# Magnetic field at an infinite distance from an infinite solenoid

I will quote Griffith's "Introduction to Electrodynamics (4th edition)":

So the magnetic field of an infinite, closely wound solenoid (...) certainly approaches zero as you go very far away.

What does he mean by "certainly"? He doesn't give any argument for this, even though it's crucial to the final conclusion (that the field outside the solenoid is constantly 0).

I'm not looking for a rigorous mathematical proof, or at least not particularly that. Any intuitive argument would be sufficient. If there is no such argument, then I will try and give meaning to the proof. All in all, any answer is welcome.

• Can you cite the page or section number? – pwf Mar 24 '18 at 0:16
• in my 3rd edition, it is Ex. 5.10 in Sec. 5.3.3 on Applications of Ampere's Law – ostrichCamel Mar 24 '18 at 5:07

One could naively remember that the B field off a long wire goes to zero with distance and intuit that the field off a solenoid should do the same, since they're both one dimensional distributions in that limit.
Maybe more convincingly, you would expect that the field off the solenoid should fall off faster than a line carrying the same current, since each current 'piece' in a loop on the solenoid can be paired with a piece 180 degrees away that produces an oppositely directed B field contribution.

Mentally taking the limit as radius goes to zero (rather than looking a large distance away) might make the current cancellation seem more intuitive.

You're right to notice that this is a shaky step in Griffiths' derivation. One argument that B goes to 0 at infinity might be as follows: First imagine that, instead of a very long solenoid, all the turns are compressed together into a ring of N turns, with N very large. The field from that ring "certainly" falls off with distance as you move away in the plane of the ring (in fact, it falls off like $1/r^3$). So, you can find a distance that makes B arbitrarily small. Now imagine spreading some of the turns out a short distance. Your distance from those spread turns is now greater than your distance from them when they were together in the ring, so their contribution to the magnetic field where you are has decreased. Now continue to spread the turns out until you have a very long solenoid. The more you spread the turns out, the weaker B gets where you are. In other words, even though the solenoid is infinitely long, you get so far away from any part of it that the magnetic field contribution is negligible, and even if you add up the contributions from the whole (infinite) length, the infinitesimal contributions are so small that even an infinite number of them is not enough to make a finite $B$.

A related argument says that for a long but finite solenoid, the field lines must return and close on themselves, but that the longer the solenoid, the farther away the returning lines are, and the farther away they are, the larger an area they cover, and the larger an area they cover, the weaker $B$ is out there, so that in the limit as the solenoid becomes infinitely long, $B \rightarrow 0$.

A better argument than Griffith's does not assume that B=0 at infinity at the beginning. Instead, you can use Ampère's law to show that B is constant outside the solenoid, all the way out to infinity; call this value $B^\infty$. Then the magnetic field inside the solenoid is not $\mu_0 n I$, but rather $\mu_0 n I + B^\infty$. If you then assume that $B=0$ at infinity, you get the usual result, but $B^\infty$ could, for example, be nonzero if, out there at infinity, there were currents that were making a uniform magnetic field throughout space. (You could imagine, for instance, that your whole laboratory was wrapped in another solenoid, a huge one, that was creating its own uniform field of strength $B^\infty$.) It's generally assumed as a boundary condition that your charges and currents are localized, i.e. there's nothing infinitely far away producing electric or magnetic fields, but that assumption is problematic when you start to analyze "infinite" solenoids, capacitors, etc., since by definition you're putting sources at some parts of infinity!

Certainly refers to an assumed condition that at infinite distance from the solenoid the magnetic field is so negligible that it is considered to be 'ZERO'.

In fact we generally do not read the physics of reality instead we assume certain conditions which are very close to the reality to reach at a conclusion which is nearly the approximate value of the solution. For e.g. Consider this question : A coin is dropped in a well and sound is heard after 1.4sec. Find the depth of the well.(g=9.8m/s², speed of sound=343m/s).

CONTRADICTION - Who observed the sound, a man or anything else. If a man heard the sound then the reaction time for a normal human being is 0.2sec. So to be more accurate why not we take the Time period(2T) as 1.4-0.2=1.2sec.

There are plenty of them in Newtonian Physics.

• It's actually not an assumed condition. There are two fundamental conditions in the problem: the Biot-Savart law and the principle of superposition. The rest are assumptions about the solenoid itself. – sbs95 Oct 14 '16 at 5:19
• how can you know if the magnetic field is negligible at an infinite distance? This is a theoretical problem, and the solenoid is actually infinite. Could you answer that? – sbs95 Oct 14 '16 at 5:20
• Yes, Biot-Savarts law involves in this theoretical problem. You must be knowing that the Magnetic field is inversely proportional to the square of distance between the point concerned and the current element. As far as the INFINITY is concerned it is an assumed mathematical term in itself. – Mihir Oct 14 '16 at 6:04
• An infinite plane with constant current (to put an example) does have a magnetic field at infinity @Maverick. – sbs95 Oct 14 '16 at 15:58
• CORRECTION: Not constant current. If it's a plane then it is assumed as constant current per unit length or per unit area, I.e.in short uniform charge density. Undoubtedly it will have magnetic field at infinity as the law indicates. In fact in theoretical physics we generally assume A UNIFORM CHARGE DENSITY. – Mihir Oct 15 '16 at 15:44

Say the solenoid is aligned along the $z$ axis and you go out to infinity along the $x$ axis. The small segment of the solenoid's circumference that is closest to you has current in the $+y$ direction; the segment furthest from you has current in the $-y$ direction. At infinity, these two segments are equidistant from you, so they make equal and opposite contributions to the field. A similar cancellation argument works for any pair of diametrically opposite current elements along the circumference.

• and, of course, I've assumed you know the direction of the field contributions of each segment along the circumference from Biot-Savart... – ostrichCamel Mar 24 '18 at 1:32
• also, I've neglected the net axial current that would exist in a real solenoid, a step that Griffiths justifies well. – ostrichCamel Mar 24 '18 at 5:12

Your problem is quite basic. The essence of the answer has been mentioned in the other answers. But they have not emphasized on the main points.

We know that the magnetic field lines for a cylindrical-shaped long wire are circles with centres on the axis of the wire with current. The direction of the field lines, that is, the sense of rotation of the circles (counterwise or clockwise) depends on the direction of the current (forth or back). Now, you have to find what happens to the magnetic field when you curve the wire into a circle. Then, you may also stack more such wires, each coiled into a circle, side by side, with centres along a common axis, to form the solenoid you want.

Now, consider the picture above. We have to find the magnetic field in region $2$. So, let me call the upper ends of the wires the "dots" and the lower ends the "crosses".

The diameter of the coil is $\mathrm{a}$ and the current flowing through the wire is $\mathrm{I}$. Keeping in mind, the direction of current flow, "dots" have current $\mathrm{I}$ and "crosses" have current $\mathrm{-I}$.

However, as mentioned in the comment, we have to use the theory for finite wires, assuming that the curved region of the wire on the top and bottom sides are more or less "straight".

So, say the point at $2$ makes an angle with the 'dot' wire; $\theta_1$ on one side and $\phi_1$ on another side. And, with the 'cross' wire, it makes an angle $\theta_2$ on one side and $\phi_2$ on another side.

Using Ampere's Law, we have that

The magnetic field in $2$ due to a "dot" wire, $$B_{dot}=\frac{\mu_0I}{4\pi r_1}(\sin \theta_1 + \sin \phi_1)$$ And the magnetic field in $2$ due to a "cross" wire, $$B_{cross}=\frac{\mu_0(-I)}{4\pi (r_2)}(\sin \theta_2 + \sin \phi_2)=-\frac{\mu_0I}{4\pi (r_1+a)}(\sin (\theta_1+\epsilon) + \sin (\phi_1+\epsilon))$$ where $\mathrm{r_1}$ and $\mathrm{r_2}$ respectively distances of the field point from the axis of the "dot" wire and "cross" wire.

Therefore' net magnetic field in $2$ is $$B_{net}=\frac{\mu_0I}{4\pi r_1}(\sin \theta_1 + \sin \phi_1)-\frac{\mu_0I}{4\pi (r_1+a)}(\sin (\theta_1+\epsilon) + \sin (\phi_1+\epsilon))$$

As $\mathrm{r_1} \to \infty$, we have $\epsilon\to 0$ and finally we get that $$\lim_\limits{\mathrm{r_1} \to \infty} B_{net}=0$$

This is the basic reason for Griffith's certainty of the zero magnetic field very far away from the axis of the solenoid.

However, very close to the axis of the "dot" wire, $\mathrm{r_1} to 0$.

And then, we get $$\lim_\limits{\mathrm{r_1} \to 0} B_{net}=\infty$$

Hence, the field blows up very close to the boundary of the solenoid in region $2$. That is why Grifith's was careful to mention that the field was negligible far away from the solenoid.

That is, when the diameter of the coil was very small compared to the length of the solenoid.

The following picture is a good depiction of what is happening to the final magnetic field due to a solenoid.

Hope this helps you.

• Your argument is in the right spirit, but your $B_{dot}$ and $B_{cross}$ are wrong - they are the expressions if the dot and cross wires were straight and infinite. For little pieces of the circle, the contributions should scale like $1/r_1^2$ and $-1/(r_1 + a)^2$, respectively, so the combined contribution for $r_1 >> a$ is proportional to $1/r_1^3$, not $1/r_1^2$ as it would be with your expressions. What's not obvious is that an infinite number of these should still add up to zero if you get "far enough" away, since $r_1$ is always short compared to the length of the solenoid. – pwf Mar 30 '18 at 4:05