# Magnetic field outside a solenoid

I couldn't picture the idea behind the zero magnetic flux outside the solenoid properly. My teacher explained it as

"Components of the magnetic field in other directions are cancelled by opposing fields from neighbouring coils. Outside the solenoid the field is also very weak due to this cancellation effect and for a solenoid which is long in comparison to its diameter, the field is very close to zero."

But I didn't understand anything at all. First he said about the components of magnetic field in other direction being cancelled by magnetic field, but didnt further explain and said nothing to support his argument. If somebody has some simple, intuitive way of understanding this phenomenon then please explain it.

I tried to address this question in a short article that recently got published in the European Journal of Physics. https://arxiv.org/abs/1610.07876

You can also look at this article by Farley, Price, which uses a nice argument and physically motivates the problem (American Journal of Physics 69, 751 (2001); doi 10.1119/1.1362694): https://doi.org/10.1119/1.1362694

Also, there is an exact calculation done in this old NASA Technical Note D-465: http://paginas.fe.up.pt/~ee08173/wp-content/uploads/2014/03/finite-solenoid.pdf

• This is the correct answer. It is really frustrating that nobody addresses the complete lack of rigor in all other answers of the type "the field on the outside is zero because the solenoid is infinite and magnetic lines reach zero density, as their flux has to be conserved". Commented Jul 24, 2017 at 21:57
• Could you consider writing a short explanation of all of this in your answer. Referencing own work is not against stackexchange rules, but the goal is to have somewhat of a self-contained answer before the references. Commented Jul 24, 2017 at 21:59
• @Krastanov: I'm not sure I can write down the arguments of either my paper or Richard Price's paper in a very concise way. I did try to write my argument on Quora once, and it became quite long. Commented Jul 25, 2017 at 0:07
• The answers of the type you mention are really problematic. It leads to another fundamental thing ; that the correlation of density of field lines with field strength is due to the divergence of the field being 0. It still shocks me that so many people haven't been taught this/haven't thought about it very much/stubbornly refuse to accept it once told. There is this answer that addresses it on another thread: physics.stackexchange.com/a/267589/78842 , which is correct, and then my answer in the same thread. But many answers there and elsewhere are not very good. – Commented Jul 25, 2017 at 1:20
• @Yakk, sure, you are right that this should be expanded upon, but instead of whining, just do the work: vote to convert this to a community wiki answer and expand upon it. This is still the only answer that rigorously addresses the question. This is supposed to be a community resource, not something where you just demand other people's work. Arxiv and AIP probably will exist long after scackexchange is dead, but the last link is indeed sketchy: it is NASA Technical Note d-465 that can also be found here ntrs.nasa.gov/api/citations/19980228117/downloads/… Commented Aug 10, 2021 at 20:52

You first should understand that magnetic field are the circles (of magnetic force) around the wire with current. The direction of circles (counterwise or clockwise) depends on the direction of the currend (forth or back). Now, you can guess what happens to the magnetic field when you curve the wire into a circle. Then, you may also stack cricles of current next to each other to form a cylinder. Now, try to imagine what will be the magnetic field in each of these cases.

The magnetic fields from different currents add up (total field in every point is the sum of fields from different currents at that point).

I agree that symmetry, which tells that cancel is complete, is not obvious here. Because outside the loop, one wire is closer is to the outer region than the wire from the opposite side, which cancels out the first. But, this should give a qualitative picture, what is happening. The complexity of symmetry might be the same as zero gravity inside a massive sphere.

edit I am coming to conclusion that it is impossible to cancel out the outer field completely just because the field approaches infinity as we come closer to the wire from outside and there is nothing close to infinity from the opposite piece of the loop that can cancel this out. Solenoid just cancels out componets of the field perpendicular to its axis and parallel to the plain of the windings since windings are very close and, though current in them flows in one direction, say forward, upper wire induces left magnetic field between the wires whereas bottom wire produces the same right field. So, all the winds that blow in the planes between windings are eliminated. The axial fields inside the solenoid are added together. Axial outer are cancelled to some extent but not completely. But, this is not bad because, though they create the axial field, opposite to what solenodi should produce from its internal part, the irregularities at the edge of solinoid wrap it around and eliminate the negative effect.

See that though the major field is produced from left to right, some field blows from right from right to left outside. Yet, this does not harm the major flow because edges isolate the negative flow.

• Please check physics.stackexchange.com/a/292696/10134 as it gives a more rigorous answer addressing the issues you have mentioned. Commented Jul 24, 2017 at 21:58
• "magnetic field are the circles (of magnetic force) around the wire with current." No. Commented Sep 28, 2021 at 17:49
• The thing is that it is possible to see that the z-component gets neutralized as the coil appoaches infinity. It is laid out very nicely in Aritro Pathak's article that he linked (see above). It's in figure 2 where you can see that windings beyond a certain break-even distance from the point of interest contribute negatively. In the following calculation he then shows that the resulting z-component approaches 0 as the coil length approaches infinity. Commented Aug 7, 2022 at 21:52

What goes through the middle must go out one end and back in the other. The magnetic field on the outside is free to spread out and in doing so the Wb/m^2 goes low, but is not zero. You must distinguish between assumptions to prove a theory and real life examples. An infinitely long solenoid (for the theory) would have infinite cross-sectional area outside the solenoid (for the flux return path) so the Wb/m^2 would reduce to zero in math terms. Magnetic fields lines from each coil of solenoid will add, there's no cancelling going on here, or anywhere (just the math with infinity term as denominator means whole term becomes zero.)

Imagine a loop of current carrying wire and consider two small elements dl which are diametrically opposite.Try to find the magnetic field at a point outside this loop due to these two small elements which lies on a line which passes through both these small elements using Biot Savarts Law.When you try to find the direction of the magnetic field at this point using right hand rule you will find that they are in opposite directions and hevce cancel.Even after this cancellation there will be a small field present because the magnetic field due to the element which lies closer to the point will be greater than from the element which is further away.The distance between them is the diameter of the loop. Now imagine that this loop is part of solenoid which is infinitely long and try to find the contribution to the magnetic field to due the same two elements at a point outside the solenoid but located an infinite distance away along the length of the solenoid.Since it is an infinite distance away the angle between the elements and the position vector can be cosidered to be the same.Also the distances of the elements from the point can now be considered equal since the diameter is negligible in comparison and hence the field due to both can be considered be of equal magnitude and cancel completely giving rise to zero magnetic field. Hope this helps.

The magnetic field outside an infinitely long AC solenoid is NOT zero, as shown by explicit calculation from Maxwell's equations by Abbot & Griffiths, American Journal of Physics 53, 1203 (1985), and also independently by Jacque Templin, American Journal of Physics 63, 916 (1995).

Every demonstration I've seen that shows the magnetic field outside an infinitely long DC solenoid is zero uses the approximation that the current of the solenoid is everywhere perpendicular to its axis, which, of course, is not true for any solenoid whose current is carried by a wire of non-vanishing dimensions. If the diameter of the wire is D and the circumference of the solenoid is C then there is a component of the current parallel to the axis of the solenoid, equal to (D/C)*I where I is the current in the wire. That component of the current makes a magnetic field outside the solenoid and though it may be a very weak field compared to that inside the solenoid, it doesn't vanish when the solenoid is infinitely long.

A long solenoid can be approximated to an infinite solenoid, if the ratio between its radius and length is very small. The magnetic field due to an infinite solenoid, on it's axis is one of the very few situations in which the Biot-Savart Law is useful. Consider the field at a distance $$z$$ away from the centre of a ring current, of radius $$R$$. It isn't much work to show that the field due to this ring is $$B(z)=\frac{\mu_{0}IR^2}{2(R^2+z^2)^{3/2}}$$ If a solenoid has n turns per unit length, the current through an element of thickness $$dz$$ would be $$In(dz)$$

Therefore, $$\int_{-\infty}^{+\infty} dB(z)=\int_{-\infty}^{+\infty}\frac{\mu_{0}nIR^2dz}{2(R^2+z^2)^{3/2}}$$

Integrating this gives, $$B(z)=\mu_{0} n I$$

Ampere Law reads, $$\oint\vec B\cdot d\vec l=\mu_{0}I_{enc}$$ Using symmetry arguments, one can eliminate the radial and tangential components of the fields (along $$\hat r$$ and $$\hat \phi$$). Considering an amperean square loop of one length $$l$$ , one side of which coincides with z axis, and the other outside the solenoid. Using amperes Law,

$$B_{in} - B_{out} = \mu_{0} n I$$ but since $$B_{in}=\mu_{0} n I$$ we can conclude that $$B_{out}$$ is 0. Very simple.

This result holds for only infinite case, and analysis on non infinite solenoid is more involved.

If for some reason one doesn't like this analysis, one can also do an analysis on a toroid, for which, one can systematically show that the field goes to 0 outside the toroid. Then, one can take the infinite radius limiting case of a toriod, which is nothing but a solenoid. Again, this is the infinite case.

Magnetic field outside a long solenoid can never be zero. Because a solenoid of finite length will have edges and field will come out. However, if we imagine a straigth solenoid of infonite length or a toroid, it will not have edges from where lines of force can come out. therefore field would be zero outside a toroid or a strigth solenoid of infilite length.

First, I assume that the solenoid is infinitely long. If you choose a Polar coordinate with $z$ being along the solenoid's axis, based on symmetry you can argue that $\vec{B} = B(r) \hat{z}$. Also, at infinity the magnetic field should be zero.

Now we use Ampre's law for the following path: assume a rectangular path with one side parallel to the $z$ axis out side of the solenoid $C_1$, one side parallel to the axis again but at infinity $C_3$, and the other two sides of this rectangle connecting these two lines together $C_2 \& C_4$. Now if you write the Ampre's law, you see that the $\int \vec{B}.ds$ is zero along $C_3$, since $B$ at infinity is zero. $\int \vec{B}.ds$ is also zero along $C_2,C_4$ because the magnetic field is perpendicular to this path. So $\int \vec{B}.ds$ should also be zero along $C_1$, since there is no current passing through this closed path. So $\vec{B}(r) L = 0$, so $\vec{B}(r) = 0$.

For an infinite solenoid, Ampère's theorem makes it possible to show very simply that the magnetic is uniform outside. If the magnetic field is zero to infinity, it is zero everywhere outside.

One can attempt a dimensional analysis to show that the field is zero at infinity. For a solenoid of radius $$R$$ , if the surface current density is $${{j}_{s}}$$, dimensional analysis shows that the magnetic field will be of the form $${{B}_{z}}={{\mu }_{0}}{{j}_{s}}f(r/R)$$

If, the radius of the solenoid is reduced to $$0$$ the currents cancels each over and the field must tend to $$0$$ : $$\underset{R\to 0}{\mathop{\lim }}\,f(r/R)=0$$ or $$\underset{u\to \infty }{\mathop{\lim }}\,f(u)=0$$

So $$\underset{r\to \infty }{\mathop{\lim }}\,f(r/R)=0$$ the field tends to 0 at infinity.

Rather formal proof ! Among my friends, some find it convincing, others do not!

• "Ampère's theorem makes it possible to show very simply that the magnetic is uniform outside." How can I do so? Commented Oct 29, 2019 at 10:47

The current flow on one side of a solenoid is in one direction. On the other side it is in the opposite direction. Inside the solenoid the contributions to the magnetic field are both in the same direction, but on the outside, the contributions tend to cancel out (more so if the length is large compared to the diameter).