# Magnetic fields of infinite solenoid and infinite current carrying plane

1. Apr 18, 2007

### s.gautam

I am following the book 'Introduction to Electrodynamics' by Griffith. In that book,the magnetic field outside an infinite solenoid has been found as follows:

First of all,it is proved that the magnetic field is along the axis. (I have no problems with that proof).
Then,a rectangular loop is taken outside the loop,and since the current passing the loop is 0,it can be said that the magnetic fields along the two sides of the loop parallel to the axes are same.
By this,it can be inferred that magnetic field outside the solenoid is constant.

Now,the problem I have is with the next argument which says that since the magnetic field will decrease to 0 at very large distances from the solenoid,and since the magnetic field outside the solenoid is constant,that means the magnetic field everywhere outside the solenoid is 0.

Now why can't we make the same assumption in case of an infinite current carrying plane (that is,the magnetic field at infinite height from the plane will decrease to 0) ?
(This assumption is wrong because using this we get the wrong answer.)

2. Apr 18, 2007

### arunma

Gautam, what page are you looking at? I own the Griffiths textbook, so I can take a look at it and provide an explanation.

3. Apr 18, 2007

### s.gautam

For the third edition,Pg 228. The line "But we know that it goes to zero for large s".

4. Apr 18, 2007

### NeoDevin

I had the same issue when reading it, when I asked my prof about it, he didn't give a straight answer. If anyone can give a better reason that the field should be 0 outside, that would be awesome.

5. Apr 19, 2007

### arunma

OK, I checked it out. Basically it's an argument by symmetry. The system has symmetry along a single axis, and we can intuitively tell that the B field must go down as you get further away from the axis of symmetry. That's why it goes to zero for large s.

Here's a qualitative explanation. Solenoids have the property of containing magnetic fields entirely within their boundaries. For a finite solenoid, the B field will come out of the ends, and so there will be a small magnetic field outside the solenoid. But if you wrap the solenoid into a toroid (=doughnut), the field will stay completely inside. An infinite solenoid can be thought of as a toroid of infinite radius.

...hope that makes sense.

6. Apr 20, 2007

### NeoDevin

But why should it go to zero for large s? Especially since it's right after the example where it doesn't go to zero (infinite plane current).

I'm not arguing with the result, just the logic used to get there. It seems silly to me that right after doing an example where the field doesn't go to zero at infinity, he assumes that for a given configuration (which also extends to infinity) the field must go to zero.

7. Apr 20, 2007

### arunma

Yes, I can see the cause of your confusion. The logic here has to do with the symmetry of the problem. The solenoid is essentially a surface current going around a cylinder; thus the problem has symmetry along the axis of the cylinder. But a plane current has an entire plane of symmetry. Furthermore, the cylinder is a three dimensional object, whereas the current sheet is two dimensional. No matter how far you get away from an infinite sheet, the sheet still looks the same size. But a cylinder gets smaller as you go away from it. That's why the magnetic field needs to reduce as you get further away from the cylinder, but not the current sheet.

Make sense this time?

8. Apr 20, 2007

### NeoDevin

It makes sense, I just don't think it's very rigorous.

9. Apr 20, 2007

### Coto

As a relation to electrostatics,

Consider an infinitely long and wide charged plane. Clearly (by Gauss's law) the electric field is: $$\frac{\sigma}{2\epsilon_{0}}$$. A result that tells us that E is independent of distance from the plane (the reason of which was explained by arunma.)

Now the infinite line charge (similar to say the solenoid now,) has a field given as: $$\frac{1}{2\pi\epsilon_{0}}\frac{\lambda}{s}$$. A 1/s dependence that causes it to go to zero as it approaches infinite. This is again explained by arunma's argument that as you approach infinite you'll no longer "see" the line charge.

Last edited: Apr 20, 2007
10. Apr 21, 2007

### s.gautam

Yes,I agree with NeoDavin. It makes sense,but its not that rigorous.

11. Apr 24, 2007

### lpfr

I arrive late in this thread.
I do not see why the magnetic field outside an infinite solenoid would be zero. In fact, is it?
If you see from a distance an infinite solenoid, you see an infinite straight wire (you do not see the wire turning). The symmetry is that of an infinite wire and the magnetic field is the same as that of an infinite wire... and different from zero.

As I write his post I see a way to obtain a zero field. You use a solenoid made of two layers of wire. One going and the other coming. Then the net current seen from far away is zero.
In this case, symmetry is respected and you can use Ampere's law. To find B=0 outside the solenoid.

As stated in precedent posts, you cannot argue that "at infinity you don't see an object of infinite dimensions". This is only true for localized charges or currents.

I don't remember seeing a calculus for an infinite solenoid. The catch has been to compute the field inside a toroidal solenoid and, then to say that the field inside is the same if you open and straighten the solenoid.

12. Apr 24, 2007

### Coto

What's the magnetic field of an infinite line charge moving at a speed v? A quick calculation by Ampère's law tell us that it has a 1/s dependence. So what does a solenoid look like out at infinite? Definitely something similar to a moving line charge at infinite.. we could easily guess that at the very least for the solenoid, that it also has a 1/s dependence.

So lets assume then that the magnetic field has a 1/s dependence for the solenoid, well at infinite if it's similar to a moving line charge, it should be zero due to the 1/s dependence. But Ampere's law for a solenoid tells us that the magnetic field at a point a away from the solenoid has to be the same as the magnetic field at a point b away from the solenoid, so the logical progression is 0 everywhere.

"The symmetry is that of an infinite wire and the magnetic field is the same as that of an infinite wire... and different from zero." (i.e. is non zero with a 1/s dependence) is correct, but incomplete.

This is the exact argument used in the textbook.

EDIT: One last thing as a conceptual question. What do you see if you go out a distance infinite from an infinitely long cylinder? The answer is not an infinite line as you suggested.

Last edited: Apr 24, 2007
13. Apr 24, 2007

### lpfr

Would you please tell me what is "s". I do not have the book.

14. Apr 24, 2007

### Coto

s is the radial component in cylindrical coordinates. It's a convention that Griffith's seems to use in all his books.

15. Apr 25, 2007

### lpfr

Thanks for "s".

Maybe I am stupid, but I still see an infinite line. Will you please tell me what should I see?

16. Apr 25, 2007

### Coto

Since the cylinder is finite radially (in cylindrical coordinates) as you leave it the infinitely long cylinder looks like an infinitely long line, but as you keep moving away, the thickness of the cylinder continues to get smaller and smaller (i.e. the radius of the cylinder continues to "shrink" from your perspective) until at infinite you no longer see the the radial component at all. In other words, the perspective on the cylinder will make it seem like the radius of the cylinder approaches 0 as you approach infinite. If radius = 0 for a cylinder, you have no cylinder, you won't see anything. An infinite plane is different in that it spreads in all directions so that if you're perpindicular to the plane, the "radial component" no longer shrinks to 0.

17. Apr 25, 2007

### lpfr

Sorry, but I still just see a line. A cylinder of radius zero has the same "radius" as a line. By convention we accept that we can see a line, even if its width is zero.

Yes, I agree and so I wrote. When you have things that span to infinity, you can't assume that their influence vanishes with distance.

18. Apr 25, 2007

### Coto

Hmm. I thought about it, lpfr, today and I agree with you. I've always set the quick justification in my mind of field going to zero at infinite for an infinite line charge first by Gauss's/Ampere's 1/s dependence relation, and secondly because it seemed to me logical that if that were the case, at some point later the 'cylinder' would cease to be seen as the radius-->0.

As a strengthening of the argument that you still see the line itself, it's just that the field is extremely weak at extremely distant points, occurs if you calculate potential at infinite for an infinite line charge (potential being relative to some finite point). You'll get infinite.. which would make sense if you continually saw the line charge since you'll continually move in its field, increasing your potential.

Sorry for the stubborness.. I've never had to fully justify the image of it dissappearing.

EDIT: Sorry for the late reply, I just finished up my last exam for the semester this morning.

Last edited: Apr 26, 2007