while the axial ones add. So only the axial part survives, and the tilt factor is $R/\sqrt{R^2+z^2}$. Multiply the total magnitude $\mu_0 I (2\pi R)/4\pi(R^2+z^2)$ by that factor:

Set $z = 0$ and it returns $\mu_0 I/2R$, as it must. Let $z \gg R$ and the $R^2$ under the root drops out, leaving $B \approx \mu_0 I R^2/2z^3$: the $1/z^3$ falloff of a magnetic dipole. For $N$ turns wound close together, multiply by $N$: every turn does the same job.

Stack loops and you get a solenoid. Treat a length $dz$ of a solenoid with $n$ turns per metre as a loop carrying current $n I\,dz$, integrate the axial formula over the length, and the result for a solenoid of length $L$ and radius $R$, at its midpoint, is

For $L = 0.30$ m, $R = 0.020$ m, $n = 2000$ per metre and $I = 3.0$ A: $\mu_0 n I = (4\pi\times10^{-7})(2000)(3.0) = 7.54\times10^{-3}$ T, and the geometric factor is $0.30/\sqrt{0.09+0.0016} = 0.30/0.3027 = 0.991$. So $B = 7.47\times10^{-3}$ T. The factor being close to one says this solenoid is long enough to count as infinite, which is what the next lesson will assume.

Currents that are not wires

A current is charge passing a point per second, and nothing in the Biot-Savart law cares whether that charge is in a wire. Spin a charged object and you have a current.

Worked example 4. A flat insulating disk of radius $R = 5.0$ cm carries $q = 2.0\ \mu$C spread evenly over its face and spins at $\omega = 100$ rad/s. Find $B$ at its centre.

You cannot use the ring answer directly, because different parts of the disk sit at different radii. So build it from rings. The surface charge density is

A ring between radius $r$ and $r+dr$ holds charge $dq = \sigma\,2\pi r\,dr$. That charge goes round once every $2\pi/\omega$ seconds, so the current it represents is charge over period:

Each ring is a full loop of radius $r$, so it puts $\mu_0\,dI/2r$ at the centre. The $r$ cancels, which is the small miracle that makes this problem easy:

Every ring contributes equally, so integrating from $0$ to $R$ just multiplies by $R$:

A ring is even easier: all the charge is at one radius, so the current is $I = q f$ with $f$ the revolutions per second, and $B = \mu_0 I/2R$. The same reasoning handles a sheet of charge sliding sideways, which is where the surface current density $K$ (amperes per metre of width) comes from; an infinite sheet gives $B = \mu_0 K/2$ on each side, pointing opposite ways.

Knowing when to reach for it

The next lesson, Ampere's law, gets the same fields far faster whenever the geometry is symmetric enough: infinite wires, infinite solenoids, coaxial cables. So the real skill is telling the two apart.

Reach for Biot-Savart when the problem names a shape: an arc, a square, a hexagon, a finite segment, a loop seen from a point on its axis, a pair of coils. Any time there is a length or a corner or an angle in the statement, the field will not be constant along any convenient closed loop, and Ampere's law has nothing to grip.

Then run the same four steps every time. One: decide which pieces contribute nothing, because a piece pointing along its own line to the field point is free. Two: get the direction of each surviving piece with your right hand, and draw the arrows. Three: add what is left, as scalars if the arrows agree, as vectors if they do not. Four: check the size against a familiar anchor, such as $\mu_0 I/2R$ for a loop's centre or $2\times10^{-7} I/d$ for a long wire. A bent-wire answer that comes out ten times either of those is a sign you dropped a $\pi$ or a factor of two.