Skip to main content
Physics LibreTexts

12.4: Ampère's Law In Differential Form (Optional)

  • Page ID
  • 11.4.1 The curl operator

    The differential form of Gauss' law is more physically satisfying than the integral form, because it relates the charges that are present at some point to the properties of the electric field at the same point. Likewise, it would be more attractive to have a differential version of Ampère's law that would relate the currents to the magnetic field at a single point.

    a / The div-meter, 1, and the curl-meter, 2 and 3.

    Intuitively, the divergence was based on the idea of the div-meter, a/1. The corresponding device for measuring the curliness of a field is the curl-meter, a/2. If the field is curly, then the torques on the charges will not cancel out, and the wheel will twist against the resistance of the spring. If your intuition tells you that the curlmeter will never do anything at all, then your intuition is doing a beautiful job on static fields; for nonstatic fields, however, it is perfectly possible to get a curly electric field.

    Gauss' law in differential form relates \(\rm div​ \mathbf{E}\), a scalar, to the charge density, another scalar. Ampère's law, however, deals with directions in space: if we reverse the directions of the currents, it makes a difference. We therefore expect that the differential form of Ampère's law will have vectors on both sides of the equal sign, and we should be thinking of the curl-meter's result as a vector. First we find the orientation of the curl-meter that gives the strongest torque, and then we define the direction of the curl vector using the right-hand rule shown in figure a/3.

    To convert the div-meter concept to a mathematical definition, we found the infinitesimal flux, \(d\Phi\) through a tiny cubical Gaussian surface containing a volume \(dv\). By analogy, we imagine a tiny square Ampèrian surface with infinitesimal area \(d\mathbf{A}\). We assume this surface has been oriented in order to get the maximum circulation. The area vector \(d\mathbf{A}\) will then be in the same direction as the one defined in figure a/3. Ampère's law is

    \[\begin{equation*} d\Gamma = \frac{4\pi k}{c^2}\,dI_{through} .\end{equation*}\]
    We define a current density per unit area, \(\mathbf{j}\), which is a vector pointing in the direction of the current and having magnitude \(\mathbf{j}=dI/|d\mathbf{A}|\). In terms of this quantity, we have
    \[\begin{align*} d\Gamma = \frac{4\pi k}{c^2}\,{j} |\mathbf{j}|\,|d\mathbf{A}| \\ \frac{d\Gamma}{|d\mathbf{A}|} = \frac{4\pi k}{c^2}\, |\mathbf{j}| \end{align*}\]
    With this motivation, we define the magnitude of the curl as
    \[\begin{align*} |\text{curl} \ \mathbf{B}| &= \frac{d\Gamma}{|d\mathbf{A}|} .\end{align*}\]
    Note that the curl, just like a derivative, has a differential divided by another differential. In terms of this definition, we find Amp To convert the div-meter concept to a mathematical definition, we found the infinitesimal flux, \(d\Phi\) through a tiny cubical Gaussian surface containing a volume \(dv\). By analogy, we imagine a tiny square Ampèrian surface with infinitesimal area \(d\mathbf{A}\). We assume this surface has been oriented in order to get the maximum circulation. The area vector \(d\mathbf{A}\) will then be in the same direction as the one defined in figure a/3. Ampère's law is {e}re's law in differential form:
    \[\begin{align*} \text{curl} \ \mathbf{B} &= \frac{4\pi k}{c^2} \,\mathbf{j} \end{align*}\]

    The complete set of Maxwell's equations in differential form is collected on page 914.

    11.4.2 Properties of the curl operator

    The curl is a derivative.

    b / The coordinate system used in the following examples.

    As an example, let's calculate the curl of the field \(\hat{\mathbf{x}}\) shown in figure c.

    c / The field \(\hat{\mathbf{x}}\).

    For our present purposes, it doesn't really matter whether this is an electric or a magnetic field; we're just getting out feet wet with the curl as a mathematical definition. Applying the definition of the curl directly, we construct an Ampèrian surface in the shape of an infinitesimally small square. Actually, since the field is uniform, it doesn't even matter very much whether we make the square finite or infinitesimal. The right and left edges don't contribute to the circulation, since the field is perpendicular to these edges. The top and bottom do contribute, but the top's contribution is clockwise, i.e., into the page according to the right-hand rule, while the bottom contributes an equal amount in the counterclockwise direction, which corresponds to an out-of-the-page contribution to the curl. They cancel, and the circulation is zero. We could also have determined this by imagining a curl-meter inserted in this field: the torques on it would have canceled out.

    It makes sense that the curl of a constant field is zero, because the curl is a kind of derivative. The derivative of a constant is zero.

    The curl is rotationally invariant.

    d / The field \(\hat{\mathbf{y}}\).

    Figure d looks just like figure c, but rotated by 90 degrees. Physically, we could be viewing the same field from a point of view that was rotated. Since the laws of physics are the same regardless of rotation, the curl must be zero here as well. In other words, the curl is rotationally invariant. If a certain field has a certain curl vector, then viewed from some other angle, we simply see the same field and the same curl vector, viewed from a different angle. A zero vector viewed from a different angle is still a zero vector.

    As a less trivial example, let's compute the curl of the field \(\mathbf{F}=x\hat{\mathbf{y}}\) shown in figure e, at the point \((x=0,y=0)\).

    e / The field \(x\hat{\mathbf{y}}\).

    The circulation around a square of side \(s\) centered on the origin can be approximated by evaluating the field at the midpoints of its sides, \begin{alignat*}{4} x&=s/2\qquad & y&=0\qquad\quad & \mathbf{F}&=(s/2)\hat{\mathbf{y}}\qquad & \mathbf{s}_1\cdot\mathbf{F}&=s^2/2\\
    x&=0 & y&=s/2 & \mathbf{F}&=0 & \mathbf{s}_2\cdot\mathbf{F}&=0\\
    x&=-s/2 & y&=0 & \mathbf{F}&=-(s/2)\hat{\mathbf{y}} & \mathbf{s}_3\cdot\mathbf{F}&=s^2/2\\
    x&=0 & y&=-s/2 & \mathbf{F}&=0 & \mathbf{s}_4\cdot\mathbf{F}&=0 ,
    \end{alignat*} which gives a circulation of \(s^2\), and a curl with a magnitude of \(s^2/\text{area}=s^2/s^2=1\). By the right-hand rule, the curl points out of the page, i.e., along the positive \(z\) axis, so we have
    \[\begin{equation*} \text{curl} \ x\hat{\mathbf{y}} = \hat{\mathbf{z}} . \end{equation*}\]

    Now consider the field \(-y\hat{\mathbf{x}}\), shown in figure f.

    f / The field \(- y\hat{\mathbf{x}}\).

    This is the same as the previous field, but with your book rotated by 90 degrees about the \(z\) axis. Rotating the result of the first calculation, \(\hat{\mathbf{z}}\), about the \(z\) axis doesn't change it, so the curl of this field is also \(\hat{\mathbf{z}}\).


    When you're taking an ordinary derivative, you have the rule

    \[\begin{equation*} \frac{d}{dx}[cf(x)] = c\frac{d}{dx}f(x) . \end{equation*}\]

    In other words, multiplying a function by a constant results in a derivative that is multiplied by that constant. The curl is a kind of derivative operator, and the same is true for a curl.

    Example 14: Multiplying the field by \(-1\).

    \(\triangleright\) What is the curl of the field \(- x\hat{\mathbf{y}}\) at the origin?

    g / Example 14.

    \(\triangleright\) Using the scaling property just discussed, we can make this into a curl that we've already calculated:

    \[\begin{align*} \text{curl} (- x\hat{\mathbf{y}}) &= -\text{curl}​ ( x\hat{\mathbf{y}}) \\ &= -\hat{\mathbf{z}} \\ \end{align*}\]

    This is in agreement with the right-hand rule.

    The curl is additive.

    We have only calculated each field's curl at the origin, but each of these fields actually has the same curl everywhere. In example 14, for instance, it is obvious that the curl is constant along any vertical line. But even if we move along the \(x\) axis, there is still an imbalance between the torques on the left and right sides of the curl-meter. More formally, suppose we start from the origin and move to the left by one unit. We find ourselves in a region where the field is very much as it was before, except that all the field vectors have had one unit worth of \(\hat{\mathbf{y}}\) added to them. But what do we get if we take the curl of \(- x\hat{\mathbf{y}}+\hat{\mathbf{y}}\)? The curl, like any god-fearing derivative operation, has the additive property

    \[\begin{equation*} \text{curl} (\mathbf{F}+\mathbf{G})= \text{curl}​\mathbf{F}+\text{curl}​\mathbf{G} , \end{equation*}\]


    \[\begin{equation*} \text{curl}​\ (- x\hat{\mathbf{y}}+\hat{\mathbf{y}}) = \text{curl}\ (- x\hat{\mathbf{y}})+\text{curl}​\ (\hat{\mathbf{y}}) . \end{equation*}\]

    But the second term is zero, so we get the same result as at the origin.

    Example 15: A field that goes in a circle

    \(\triangleright\) What is the curl of the field \(x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\)?

    h / Example 15.

    \(\triangleright\) Using the linearity of the curl, and recognizing each of the terms as one whose curl we have already computed, we find that this field's curl is a constant \(2\hat{\mathbf{z}}\). This agrees with the right-hand rule.

    Example 16: The field inside a long, straight wire

    \(\triangleright\) What is the magnetic field inside a long, straight wire in which the current density is \(j\)?

    \(\triangleright\) Let the wire be along the \(z\) axis, so \(\mathbf{j}= j\hat{\mathbf{z}}\). Ampère's law gives

    \[\begin{equation*} \text{curl} \ \mathbf{B}= \frac{4\pi k}{ c^2} \, j\hat{\mathbf{z}} . \end{equation*}\]

    In other words, we need a magnetic field whose curl is a constant. We've encountered several fields with constant curls, but the only one that has the same symmetry as the cylindrical wire is \(x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\), so the answer must be this field or some constant multiplied by it,

    \[\begin{equation*} \mathbf{B}= b\left( x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\right) . \end{equation*}\]

    The curl of this field is \(2 b\hat{\mathbf{z}}\), so

    \[\begin{align*} 2 b &= \frac{4\pi k}{ c^2} \, j ,\\ \text{and thus} \mathbf{B}&= \frac{2\pi k}{ c^2} \, j\left( x\hat{\mathbf{y}}- y\hat{\mathbf{x}}\right) .\\ \end{align*}\]

    The curl in component form

    Now consider the field

    \[\begin{align*} F_x &= ax+by+c \\ F_y &= dx+ey+f ,\ \text{i.e.,} \\ \mathbf{F} &= ax\hat{\mathbf{x}}+by\hat{\mathbf{x}}+c\hat{\mathbf{x}}+dx\hat{\mathbf{y}}+ey\hat{\mathbf{y}}+f\hat{\mathbf{y}} . \end{align*}\]

    The only terms whose curls we haven't yet explicitly computed are the \(a\), \(e\), and \(f\) terms, and their curls turn out to be zero (homework problem 50). Only the \(b\) and \(d\) terms have nonvanishing curls. The curl of this field is

    \[\begin{align*} \text{curl} \ \mathbf{F} &= \text{curl} \ (by\hat{\mathbf{x}})+\text{curl}​ \ (dx\hat{\mathbf{y}}) \\ &= b\, \text{curl} \ (y\hat{\mathbf{x}})+d\, \text{curl} \ (x\hat{\mathbf{y}}) \ \text{[scaling]}\ \\ &= b(-\hat{\mathbf{z}})+d(\hat{\mathbf{z}}) \ \text{[found previously]}\\ &= (d-b)\hat{\mathbf{z}} \end{align*}\]

    But any field in the \(x-y\) plane can be approximated with this type of field, as long as we only need to get a good approximation within a small region. The infinitesimal Ampèrian surface occurring in the definition of the curl is tiny enough to fit in a pretty small region, so we can get away with this here. The \(d\) and \(b\) coefficients can then be associated with the partial derivatives \(\partial F_y/\partial x\) and \(\partial F_x/\partial y\). We therefore have

    \[\begin{equation*} \text{curl}​\mathbf{F} = \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}} \end{equation*}\]

    for any field in the \(x-y\) plane. In three dimensions, we just need to generate two more equations like this by doing a cyclic permutation of the variables \(x\), \(y\), and \(z\):

    \[\begin{align*} (\text{curl}​\mathbf{F})_x &= \frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z} \ (\text{curl}\mathbf{F})_y = \frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x} \ (\text{curl} \mathbf{F})_z = \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} \end{align*}\]

    i / A cyclic permutation of \(x\), \(y\), and \(z\).

    Example 17: A sine wave

    \(\triangleright\) Find the curl of the following electric field

    \[\begin{equation*} \mathbf{E} = (\text{sin}\, x)\hat{\mathbf{y}} , \end{equation*}\]

    and interpret the result.

    j / Example 17.

    \(\triangleright\) The only nonvanishing partial derivative occurring in this curl is

    \[\begin{align*} (\text{curl}​ \mathbf{E})_z =\frac{\partial E_{y}}{\partial x}= \text{cos}\, x ,\ \text{so} \quad \text{curl}\ \mathbf{E} = \text{cos}\,\hat{\mathbf{z}} \end{align*}\]

    This is visually reasonable: the curl-meter would spin if we put its wheel in the plane of the page, with its axle poking out along the \(z\) axis. In some areas it would spin clockwise, in others counterclockwise, and this makes sense, because the cosine is positive in some placed and negative in others.

    This is a perfectly reasonable field pattern: it the electric field pattern of a light wave! But Ampère's law for electric fields says the curl of E is supposed to be zero. What's going on? What's wrong is that we can't assume the static version of Ampère's law. All we've really proved is that this pattern is impossible as a static field: we can't have a light wave that stands still.

    Figure k is a summary of the vector calculus presented in the optional sections of this book. The first column shows that one function is a related to another by a kind of differentiation. The second column states the fundamental theorem of calculus, which says that if you integrate the derivative over the interior of a region, you get some information about the original function at the boundary of that region.

    k / A summary of the derivative, gradient, curl, and divergence.


    Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.