The Biot-Savart Law

Beyond AP Physics 1

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Section 1

The Biot-Savart Law

What would you expect to happen when you stick a compass next to a current carrying wire? Performing the experiment we would find that the compass needle would actually move as you neared the wire.

In fact we can perform other experiments. Let’s consider two separate current carrying wires. If we take two wires and place them parallel to one another we would find several things.

Our first observation would be that depending upon the orientation of the current the wires would either attract or repel. If the currents in the wires both point in the same direction the wires will attract one another. If the currents are opposite one another the wires would repel.

Next we would observe that by varying the currents of the wire we would get results that match the forces you’d expect to be cause by magnetic fields.

This is no coincidence, currents generate magnetic fields!

As with other fundamental notions in electromagnetism we have a corresponding law which describes the effect. For currents this is the Biot-Savart Law

$\vec{B}=\frac{\mu_0}{4\pi}\mu_0\int\frac{I\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

which gives us the magnetic field at a particular point away from a thin wire described by a curve C.

The constant is what we call the magnetic permeability of free space and has the value

$\mu_0\approx1.26\times10^-6\frac{N}{A^2}$

The Biot-Savart is probably best demonstrated using examples.

Example 6.1.1

You are given an infinitely long piece of straight wire which carries a constant current. Determine the magnitude and direction of the magnetic field a distance d from the wire.

To begin this problem we place the wire in an appropriate coordinate system. We chose to place the wire along the x-axis and have the point where we wish to study the magnetic field be in the xy-plane as in fig#blah. The current of our wire points in the positive x direction.

Now we must determine the values of the various quantities within the Biot-Savart law. We begin with $\vec{dl}$ . Since the wire alongs only along the x-axis $\vec{dl}=dx\hat{x}$ .

Next we look at $\vec{r}$ , the vector which points from the wire to the point we are studying. In order to get this vector to point in the correct direction we must subtract the x component so that

$\vec{r}=-x\hat{x}+d\hat{y}=(-x,\ y)$

and $|\vec{r}|=\sqrt{(-x)^2+d^2}=\sqrt{x^2+d^2}$

Using $\vec{r}$ we can also get $|\vec{r}|$ .

$|\vec{r}|=\sqrt{(-x)^2+d^2}=\sqrt{x^2+d^2}$

The final component we need is $|\vec{dl}\times\vec{r}|$ . In order to find this all we must do is look at the appropriate matrix.

$|\vec{dl}\times\vec{r}| = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ dx & 0 & 0\\ -x & d & 0 \end{vmatrix} = ddx\hat{k}$

Now using the above information and the fact that I is constant we can change the Biot-Savart law to get something a little more manageable.

$\vec{B}=\frac{\mu_0}{4\pi}\int\frac{I\vec{dl}\times\vec{r}}{|\vec{r}|^3} = \frac{\mu_0I}{4\pi}\int\frac{d}{\sqrt{x^2+d^2}^3}dx\hat{k}$

which is nothing more than an ordinary integral. To solve this we must use trigonometric substitution. Therefore we let $x=d\tan\theta$ and get that $dx=d\sec^2\theta d\theta$ . Making these substitutions leads to the integral

$\frac{\mu_0I}{4\pi}\int\frac{d^2\sec^2\theta}{\sqrt{d^2\tan\theta +d^2}^3}d\theta\hat{k}$

Simplifying our integral leads us to the following

$\frac{\mu_0I}{4\pi d}\int\cos\theta d\theta\hat{k}$

We’re almost done! The final piece of information to solve this integral is the bounds. Since we are determining the field produced by an infinitely long wire we must integrate over the domain $(-\infty,\infty)$ . Using our substitution we can convert this into radians.

Since $x=d\tan\theta$ we have $\theta=\tan^{-1}\frac{x}{d}$ so

$\theta_2 =\lim_{x \to \infty}\tan^{-1}\frac{x}{d}=\frac{\pi}{2}$

Similarly

$\theta_1=\lim_{x \to -\infty}\tan^{-1}\frac{x}{d}=-\frac{\pi}{2}$

Finally for our magnetic field we get that

$\vec{B}=\frac{\mu_0I}{4\pi d}\int\limits_{\frac{/pi}{2}}^{\frac{\pi}{2}}\cos\theta d\theta=\frac{\mu_0I}{2\pi d}\hat{k}$

Example 6.2.2

A circular loop of wire with radius R is placed flat on a table as in fig#blah. Determine the magnitude and direction of the magnetic field a height h above it’s center.

In this problem we will take the surface of the table to be the xy-plane and z to point upwards from it. As in the earlier example we will first determine values for each part of the Biot-Savart law

$\vec{B}=\frac{\mu_0}{4\pi}\mu_0\int\frac{I\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Before we start though it should be noted that we will be make trigonometric substitutions for cartesian coordinates so that at the end we will simply have to integrate from 0 to 2π minimizing the overall amount of math.

Let us now begin by determining a value for $\vec{dl}$ . In terms of cartesian coordinates we have that $\vec{dl}=dx\hat{i}+dy\hat{j}+dz\hat{k}$ . In terms of of an angle θ from the x-axis we can convert these dx and dy since if $x=R\cos\theta$ then $dx=-R\sin\theta d\theta$ and similarly $y=R\sin\theta$ then $dy=R\cos\theta d\theta$ .

Now z is held constant at h we have that $dz=0$ because since it never changes and hence we have for $\vec{dl}$

$\vec{dl}=-R\sin\theta d\theta\hat{i}+R\cos\theta d\theta\hat{j}$

Next let us examine $\vec{r}$ . In order to point in the correct direction we must have

$\vec{r}=h\hat{k}-(x\hat{i}+y\hat{j})=-x\hat{i}-y\hat{j}+h\hat{k}$

Using our trigonometric substitutions we get that

$\vec{r}=-R\cos\theta\hat{i}-R\sin\theta\hat{j}+h\hat{k}$

Since we have both $\vec{dl}$ and $\vec{r}$ we can go ahead and determine $\vec{dl}\times\vec{r}$ .

$\begin{aligned} \vec{dl}\times\vec{r} &= \left| \begin{aligned} \hat{i} && \hat{j} && \hat{k}\\ -R\sin\theta d\theta && R\cos\theta d\theta && 0\\ -R\cos\theta && -Rsin\theta && h \end{aligned} \right|\\ &=hR\cos\theta d\theta\hat{i}+hR\sin\theta d\theta\hat{j}+…\\ &\ \ \ \ (R^2\sin^2\theta d\theta+R^2\cos^2\theta d\theta)\hat{k}\\ &=hR\cos\theta d\theta\hat{i}+hR\sin\theta d\theta\hat{j}+R^2d\theta\hat{k} \end{aligned}$

The final piece we need is $|\vec{r}|^3$ . Luckily the vector pointing from our loop to our point of interest has constant length. Therefore we have that

$|\vec{r}|^3=\sqrt{h^2+R^2}^3=(h^2+R^2)^{\frac{3}{2}}$

Since we have all the needed bits of information we can go ahead and calculate our integral.

For the x component we have that

$B_x=\frac{\mu_0I}{4\pi}\int\limits_{0}^{2\pi}\frac{hR\cos\theta}{(h^2+R^2)^\frac{3}{2}}d\theta\hat{i}=0$

Similarly, for the y component

$B_y=\frac{\mu_0I}{4\pi}\int\limits_{0}^{2\pi}\frac{hR\sin\theta}{(h^2+R^2)^\frac{3}{2}}d\theta\hat{j}=0$

Finally, for the z component we get that

$B_y=\frac{\mu_0I}{4\pi}\int\limits_{0}^{2\pi}\frac{R^2}{(h^2+R^2)^\frac{3}{2}}d\theta\hat{k}=\frac{\mu_0I}{2}\frac{R^2}{(h^2+R^2)^{\frac{3}{2}}}\hat{k}$

Section 1

The Biot-Savart Law

What would you expect to happen when you stick a compass next to a current carrying wire? Performing the experiment we would find that the compass needle would actually move as you neared the wire.

In fact we can perform other experiments. Let’s consider two separate current carrying wires. If we take two wires and place them parallel to one another we would find several things.

Next we would observe that by varying the currents of the wire we would get results that match the forces you’d expect to be cause by magnetic fields.

This is no coincidence, currents generate magnetic fields!

As with other fundamental notions in electromagnetism we have a corresponding law which describes the effect. For currents this is the Biot-Savart Law

$\vec B={\mu_0\over4\pi}\int{I\,d\vec l\times\vec r\over|\vec r|^3}$

which gives us the magnetic field at a particular point away from a thin wire described by a curve C.

The constant is what we call the magnetic permeability of free space and has the value

$\mu_0\approx1.26\times10^{-6}\frac{N}{A^2}$

The Biot-Savart is probably best demonstrated using examples.

Example 6.1.1

Optional Biot–Savart calculus derivation for an infinite straight wire. Prerequisites: vector components and an improper integral; the final long-wire field can also be used directly. This preserved source example makes no AP coverage claim.

An infinite straight wire carries current I in +x. Find B at (0,d,0), d>0.

Authored clarification: The original fig#blah reference is unresolved; this explicit geometry matches the worked parametrization.

Reviewed worked solution:

Optional calculus evaluation. The antiderivative is supplied below, and the limits at both ends of the infinite wire are evaluated explicitly.

Use displacement and vector cross product.
$\vec r=(-x,d,0),\quad d\vec l=dx\hat x,\quad d\vec l\times\vec r=d\,dx\hat z$
Use Biot–Savart with one μ₀.
$\vec B={\mu_0I\over4\pi}\int_{-\infty}^{\infty}{d\,dx\over(x^2+d^2)^{3/2}}\hat z$
Use the supplied antiderivative for d>0.
$\int{d\,dx\over(x^2+d^2)^{3/2}}={x\over d\sqrt{x^2+d^2}}+C$ . Differentiating the right-hand expression gives the integrand.
Evaluate both improper-integral limits.
Since $\lim_{x\to\pm\infty}{x\over\sqrt{x^2+d^2}}=\pm1$ , the upper limit minus the lower limit is $1/d-(-1/d)=2/d$ . Thus $\vec B={\mu_0I\over2\pi d}\hat z$ .

Answer: $\vec B={\mu_0I\over2\pi d}\hat z$

Check: The right-hand rule gives +z for current +x at the +y observation point.

Simplifying our integral leads us to the following

$\frac{\mu_0I}{4\pi d}\int\cos\theta d\theta\hat{k}$

Since $x=d\tan\theta$ we have $\theta=\tan^{-1}\frac{x}{d}$ so

$\theta_2 =\lim_{x \to \infty}\tan^{-1}\frac{x}{d}=\frac{\pi}{2}$

Similarly

$\theta_1=\lim_{x \to -\infty}\tan^{-1}\frac{x}{d}=-\frac{\pi}{2}$

Finally for our magnetic field we get that

$\vec B={\mu_0I\over4\pi d}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta\,\hat z={\mu_0I\over2\pi d}\hat z$

Example 6.2.2

Optional advanced on-axis current-loop method and result. Prerequisites: parametrized vectors and Biot–Savart integration. This is outside the required AP calculation model. This preserved source example makes no AP coverage claim.

A radius-R circular wire loop in the xy plane carries current I counterclockwise viewed from +z. Find B at (0,0,h).

Authored clarification: The source’s figure reference and current value/orientation are missing; these explicit symbolic conditions supply them.

Reviewed worked solution:

Parametrize source elements at z=0.
$d\vec l=(-R\sin\theta,R\cos\theta,0)d\theta,\quad\vec r=(-R\cos\theta,-R\sin\theta,h)$
Cross and cancel transverse integrals.
$d\vec l\times\vec r=(hR\cos\theta,hR\sin\theta,R^2)d\theta$
Integrate the axial component through 2π.
$B_z={\mu_0I\over4\pi}{2\pi R^2\over(R^2+h^2)^{3/2}}$

Answer: $\vec B={\mu_0IR^2\over2(R^2+h^2)^{3/2}}\hat z$

Check: At h=0, B=μ₀I/(2R); reversing current reverses B.

The final piece we need is $|\vec{r}|^3$ . Luckily the vector pointing from our loop to our point of interest has constant length. Therefore we have that

$|\vec{r}|^3=\sqrt{h^2+R^2}^3=(h^2+R^2)^{\frac{3}{2}}$

Since we have all the needed bits of information we can go ahead and calculate our integral.

For the x component we have that

$B_x=\frac{\mu_0I}{4\pi}\int\limits_{0}^{2\pi}\frac{hR\cos\theta}{(h^2+R^2)^\frac{3}{2}}d\theta\hat{i}=0$

Similarly, for the y component

$B_y=\frac{\mu_0I}{4\pi}\int\limits_{0}^{2\pi}\frac{hR\sin\theta}{(h^2+R^2)^\frac{3}{2}}d\theta\hat{j}=0$

Finally, for the z component we get that

$B_z={\mu_0IR^2\over2(R^2+h^2)^{3/2}},\qquad\vec B=B_z\hat z$

Review 1.1

This review preserves the author’s recorded prompts, choices, and answer states. Corrected items have separate reviewed explanations and keys; unchanged draft keys are marked unreviewed. Unfinished items are not scored.