# How to find the magnetic field outside an infinite solenoid without Biot-Savart law?

Let's consider an infinite solenoid, with $n$ loops per unit length, radius $a$, which carries a current $i$. Using symmetries and Ampère's law, one can find that the magnetic field is ortho-radial, and that

$$\begin{cases} B_{int} = \mu_0ni + A \\ B_{ext} = A \end{cases}$$

where $A$ is a constant that we must now determine.

Using Biot-Savart law, I find that $A=0$, however I'd like to do it without this law. Is there a way to do so ?

• Ampere's law should show that A = 0 as well. An infinite solenoid has no return field outside of it precisely because it's infinite, i.e., the fields that exist outside for finite solenoids result from the ends of the solenoid. – honeste_vivere Jul 15 '17 at 18:32
• @vs_292 I've already used it as I stated on the question. – Spirine Jul 16 '17 at 9:35
• Please include your derivation. – garyp Jul 16 '17 at 11:57

# No.

There is no way (either with or without the Biot-Savart law!) to argue that $A=0$. In fact $A=0$ is a boundary condition, and it therefore does not have to be true.

# Details.

The Maxwell equations can generally be applied within any big volume $V$ as long as you specifiy what $\vec E$ and $\vec B$ are on the boundary $\partial V$, and the Biot-Savart law is a result of the Maxwell equations.

If you imagine that your solenoid is oriented "up-down" and we put a North pole of a very big (much bigger than our room) round magnet somewhere underneath the floor, and a South pole of a very big round magnet somewhere above the ceiling, then (still talking in the case where V is the infinite 3D space!) we will essentially be adding a field $\vec B = B_0~\hat z$ to the field of the solenoid within the room, without messing with the existing rotational symmetry. Therefore all of your arguments about rotational symmetry still go through, but we find out that $A \ne 0$ in the room itself, rather $A = B_0$. Now if the room is much much larger than the solenoid therein, we see that the field is approximately $B_0~\hat z$ out on the boundary of this room and the Maxwell equations will give us this same solution if we remove the magnets but insist that we're only concerned with the field within this room, subject to this boundary condition that the field is $B_0~\hat z$ out on the boundary.

As we systematically allow increase the size of this room with those boundary conditions, we find a valid limit where the field "out at infinity" is $B_0~\hat z$. In fact the Maxwell equations allow us to fill space with any field which satisfies those equations in vacuum: so there are solutions where there is in fact a homogeneous plane wave of light permeating all of space, oscillating in the $x,y-$directions and travelling in the $z$-direction, too. The Maxwell equations don't say that this is invalid, because they cannot say that this is invalid. All they can say is that these do not have zero fields at the boundary.

However, if we do force the fields to be zero at infinity, the existence and uniqueness theorems tell us that there is just one solution of the Maxwell equations which doesn't have this pathological behavior way out there. Even better, that boundary condition is composable, so that we can take two point charges and their individual fields (assuming that said fields go to 0 at infinity) and construct the field for both charges together, by simply vector-summing the two fields. This is only possible because at the boundary we have that this reduces to $0 + 0 = 0$, so we now have a solution for both point charges when the field goes to 0 out at infinity.

And in fact this prescribes also the best way to approach these problems for large volumes $V$: get a solution which, with no charges inside $V$, enforces the (non-zero) boundary conditions; then superimpose all of the zero-boundary-condition results for all of your point charges and currents and whatnot on this background field to find the "final field" that you want.

• Since we can measure the magnetic field (for example, a moving charge will experience a force $q\,\mathbf{v}\times\mathbf{B}$), why couldn't we find its value theoretically ? – Spirine Jul 18 '17 at 13:40
• We indeed have a great theory which will predict the electromagnetic field inside of a closed volume if you tell it the field's value on the boundary of that volume and the charges $\rho$ and the currents $J$ inside that volume. It has a wonderful existence and uniqueness theorem, "I always predict some field and I only predict one." The reason that the fields on the boundary of that volume are important is precisely what I mentioned, that the theory has a large family of vacuum solutions $\rho=0,J=0$ corresponding to light waves propagating through the vacuum. You have to choose one. – CR Drost Jul 18 '17 at 14:15
• You can either say "Yay, I don't have to specify boundary conditions" but you lose this wonderful uniqueness theorem, or you can gain this wonderful uniqueness theorem but "Boo, I have to specify boundary conditions so that this theory knows which vacuum solution to superimpose on my field." Because every field can be viewed as the zero-on-the-boundary solution for each of the charges and currents plus the no-charge-no-current field which enforces the boundary condition, and that freedom is irreducible. – CR Drost Jul 18 '17 at 14:18
• Last question: then, how can we use this theory to describe reality ? Even though the electromagnetic field is an abstract, mathematical object, its effects are real: charges move in a unique manner. Thus, the electromagnetic field must be definite and unique, and then how do we know what it is except through experiments ? – Spirine Jul 18 '17 at 19:38
• We don't. (That is, we don't know how the electromagnetic field works except through experiments.) The Maxwell Equations are important because they compactly summarize the results of many, many experiments and strongly hint at a couple more (b/c the electromagnetic field is a $[0,2]$-valence tensor in Minkowski spacetime, the Maxwell Eq's led us toward special relativity). They also suggest a couple unphysical things like the Rayleigh-Jeans catastrophe, which have ultimately been solved with finer theories since then. – CR Drost Jul 18 '17 at 20:30

We typically require that the field approaches zero at an infinite distance from the solenoid. This leads immediately to $A=0$.

• And why do you require this ? Using the same argument you could show that a charged plate creates a null electric field everywhere... – Spirine Jul 13 '17 at 9:49
• @Spirine Because in the limit of a very long finite solenoid, the field approaches zero. See aapt.scitation.org/doi/abs/10.1119/1.1362694. – probably_someone Jul 13 '17 at 9:50
• @Spirine Also because that assumption is built into the Biot-Savart law. – probably_someone Jul 13 '17 at 9:50
• I don't have access to the linked pdf, but does the author use the forbidden law ? I'm solely interested in a proof that doesn't require Biot-Savart law – Spirine Jul 13 '17 at 10:00
• @Spirine Doesn't seem like it. Just why are you so intent on avoiding the Biot-Savart law, anyway? – probably_someone Jul 14 '17 at 5:52

Consider a rectangular amperian loop abcd. Along cd the field is zero. Along transverse sections bc and ad, the field component is zero. Thus these two sections make no contribution. Let the field along ab be B. Thus,the relevant length of the amperian loopis, L=H. let n be number of turns per unit length then the total number of turns is nh. The enclosed current is I(nh), where I is the current in the solenoid. From ampere circuital law BL=uI Bh=uI(nh) B=unI (Here u is free space permeability) • I wonder why the field is null along $CD$. This is tricky part of the question. – Spirine Jul 15 '17 at 15:45
• Along CD the magnetic field is not zero but as we have taken infinitely long solenoid the magnetic field along CD will be zero. This can be explained as, the 2 end faces of solenoid acts as North pole and South Pole and as the field lines outside the solenoid is parallel along its length coming out and entering from the 2 faces. The infinite long solenoid suggest that North Pole and South Pole are far apart thus the strength of the field line decreases and we approximately take it zero – MANSI Aug 4 '17 at 15:15

Much of what I write can be read in greater detail in Griffiths. Let solenoid be along z axis. B can only have a z component throughout space. By symmetry you can kill radial component and by using ampere's law you can kill the tangential component. Look at loop 1. Ampere's law applied here easily gives $B(a)=B(b)$. Now, its easy to say why $B(a)= B(b) = 0$. Let b tend to infinity.

If field at infinity is finite, it means, for every finite length of solenoid, there is infinite energy stored in magnetic field, Given by $$\frac{1}{2\mu_0}\int B^2\; dV$$

Comparing energy in whole space due to infinitely long solenoid with this energy is like comparing a plane to a 3D space. They are totally different.

• The issue here is that you're assuming that the field in the volume $[z, z+\Delta z]$ is only caused by the finite length of solenoid $\Delta z$, which is wrong – Spirine Jul 16 '17 at 9:39
• No, I didn't assume that. Of course it isn't so. You seem to have misinterpreted it. The focus is on point that some infinities can be larger than other infinities. – physicophilic Jul 16 '17 at 10:58
• And why should the energy "due to the finite part" of the solenoid be finite? – Spirine Jul 16 '17 at 11:01
• Its physically impossible to produce infinite energy out of finite energy (carried by finite part of solenoid). Obviously. – physicophilic Jul 16 '17 at 11:05
• Now read my first comment ;) – Spirine Jul 16 '17 at 11:24