# Can we use Ampere's law without requiring help from Biot-Savart law?

Consider the case of a straight, infinite wire carrying uniform current. From reversing the direction of the current we can rule out radial dependence of the magnetic field but are left with two possibilities: a circumferential field and a longitudinal field. Can we proceed further from here without using Biot-Savart law and only applying Ampere's law?

• Commented Nov 22, 2022 at 14:56
• What do you mean, in this context, by longitudinal field? By the circumferential field, I guess you mean that field lines are circumferences orthogonal to the wire direction. Commented Nov 24, 2022 at 8:34
• @GiorgioP in the direction of current Commented Nov 24, 2022 at 10:45
• "reversing the direction of the current" is not a symmetry. However, reflection about a perpendicular plane together with a $\pi$-rotation about an axis within that plane is a symmetry of the source wire. If you consider a hypothetical longitudinal $\vec{B}$, the reflection will leave it unchanged (it reverses in the reflection but picks up a factor of $-1$ because it is a pseudovector), but the rotation then reverses it. So $\vec{B}$ cannot have a longitudinal component because it would be altered by a symmetry transformation of the source. See my fuller answer below. Commented Nov 24, 2022 at 16:38

Yes. Because Maxwell's Equations are an axiomatic theory of E&M, it must be true that one can use them without appealing to the "laws" that are derived from them. The key is to make a symmetry argument to determine the some properties of the unknown field, before using one of Maxwell's equations to calculate.

It is easiest to demonstrate the idea with Gauss's Law first. Just as Biot-Savart is often used to specify the direction of $$\vec{B}$$ when applying Ampere's Law in introductory physics, we similarly appeal to Coulomb's law to determine the direction of the unknown $$\vec{E}$$ field due to a given source. Then Gauss's law is used only to determine the magnitude $$\left|\vec{E}\right|$$. Let's instead solve this problem without using Coulomb's law.

Consider a spherically symmetric positively-charged object of uniform density. We make a symmetry argument:

A rotation of this object about any axis passing through its center, by any angle, leaves the object unchanged (because it is spherically symmetric). Therefore, the electric field that the charged object generates must also be unchanged by the same transformation (because the source has not changed). This has two consequences: (1) If we choose a point in space on the axis of rotation, we see that the electric field vector cannot point off axis (because if it did, its direction would be changed under rotation). Therefore, that electric field vector must point toward or away from the center of the source. For any point in space one can find such a symmetry axis passing through it, so all electric field vectors are radial. (2) A point in space that is off-axis, located a distance $$d$$ from the center of the source, will be mapped under the rotation to another point in space that also lies on the sphere of radius $$d$$. Therefore all electric field vectors on a sphere about the center of the source must have the same magnitude.

These two facts are required to determine $$\vec{E}$$ using Gauss's Law. Choosing the gaussian surface, $$S$$, to be a sphere of radius $$r>R_{\rm source}$$ about the charge center, we have \begin{align} \int_S \vec{E} \cdot \hat{n} \, {\rm d}A &= \frac{Q_{\text{enc by } S}}{\varepsilon_0} \\ \int_S \left|\vec{E}\left(\vec{r}\right)\right| \, {\rm d}A &= \frac{\rho V_{\rm source}}{\varepsilon_0} \\ \left| \vec{E}\left(r\right) \right| \, \int_S {\rm d}A &= \frac{q_{\rm source}}{\varepsilon_0} \\ \left| \vec{E}\left(r\right) \right| \, 4\pi r^2 &= \frac{q_{\rm source}}{\varepsilon_0}\\ & \rightarrow E(r) = \frac{1}{4 \pi \epsilon_0} \frac{q_{\rm source}}{r^2} \end{align} where fact (1) was used in going from line 1 to line 2 ($$\vec{E}$$ is parallel to the normal $$\hat{n}$$ of the spherical surface at every point on the surface), and fact (2) was used going from line 2 to line 3 ($$E\left(\vec{r}\right)= E\left(r\right)$$, i.e., the magnitude is the same at all points on the sphere, so it is a constant of the integral). There is one more subtle point about whether $$\vec{E}$$ points toward or away from the source charge: that will determine the sign of $$\vec{E} \cdot \hat{n}$$, which must agree with $$q_{\rm source}$$.

Now that we derived Coulomb's Law from Maxwell's Equations (without using Coulomb's Law to do so!), we could use it to construct electric fields of non-symmetric continuous charge distributions. But there are other symmetric charged objects (infinite cylinder, infinite plane) for which the above technique works directly.

For Ampere's law applied to the long straight wire, the method is exactly the same: use symmetry arguments to determine the facts needed to apply the law in the familiar way (using a circle for the "amperian loop"). There is a cylindrical rotational symmetry of the source, and a symmetry of reflection about a perpendicular plane plus a rotation by $$\pi$$ radians (since the source current direction is reversed by either of the two alone). There is one additional fact that must be taken into account, however: the magnetic field is a "pseudovector" rather than a vector, which means its direction reverses under a reflection transformation. Using that information you should be able to make a set of symmetry arguments that provide all of the information needed to perform the Ampere's law calculation without appealing to Biot-Savart.

• Hi Ben, I apologize for leaving your answer unattended for so long. Can you please expand on the perpendicular plane symmetry and $\pi$ radian rotation? Thank you. Commented Nov 27, 2022 at 21:32
• @GedankenExperimentalist I put a bit more specific information about them in the comments to your question, did you see that? Commented Nov 28, 2022 at 3:00
• The idea is that you can get a symmetry of the current source by performing a sequence of two transformations: reflection through a plane perpendicular to the wire (reverses the current), and then rotation by $\pi$ to bring it back to its original position (current same as original). Under those transformations a longitudinal $\vec{B}$ at a point within the plane of reflection will reverse in direction, because the reflection will first reverse it and then reverse it again because it is a pseudovector, and the rotation will reverse it a third time. Commented Nov 28, 2022 at 3:08

This answer is wrong, for reasons mentioned at the bottom of the post.

After posting this question I realized that we do not need Biot-Savart Law to rule out a longitudinal dependence. Construct a rectangular loop perpendicular to the wire in the direction of the current as shown. Ampere's law along with our ruled-out radial direction of the field and with the fact that any circumferential contribution will give $$\vec B_c\cdot d\vec l =0$$ ($$\vec B_c$$ being circumferential magnetic field) gives $$\oint Bdl=0\rightarrow B=0$$ Since no current is passing through the loop.

But as Triatticus kindly pointed out in the comments "... integral is ultimately zero because the longitudinal field gives two contributions at the top and bottom that cancel, leaving you with just zero and no statement that the longitudinal part is zero, in other words, the top contributes $$\int B_zdz$$ and the bottom yields −$$\int B_zdz$$, so the integral is zero with no info on the longitudinal field", makes it clear that the solution I provided is wrong.

• Unfortunately that statement isn't true, that an integral is zero doesn't immediately imply that the integrand is zero, for instance it may just be that the scalar product of B and $d\ell$ is zero. Commented Nov 19, 2022 at 22:26
• @Triatticus true, but when you construct the loop, the perpendicular component has already been proven to be having a zero magnetic field, since it's radial. The only contribution left is by the longitudinal components where the angle is zero. Commented Nov 19, 2022 at 22:47
• Unfortunately no, the curcumferencial component also contributes to your loop, though I haven't really a good idea how your loop is constructed in the first place. This already shows that B is in fact nonzero (it's nonzero anyways just from Biot-Savart). Commented Nov 20, 2022 at 0:58
• @Triatticus I have added an image in the post, please refer to it and confirm if my logic seems right to you. Commented Nov 21, 2022 at 12:50
• Regardless of the loop you choose and how you orient it, the field $B_{\phi}$ contributes, so the integrand can not be identically zero either because of cancellation or a zero scalar product (recall that amperes law includes a scalar product and so should be $\oint \vec{B}\cdot d\vec{\ell}$). You will have to apply other logic to obtain that the longitudinal field is also zero. Commented Nov 22, 2022 at 16:29