When considering the same setup as in this question, i.e. a straight, infinitely long wire carrying the current $I$, Ampère's circuital law

$$\oint_C \vec{B} \cdot \mathrm{d}\vec{r} = \mu_0 I_\text{enc}$$

is often used to calculate $\vec{B}$, as this is a cylindrically symmetric problem. Due to the symmetry, $\vec{B} = B(\rho) \vec{e}_\varphi$ (cylinder coordinates $\rho, \varphi, z$) is assumed and, by integrating over a circle with radius $\rho$, one finds

$$B(\rho) \cdot 2 \pi \rho = \mu_0 I \Leftrightarrow B(\rho) = \frac{\mu_0 I}{2 \pi \rho}$$

My question is whether the assumption $\vec{B} = B(\rho) \vec{e}_\varphi$ is actually valid. I can accept that $B$ has to be independent of $\varphi$ and $z$ due to the symmetry, but I'm not so sure about the direction of $\vec{B}$. In particular, $\vec{B}$ does depend upon $\varphi$, as $\vec{e}_\varphi = \begin{pmatrix} -\sin(\varphi)\\ \cos(\varphi) \\ 0\end{pmatrix}$, so the assumption that $\vec{B}$ is independent of $\varphi$ (as I've heard or seen in books) really only applies to the absolute $|\vec{B}|$, doesn't it? In fact, with this rationale, could $\vec{B}$ not also have components in other directions which are not "seen" by the line integral? Assuming

$$\vec{B} = B_\rho(\rho) \vec{e}_\rho + B_\varphi(\rho) \vec{e}_\varphi + B_z(\rho) \vec{e}_z,$$

the above would only calculate $B_\varphi(\rho)$. Does one have to use additional reasoning to be able to assume $\vec{B} = B(\rho) \vec{e}_\varphi$?

  • $\begingroup$ That B has only $\phi$ component you can see from the Biot-Savart law en.wikipedia.org/wiki/Biot-savart $\endgroup$
    – hyportnex
    Commented Feb 4, 2014 at 19:54
  • $\begingroup$ So it is true that you cannot make the assumption based on Ampère's law alone, i.e. you have to add Biot-Savart? $\endgroup$
    – Socob
    Commented Feb 4, 2014 at 22:24
  • $\begingroup$ I believe you are correct, yes. You need input as to the actual direction of $\vec{B}$. $\endgroup$
    – BMS
    Commented Feb 5, 2014 at 3:23

1 Answer 1


As stated in the comments, the answer does depend on the Biot-Savart law. But in fact, not all the information from the Biot-Savart law is needed. The only two facts needed from the Biot-Savart law is that 1) $\vec{B}$ is a pseudo-vector, and 2) $\vec{B}$ is linear in the current $I$. In particular if I multiply $I$ by $-1$ then $\vec{B}$ also gets multiplied by $-1$. Any vector field that satisfied these assumptions would have to be purely azimuthal in this geometry.

You already understand why the field must depend only on $\rho$ and not on $\phi$ or $z$, but let's see why the two properties above imply that it cannot have a component in the $\hat{z}$ or $\hat{\rho}$ directions.

We will pick an arbitrary point $p$, and choose coordinate so $p$ lies on the $x$-axis. Then we consider three transformations that each have the effect of flipping the sign of $I$ while leaving $p$ invariant. If I write the magnetic field as $(B_\rho, B_\phi, B_z)$, then the effect of each of these symmetry transformations will be to either leave a component alone or multiply by $-1$. Therefore I will represent the effect of the transformation by a triple of plus or minus signs.

The first transformation we consider is rotation about the $x$ axis. It can be seen that this transformation reverses the $\phi$ and $z$ component, but leaves the $\rho$ component. So its effect is $(+,-,-)$.

The next transformation we will consider is simply inverting $I$. The effect of this transformation is to invert $\vec{B}$, since $\vec{B}$ is linear in $I$. So its effect is $(-,-,-)$.

The third effect we will consider is reflection through the $x$-$y$ plane. I will view this as a composition of parity inversion with a $\pi$ rotation about the $z$-axis. We know that for parity inversion $\vec{B}$ remains invariant because $\vec{B}$ is a pseudo-vector. But when we say invariant we mean that $\vec{B}(p)$ before the transformation is the same as $\vec{B}(-p)$ after the transformation. But $\hat{\rho}(p) = -\hat{\rho}(-p)$, $\hat{\phi}(p) = -\hat{\phi}(-p)$, and $\hat{z}(p) = \hat{z}(-p)$. Thus for $\vec{B}$ to be the same the effect of the transformation must be $(-,-,+)$. Then to this we compose a $\pi$ rotation about the $z$ axis, but this does not change the coordinates.

So we found our three equivalent transformations have different effects. They are $(+,-,-)$, $(-,-,-)$, and $(-,-,+)$. Let's consider the first component. The first transformation rule tells us that $B_\rho$ after the transformation is the same as $B_\rho$ before the transformation. However, the second law tells us that $B_\rho$ after the transformation is the opposite as $B_\rho$ before the transformation. The only way this is possible is if $B_\rho = 0$ after the transformation. Then $B_\rho = 0$ before the transformation as well. So $B_\rho$ must be zero no matter what. Similar logic shows $B_z$ must be zero. However, we have not shown that $B_\phi$ must be zero because we consistently saw that it gets multiplied by $-1$.

  • $\begingroup$ that B is a pseudo-vector is a consequence of the Biot-Savart law because it is the vector product of two polar vectors. $\endgroup$
    – hyportnex
    Commented Feb 7, 2014 at 16:36

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