I came across a problem that I cannot get my head around.

Consider two very small spherical metallic balls given charges $+Q$ and $-Q$. Assume that both can be approximated as point charges. Now, they are connected by a straight, finite, conducting wire. A current will flow in the wire until the charges on both balls become zero. Consider a point P on the perpendicular bisector of the wire, at a distance $r$ from the wire. My goal is to find the magnetic field at point P, when the current in the wire is $i$. The following figure illustrates the mentioned situation.

enter image description here

I will now use the Ampère-Maxwell equation to obtain an expression for the field.

I have constructed a circular loop of radius $r$ around the wire, to use the Ampère-Maxwell Law. Firstly, one must notice that the two charges produce an electric field everywhere in space. And since the balls are getting discharged, the electric field is actually changing. I have calculated the electric flux through the surface when the charges on the balls are $+q$ and $-q$ below.

enter image description here

Now, for the final substitution...

enter image description here

So I have obtained a neat result after all! But, I realized there was a problem.

Let me use the Biot-Savart Law to find the magnetic field created only due to the current in the wire. This is a relatively easier calculation since the formula for the field due a finite current carrying straight wire is already known.

enter image description here

The answer turns out to be the same.

First of all, is the answer correct? If not, where did I go wrong?

This is what I cannot understand. The Biot-Savart Law gives you the magnetic field created merely due to the current flowing in a conducting wire. On the other hand, the Ampère-Maxwell Law gives you the net field due to the current carrying wire and due to the induced magnetic field (caused by the changing electric field).

So how is it that I get the same answer in both cases? The Biot-Savart Law cannot account for induced fields, right?

Why does there seem to be an inconsistency in the two laws? Have I missed something, or used a formula where it is not applicable?

  • 2
    $\begingroup$ This looks quite similar to physics.stackexchange.com/q/267997 $\endgroup$
    – jim
    Jul 14, 2016 at 13:31
  • $\begingroup$ This is quite like : consider a uniformly charged sphere. Find the field at a distance r less than its rafdius using integration or gauss law. Both yield the same result. Using gauss law you may think that it is only due to the charge inside the sohere of radius r, while using integration, you find the field at r due to all charges. They are consistent, because gauss law actually incorporates the field due to all chsrges in space as well. Same here. $\endgroup$
    – Lelouch
    Jul 14, 2016 at 13:39
  • $\begingroup$ How does Biot Savart law account for INDUCED magnetic fields? After all, we are using it only on the current flowing in a wire. $\endgroup$
    – Newton
    Jul 14, 2016 at 13:41
  • 1
    $\begingroup$ The following may be of interest: Griffiths, Introduction to Electrodynamics (Third Edition), Prob 7.55: If the charge density is a linear function of time, $\rho({\bf r}, t) = \rho({\bf r}, 0) + \dot{\rho}({\bf r}, 0) t$ (then the current density ${\bf J}({\bf r})$ is constant but the charge density is not, conditions that might prevail during the charging of a capacitor. This is not an electrostatic or magnetostatic problem but Coulomb's Law (in the form ${\bf E}({\bf r})= \frac{1}{4 \pi \epsilon_0} \int \rho({\bf r'}) \hat{{\bf r}}/r^2 d \tau$) and the Biot Savart Law hold. $\endgroup$
    – jim
    Jul 18, 2016 at 13:04
  • $\begingroup$ In computing the electric flux, as far as I understood, you have used implicitly the Coulomb law, which is also applicable only in the static situations. the time-varying magnetic field also induces an electric field which is not accounted for when you just compute the solid angles. Maybe both answers are wrong but still it is amusing to think why they reach the same answer. $\endgroup$
    – Seyed
    Jul 21, 2016 at 13:26

10 Answers 10


The short answer is that the case of a finite wire violates one of the postulates of magnetostatics, which is that $\nabla \cdot \vec j = 0$. In cases when the current has sinks, the Biot-Savart law is not equivalent to the Ampere law, but to a "magnetostationary" Maxwell-Ampére law.

Hence, in this rather special case you get the same result from the Biot-Savart as well as the Maxwell-Ampére law. It is not by coincidence.

Sketch of a proof follows, for gory details see Griffiths' Introduction to electrodynamics. (Some details are on wikipedia)

The Biot-Savart law can be equivalently written as $$ \vec B(\vec{r}) = \frac{\mu}{4 \pi} \nabla \times \int \frac{\vec{j}(\vec{r}')}{|\vec{r}-\vec{r}'|} d^3 r' $$ It is important to remember that the curl acts only on the unprimed r. We can now take a curl of the above equation, use the curl curl formula, realize that the laplacian of $1/r$ is proportional to the delta function, and use a few other tricks to obtain $$\nabla \times \vec{B} =\frac{\mu}{4 \pi} \nabla( \int \frac{ \nabla' \cdot \vec{j}}{|\vec{r}-\vec{r}'|} d^3 r' ) + \mu \vec j$$ Where the primed nabla acts on the primed r. When the current is divergenceless, we then simply obtain the Ampere law in differential form.

However, if the current has a nonzero divergence and satisfies charge conservation we have $\nabla \cdot \vec{j}= -\partial_t \rho$. If we then assume that $\partial_t B=0$ ("magnetostationarity") the Gauss' law and other laws of electrostatics are unchanged. Then, we can use the electrostatic solution of the electric potential $\phi$ to easily derive that $$ \frac{\mu}{4 \pi} \nabla( \int \frac{-\partial_t \rho}{|\vec{r}-\vec{r}'|} d^3 r )=-\mu \epsilon \nabla( \partial_t \phi) $$ But we can also switch the order of derivatives and use $\vec E = -\nabla \phi$ to finally see that in the special "magnetostationary" case with conservation of charge the Biot-Savart law will be equivalent to the full Maxwell-Ampére law $$\nabla \times \vec{B} = \mu (\vec j + \epsilon \partial_t \vec E )$$

Note that the practical occurence of cases where the Maxwell contribution to Ampére law is non-negligible while the Faraday induction is negligible is very little. One should thus understand the "magnetostationary" validity of the Biot-Savart law as more of a curiosity.

For instance in your case the charges would have to be very large and the conductor between them a very bad conductor. As the two spheres become connected, an electromagnetic wave emerges. Only as the wave leaves the system and the stationary current is well established we can use the Biot-Savart law.

  • $\begingroup$ "practical occurence of cases where the Maxwell contribution to Ampére law is non-negligible while the Faraday induction is negligible is very little." On the contrary, I think such occurences are quite common - explanation of operation of most regular electronic components under low frequency currents does not require the Faraday effect taking place. Only components with high inductance (coils) require it for explanation of their operation - but such components are not as common in most low frequency circuits. $\endgroup$ Mar 15, 2017 at 21:36

Yes, the result is correct provided the discharge is slow enough so that the electric field in the space near the point of interest and around the wire is accurately given by the Coulomb formula:

$$ \mathbf E_C(\mathbf x)=\frac{1}{4\pi \epsilon_0}\int \rho(\mathbf x') \frac{\mathbf x-\mathbf x' }{|\mathbf x-\mathbf x'|^3} d^3\mathbf x'. $$

This seems to be a very common situation - electric field near low-frequency AC electric circuits made of resistive and capacitive elements should be almost Coulombic(plausible, since the concept of potential is used successfully to analyze their operation$^1$).

In such situations, the Biot-Savart formula is applicable and gives accurate result. Here is the reason why.

The exact Maxwell-Ampere equation is

$$ \nabla \times \mathbf B = \mu_0 \mathbf j +\mu_0 \epsilon_0 \partial_t \mathbf E.~~~(*) $$

Let us denote the field given by the Biot-Savart formula as $$ \mathbf B_{BS}(\mathbf x) = \int \frac{\mu_0}{4\pi} \frac{\mathbf j(\mathbf x') \times (\mathbf x-\mathbf x')}{|\mathbf x-\mathbf x'|^3}\,\mathrm d^3\mathbf x'. $$

We will show that this field solves the exact equation (*) if electric field is given by the Coulomb formula.

Curl of this field can be expressed as

$$ \nabla \times \mathbf B_{BS}(\mathbf x) = \frac{\mu_0}{4\pi} \int \nabla_{\mathbf x} \!\times \big(\mathbf j(\mathbf x') \times \mathbf F (\mathbf x,\mathbf x')\big) d^3\mathbf x' $$ where $$ \mathbf F (\mathbf x,\mathbf x') = \frac{\mathbf x-\mathbf x' }{|\mathbf x-\mathbf x'|^3}. $$ The integral can be transformed into

$$ \int \mathbf j 4\pi \delta(\mathbf x'-\mathbf x) d^3\mathbf x' + \int (\nabla'\!\cdot \mathbf j) \mathbf F d^3\mathbf x' - \int\nabla' \cdot (\mathbf j \mathbf F) d^3\mathbf x'. $$

Let us choose the integration region in such a way that $\mathbf j = \mathbf 0$ everywhere on its boundary (moving charges are not crossing the boundary). Then the third integral is zero (this can be shown using the Gauss theorem). The divergence in the second integral can be expressed as $-\partial_t\rho$ based on the equation of continuity. Going back to curl of $\mathbf B_{BS}$, we obtain

$$ \nabla \times \mathbf B_{BS}(\mathbf x) = \mu_0 \mathbf j(\mathbf x) + \epsilon_0\mu_0\partial_t \mathbf E_C. $$

Now we can see that $\mathbf B_{BS}$ solves the exact Maxwell-Ampere equation (*) as long as the electric field is given by the Coulomb formula.

So, the Biot-Savart formula is not limited to static currents as people sometimes assume, but applies to changing currents as well. It is therefore more general than the integral Ampere formula.

$^1$ Except for phenomena of electromagnetic induction - near coils and moving magnets the electric field is rotational which cannot be described by the Coulomb formula.

  • $\begingroup$ In your second last expression, you have written jF together. What does that mean? They are both vectors. And how do you define the divergence of jF? $\endgroup$ Mar 14, 2019 at 6:21
  • $\begingroup$ @YamanSanghavi dyadic product, the result is a tensor (3x3). Divergence is defined this way: $\nabla \cdot (\mathbf j \mathbf F)= (\nabla \cdot \mathbf j) \mathbf F$. $\endgroup$ Mar 14, 2019 at 18:30

Implicitly, what you are doing in this problem is taking the limit $\epsilon_0\rightarrow 0$ in an electrodynamics problem.

Let $q_+(t)$ and $q_-(t)$ be the two charges of the spheres at time $t$: $q_+(0)=Q$, $q_-(0)=-Q$. Since the velocity of light is (large but) finite, it will take some finite time for charge to move through the wire, so $|q_-(t)|-|q_+(t)|\geq 0$, and the difference is in the wire as a charge density $\rho(t,x)$. The current density in the wire is then $I(x)=v(x)\rho(x)$, where $v(x)$ is the velocity of the charges in the wire.

The version of Ampere-Maxwell that you have to use is w.r.t. $q_+, q_-,\rho$ and the electric fields and currents these two induce. Now, we take the limit $\epsilon_0\rightarrow 0$. What happens to the quantities in the problem? First off, at $\epsilon_0=0$, the Coulomb force is infinite, so $v(x)\rightarrow \infty$, and similarly, because the charge moves through the wire infinitely fast we get $|q_-(t)|-|q_+(t)|= 0$ and,therefore, $\rho(x)=0$. Now, the current density $I(x)$ should go to $dq/dt$ so that the continuity equation is satisfied. This explains that you use the correct current density.

The induced electric field is equal to the field by the charges $q_-,q_+$, that you use, plus radiative effects. These radiative effects (See the Lienard-Wiechert potentials for an example) vanish in the limit $\epsilon_0\rightarrow 0$, which takes $c\rightarrow \infty$. So for the electric field you only have to use the static component, like you did. This static component is infinite, but the infinity cancels: the $\epsilon_0/\epsilon_0$ that you crossed out takes care of that.

For full mathematical rigor, you have to be a bit careful about how you take all these limits, but I promise you it checks out.

Equivalently, if you look up the the Lienard-Wiechert potentials on wikipedia, you find the relativistic generalisation to the Biot-Savart law. If you take $\epsilon_0\rightarrow 0$ in this expression, you recover the Biot-Savart law. This proves, in an admittedly roundabout manner, that both approaches are equivalent.

If you read between lines of the above argument, what is hidden there is a proof that the Biot-Savart law only neglects the radiating fields. Variations in the static Coulomb part of the electric field are correctly taken into account by Biot-Savart, which is why the two methods agree.

  • $\begingroup$ I am downvoting the answer because I think when we talk about wires, we are considering 1-D conductors. A 1-D conductor is incapable of having a non-zero charge density, because the potential energy in the electric field will blow to infinity. Therefore, I think OP's question has nothing to do with the charge density of the wire. Also, note that this has nothing to do with the finite speed of electrons in the wire; it just means the velocity is constant throughout the wire at a certain time. $\endgroup$
    – Ali
    Jan 11, 2017 at 5:54

It's very interesting indeed. Normally Biot-Savart is derived under conditions of static charge density which is clearly not the case here. It would seem it is also possible to derive only assuming that current density is static and electric field (/ charge density) is allowed to grow linearly in time.

If you examine the derivation of Biot-Savart currently on wikipedia, at one point they invoke that current is divergenceless. If they did not do this, they could have used a weaker assumption and would have obtained the full Ampere-Maxwell law.

Another way to see it, in general we have Jefimenko's equations however if you examine them carefully, you will see that when current is static, Biot-Savart law is recovered exactly.

The key assumption is that in your problem the current should be constant in time. In reality, two charged metal spheres linked by a wire will discharge with a non-constant current and some radiation will occur.

  • $\begingroup$ If we neglect the radiation, does the spheres discharge with a constant current? The spheres are discharging; therefore the force of each one on the other sphere's charge will decrease with time, Why the current do not decay for example exponentially (similar to the case of a capacitor with $\sigma$ )? $\endgroup$
    – SMA.D
    Jul 14, 2016 at 17:22
  • $\begingroup$ Yes, some of the initial energy would be lost as heat as well (due to the finite resistivity of the wire). Also in practice the wire itself would charge up slightly, and the current would be non-uniform in space as well. $\endgroup$
    – Nanite
    Jul 15, 2016 at 8:35
  • 1
    $\begingroup$ However in the limit of a sufficiently resistive wire, the discharging time could be made long enough that OP's solution would apply. $\endgroup$
    – Nanite
    Jul 15, 2016 at 8:38
  • $\begingroup$ @Nanite this is a comment, can you expand it to an answer? $\endgroup$
    – rmhleo
    Jul 21, 2016 at 12:21

I believe your answer is correct. The BS law is applicable, provided the time constants are long compared with the time light takes to cross the set-up.

It is possible to include the displacement current (DC) in the BS calculation, but there is a general result that the total field calculated using BS for the DC is always zero! Tghere is a clear discussion of this in Berkely Physics Course, volume 2, pp256-263 (in my edition).

Something analagous happens when BS is used to calculate the field from the current flowing through the wire using a voltage or current source. In this case, the BS calculation must include the current flowing through the whole wire - including connections to the battery.



As others have said, the Biot-Savart method does not apply directly to this problem because you do not have constant charge densities. Since the charge density is not constant, you have a changing electric field which sources your magnetic field.

Since your electric field is changing, your magnetic field comes from two sources: current and changing electric field. The ampere's law method takes both of these into account, but the Biot-Savart law only takes the current source into account. As you say, it is surprising that the Biot-Savart law gives you the correct answer even though it shouldn't apply. I will give a simple argument as to why it gives the correct answer.


My strategy will be to construct a magnetostatics problem which has the same answer, and apply both methods to this new problem and get the same answer. (The key idea being that we are now applying the Biot-Savart law to a magnetostatics problem so that we can trust its answer.) Along the way, we will see that we carry out steps that are analogous to your steps in the original problem, and we will see why in the original problem only the current affects the magnetic and the changing charge density doesn't matter.

The idea behind the new problem we will construct is that changing electric fields source the magnetic field the same way as currents, so if we just replace all changing electric fields in the original problem with currents, we get the same magnetic field. However, the resulting problem is now a magnetostatics problem.

Let's apply this idea to your specific example. We add a converging $1/r^2$ spherically symmetric current density coming in from infinity converging to the $+q$ charge, and a outward point $1/r^2$ current density going out to infinity centered on the $-q$ charge. With this currents in place, the charge density is now constant, and since constant charge densities don't produce a magnetic field, we can safely ignore the charges.

Now that we have created our magnetostatics problem, we can apply your two methods. For the ampere's law method, we lose the term from the electric flux, but since we added electric currents to replace our changing electric fields, we pick up an extra contribution to the amount of electric current passing through our amperian loop, and by design this new contribution exactly replaces the contribution that had been made by the changing electric flux. This confirms that this new problem does indeed have the same answer as the old problem.

Now let's apply the Biot-Savart law. With this new problem, we still get a contribution from the line segment, but there may be contributions from the two converging/diverging spherical current densities. We must show these contributions are zero. Let's concentrate on the converging current density. The resulting magnetic field cannot have a radial component, because the field must be divergenceless. Also, the field at a point $\vec{r}$ relative to the center cannot have a component perpendicular $\vec{r}$ because it must be invariant under rotations about $\vec{r}$. Therefore the magnetic field contribution must be zero as expected. Consequently, in the original problem it was actually sufficient just to consider the current from the line segment when applying the Biot-Savart law.

Range of applicability

Notice this wouldn't work if your charges weren't discharging at a constant rate. My argument relied on the changes in electric field being nice and spherical. This works when the charges are changing linearly in time. However, if your current $i$ is changing with time, then there will be an induced electric field, and this field will change with time. This solenoidal field will lack the spherical symmetry we previously had, so the solenoidal field will create its own magnetic field. So only in this very special non-magnetostatic problem can you use the Biot-Savart law.


In magnetostatics the BS law and Ampere's law are equivalent as long as the current density $\mathbf{J}$ is sufficiently bounded at infinity. Using the integral form of the BS law, we have $$\mathbf{B}(\mathbf{r}) = \dfrac{\mu_0}{4\pi}\int \dfrac{\mathbf{J}(\mathbf{r}') \times \left(\mathbf{r}-\mathbf{r}' \right)}{\left|\mathbf{r}-\mathbf{r}' \right|^3}\text{d}\tau \iff \nabla\times\mathbf{B} = \mu_0 \mathbf{J}.$$

To generalize this to include the displacement current (and restore consistency with charge continuity) it suffices to replace $\mathbf{J}$ with $\mathbf{J} + \epsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$, provided that the displacement current density satisfies the same boundedness conditions as $\mathbf{J}$ at infinity, thus: $$\mathbf{B}(\mathbf{r}) = \dfrac{\mu_0}{4\pi}\int \dfrac{\left(\mathbf{J}(\mathbf{r}') + \epsilon_0\dfrac{\partial \mathbf{E}(\mathbf{r}')}{\partial t}\right) \times \left(\mathbf{r}-\mathbf{r}' \right)}{\left|\mathbf{r}-\mathbf{r}' \right|^3}\text{d}\tau \iff \nabla\times\mathbf{B} = \mu_0 \mathbf{J}+\mu_0 \epsilon_0\dfrac{\partial \mathbf{E}}{\partial t}.$$

Consider the new term in the extended BS law for your problem. Since $\mathbf{E}$ is produced by discrete point charges, you can consider the contribution from each point charge separately. It's not hard to show by symmetry that the integral from a single point charge is zero, so it must be for a collection of point charges as well. Hence the unmodified and modified BS laws give the same result.

More generally, the modified and unmodified BS laws will give the same result if and only if $$\dfrac{\partial}{\partial t}\int \dfrac{ \mathbf{E}(\mathbf{r}') \times \left(\mathbf{r}-\mathbf{r}' \right)}{\left|\mathbf{r}-\mathbf{r}' \right|^3}\text{d}\tau = 0.$$

This will always be true for $\mathbf{E}$ generated by a collection of point charges, and I think it will be true for any irrotational $\mathbf{E}$, i.e. any electrostatic displacement current, but I'm not sure of that.

  • $\begingroup$ @Emilio Pisanty: I don't want to start a religious war, and thanks for editing for me, but the "naught" after $\mu$ should be a circle ($\mu_\circ$), not a zero ($\mu_0$)! ;-) $\endgroup$
    – pwf
    Mar 15, 2017 at 23:43
  • $\begingroup$ Not at all, but if you insist on using incorrect notation you can roll it back (in which case, actual references to that choice would take your case up from "I misread some notation once" to "other people are wrong too"). There is a long tradition of 'default' values (i.e. in vacuum as opposed to in a material) labelled with a subscripted zero; there is no such equivalent for a subscripted circle. But again, if you insist you can roll it back. $\endgroup$ Mar 16, 2017 at 7:43
  • $\begingroup$ "It's not hard to show by symmetry that the integral from a single point charge is zero" The integral is zero when the electric field is potential field. Whether charge distribution is made of point charges or not does not matter. $\endgroup$ Mar 17, 2017 at 10:49

I think your both approaches are correct. Regarding your doubt that why biot savart law is yielding the same result. To understand this consider the situation as a superposition of two cases. First one the current carrying wire and the second one the presence of charges. Apply Biot Savart law to the two cases independently. Since static charges don't contribute anything to the field so what we are left with is a simple current carrying wire. The rest follows from your question.


Reduced to its simplest terms, isn't the difference between Ampere and Biot-Savart this? Imagine a point charge q moving with velocity v. This charge will generate a magnetic field at an arbitrary point in space, which can be on the perimeter of an Ampereian closed surface through which the charge passes. Ampere's law predicts that the field will be an instantaneous blip which occurs whenever the charged particle passes through the surface, which is obviously physically absurd, as the surface can be distorted so that the blip can be made to occur wherever we like along the particle's trajectory. (Alternatively, and more realistically, you can say that Ampere makes no prediction as the current is not a steady one.) However, if we model the charge density as a delta function and insert into the Biot-Savart law, we obtain an expression for B which agrees with that obtained using the Maxwell term. Hence it's hard to agree with the commonly-quoted statement that Biot-Savart and Ampere share the same region of validity.


The biot savart law is valid only for steady currents. In this case, there is no circuit, so the flow of charges is instantaneous(almost). So the current is more like a dirac delta function. Biot savart law is inapplicable in such cases and you must resort to maxwells equation in these situations.

  • $\begingroup$ So you're saying that my answer is correct because I derived it using the Maxwell equation, but wrong at the same time because it could also be obtained through Biot Savart Law? $\endgroup$
    – Newton
    Jul 14, 2016 at 14:02
  • $\begingroup$ Please explain further. $\endgroup$
    – Newton
    Jul 14, 2016 at 14:02
  • $\begingroup$ Or was my answer wrong in the first place? $\endgroup$
    – Newton
    Jul 14, 2016 at 14:03
  • 1
    $\begingroup$ An instantaneous flow of chargs do not constitute a steady current. Biot savart law is inapplicable. Whether you get the same answer is not important, when the law itself is invalid. $\endgroup$
    – Lelouch
    Jul 14, 2016 at 15:15
  • 4
    $\begingroup$ If the wire has resistance in it, there is no need to call the current flow instantaneous. It can be as slow as you want (in the limit where resistance is very large, the time constant also becomes very large, and the result should tend to the steady state solution); at that point there is no reason why B-S would not be applicable. $\endgroup$
    – Floris
    Jul 14, 2016 at 15:37

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