# Magnetic Field 'Above' a straight wire

The magnetic field around the wire at lower point is circular (blue circle), that I know. But what about a point above wire, where there is no current flowing. Will there be a circular magnetic field, just like shown by green loop? Will a tiny magnet hover and revolve along that loop?

[The wire I have shown has current flowing through it, the top point of wire isn't meant to be its 'end'. Maybe imagine that the wire bends inside the screen at its top and bottom points. Now, will the shown(red) part of wire produce the green loop field?]

Where does your wire start and end? Is it a part of a circuit, not shown here? Is it connected to some generator? Does it work as an antenna?

## Physics

It's not possible to have a closed line of the magnetic field, enclosing a surface $$S$$ across which there is no electric current or a time-varying flux of the electric displacement field $$\mathbf{d}$$.

This follows as a combination of the Gauss' law for the magnetic field $$\mathbf{b}$$ and Ampére-Maxwell's equation, whose differential forms read

$$\nabla \cdot \mathbf{b} = 0$$
$$\nabla \times \mathbf{h} = \mathbf{j} + \partial_t \mathbf{d}$$

and integral forms

$$\Phi_{\partial V}(\mathbf{b}) = 0$$
$$\Gamma_{\partial S}(\mathbf{h}) = \Phi_{S}(\mathbf{j}) + \dot{\Phi}_{S}(\mathbf{d}) = \sum_{k} i_k + \dot{\Phi}_{S}(\mathbf{d})$$,

being $$\Gamma_{\partial S}(\mathbf{h})$$ the circuitation of the magnetic field $$\mathbf{b} = \mu \mathbf{h}$$, and $$\sum_k i_k$$ the sum of all the electric current crossing $$S$$.

In the sketch you can find a qualitative picture of the field lines of the magnetic field produced by direct current in the circuit.

In the following sketch, you can see a qualitative representation of the lines of the magnetic fields passing through the points of the green circle, with a magnet of a compass oriented in the same direction of the local magnetic field.

To cut a long story short, the current flowing in the wire produces a magnetic field in the points of the green circle, that is not zero. What is identically zero is the circuitation of the magnetic field on the green circle,

$$\Gamma_{\partial S^{Green}} = \displaystyle \oint_S \mathbf{b}(\mathbf{r}) \cdot \mathbf{\hat{t}}(\mathbf{r}) = \Phi_{S^{Green}}(\mathbf{j}) = \sum_k i_k = 0$$,

since no current crosses the surface enclosed by the green loop.

If you want/need, you can separate the contributions of each wire to the magnetic field exploiting the principle of superposition. As an example, assuming that the green path is sufficiently close to the angle formed by two edges, you can approximately neglect the contributions of the other two edges of a "square" circuit.

The contributions to the magnetic field due to the magenta edge is orthogonal to the $$Z$$-direction, while the contributions of the orange edge are orthogonal to the $$X$$-direction. This two contributions combines in each point of space with the vector sum, resulting in a vector that in general is not orthogonal to any of the axis of the reference frame considered.

• direct current? Commented Dec 9, 2022 at 15:09
• So, no. With direct current, the current flowing in the wire won't produce closed loops of magnetic field lines that don't embrace the wire of the circuit itself. Commented Dec 9, 2022 at 15:23
• there is magnetic fields everywhere in the domain (even with negligible magnitude far away from the circuit), even in the points of your green circle. Without asking a new question, try to have a look at the new sketch, where I drew the lines of the magnetic fields passing through the points of the green circle, with a magnet of a compass oriented in the same direction of the local magnetic field. Commented Dec 9, 2022 at 15:49
• I should have charged many people before...your green loop is just a geometric line that you can draw in the domain. What you can do, is considering the influence of each edge of the circuit separately and then sum their effects, exploiting the principle of superposition. I'll add a new sketch. Commented Dec 9, 2022 at 16:01
• pink wire produces a magnetic field that is tangent to the green path, if this is orthogonal to the wire, and its center is on the axis of the wire itself. I think I have nothing more to say about the problem. If you're fine with one of the answers given, accept it, so that the community can know the question has been solved Commented Dec 9, 2022 at 16:52

To make the problem consistent without adding an extra wire, you can add a small conductive sphere to the end of the finite wire. You are then dealing with the charge of a conductor $$Q(t)$$ and $$I=dQ/dt$$

If this charge is slow, then you can obtain the magnetic field by applying Biot and Savart's law. The integral is the same integral as for the infinite wire but with a finite upper bound. In cylindrical coordinates $$(r,z)$$ with the origin on the upper end of the wire the magnetic field is $$\vec{B}\left(r,z\right)=\left(\mu_0I/4\pi r\right)\left(1-z/\sqrt{\left(r^2+z^2\right)}\right)\vec{e_\theta}$$

In $$z=0$$, it is half the field of the infinite wire, as expected.

Note that, maybe surprisingly, you can also get this magnetic field from the integral form of Maxwell Ampere's equation in vacuum. The electric field is simply that of a point charge and its flux is easy to calculate.

Applying the Biot-Savart law to the wire you have shown, assuming that it is carrying a steady current, does indeed give the circular magnetic field line that you have shown.

But if there is to be a steady current, the wire must be part of a circuit, the rest of which you haven't shown. The rest of the circuit will also give rise to a magnetic field in the region of your green circle, and the resultant field in that region will NOT correspond to a circular field line. As pointed out in another answer, according to Ampère's law the line integral of the magnetic field around your green circle will be zero, because it does not bound any surface pierced by a net current.

• That's exactly what is confusing me. The Biot Savart Law gives magnetic field at the green loop. So other wires(of circuit) also produce magnetic field, but if we are to just show the magnetic field by the shown 'red wire' only, is the green loop of field correct? Commented Dec 9, 2022 at 15:46
• Yes it IS correct, but of little use, and likely to be misleading, because you do have to have the rest of the circuit in order to have a steady current! Commented Dec 9, 2022 at 15:48
• Thanks a lot. I do understand the consideration of rest of the circuit, I just wanted to clarify my understanding of Biot Savart Law which implies magnetic fields at green loop, whereas Ampere's Loop Law says that there would be no magnetic field. Commented Dec 9, 2022 at 16:02
• You don't have to include the rest of the circuit to have a steady current. In real life it would be extremely hard to do, but the math can find the solution to a finite wire with a steady current ( which would require a very strange polarised charge distribution, but still mathematically possible) Commented Dec 9, 2022 at 16:04
• Amperes law can be used to say there is or isn't a magnetic field, depending on the surface chosen. However is only valid for divergenceless distributions. Biot savart does not satisfy amperes law for a finite wire Commented Dec 9, 2022 at 16:06

Note that:

$$\vec\nabla\cdot \vec j + \frac{\partial \rho}{\partial t} = 0$$

There is a serious divergence of current at the top of your wire, hence charge is piling up there. The whole thing is a growing electric dipole moment which is sending a changing electric flux through the loop up top, since in current free regions:

$$\vec \nabla \times \vec B = \frac 1 {c^2} \frac{\partial \vec E}{\partial t}$$

• Pardon my ignorance, but how and where is charge pilling up? I mean, the wire I have shown has current flowing through it, the top point of wire isn't meant to be its 'end'. Maybe imagine that the wire bends inside the screen at its top point. Now, will the shown(red) part of wire produce the green loop field? Commented Dec 9, 2022 at 15:04
• @RohitShekhawat You should edit that into your question, because it's not clear at all from context. Commented Dec 9, 2022 at 15:26
• Thanks @J.Murray I did it. Commented Dec 9, 2022 at 15:49