# Magnetic and Electric Field outside Solenoid

1. Jun 9, 2013

### *FaerieLight*

Hi,

I've seen a couple of posts in the Homework section of Physics Forums about the fields of solenoids, but neither of them seem to address the problem that I have about it. The set-up is a long solenoid that has a sinusoidally varying current going through it. We can establish that the magnetic field outside the solenoid is 0 and inside the solenoid it is varying sinusoidally with the current. The varying B field then sets up a sinusoidally varying E field inside and outside the solenoid. My question is, if there is a varying E field outside the solenoid (which I find to be in the azimuthal direction), shouldn't this induce a B field outside the solenoid as well, by Ampere's Law? If this is so, then we wouldn't have been able to establish the result that B=0, since that result assumes there is no varying E field within an Amperian loop outside the solenoid. Could someone please tell me what is happening here? It appears that Maxwell's equations can't be used in this way, or we arrive at inconsistencies. Or am I making a mistake with my reasoning?

Thanks!

2. Jul 2, 2013

### crx

In real life there is always some field leakage around a toroidal inductor (its a toroid rigt?). So the statement that magnetic field outside the coil is zero is wrong. Because the field outside the coil is weak it can be considered to be zero for calculatios. (don't take this to your teacher , I'm just telling you this as buddies ;)

3. Jul 3, 2013

### marcusl

B=0 outside of a solenoid occurs in the limit as the length L of the solenoid becomes infinite. In fact, the total flux outside is the same as that inside (lines of B are closed loops so there must always be a return path for the fields inside the solenoid). As L→∞, the return flux occupies an infinite volume so the flux density (which is B) goes to zero.

A finite-length solenoid has non-zero B outside. When energized with an AC current, it should produce both B and E fields as you surmise.

Last edited: Jul 3, 2013
4. Jul 3, 2013

### darkxponent

You mean a toroid right? because Magnetic Field outside solenoid is not zero. In a time varying Electric field, time Magnetic field will be definitely produced and which will in turn produce time varying Electric field and so on, this chain goes on. This way EM wave is produced, which will give Electric as well Magnetic Field outside toroid!

5. Jul 3, 2013

### Redbelly98

Staff Emeritus
I have heard that before, but doesn't a straightforward application of Ampere's law tell us the outside field is the same as that of a single wire running down the axis of the solenoid?

6. Jul 3, 2013

### BruceW

you mean because the current is moving in a helix, so there is a movement of charge in the axial direction? Yeah, that sounds right to me. I think the idea is that if we make it a very 'tight' helix, then the magnetic field inside the solenoid is far greater than the magnetic field outside.

7. Jul 3, 2013

### marcusl

No, the geometry is all wrong. For an infinitely long solenoid, B is uniform inside and parallel to the axis. Applying Ampere's law around a circle coaxial with the axis gives zero. For a wire along the axis, B is concentric around it and Ampere's integral is non-zero.

8. Jul 4, 2013

### BruceW

But that's not strictly true. The solenoid is effectively a coil, right? No matter how tightly you wrap it, there must still be a current in the axial direction. So Ampere's law around a circle coaxial with the axis gives nonzero. It is just that when we wrap the coil tightly, then the B field outside will be much smaller than that of the B field inside. right? I can only think of a solenoid as a limiting 'approximation'.

9. Jul 4, 2013

### Redbelly98

Staff Emeritus
But a solenoid is a helical coil. There is a current component along the solenoid axis, and this non-zero current is enclosed by the Ampere's law loop.

Yes, that is my point exactly. The solenoid wire is a helix, so there is a "current enclosed" when you do the integral in Ampere's law.

Agree that the field is much smaller outside than in the interior. Alternatively, the B=0 result would apply if we had a stack of planar current loops, all with the same current -- but instead we have a helix with its associated axial-direction component of current.

10. Jul 4, 2013

### marcusl

Ah, you are talking about a real (not idealized) solenoid. Then I agree, there is a small component of axial current outside. For that matter, the field inside has a radial component that spirals as one moves in length.

11. Jul 4, 2013

### BruceW

@marcusl - agreed. The B field inside of a helix-type solenoid will also not be perfectly in the axial direction. Also, unrelated, I am not sure what the OP means here:
I don't know why faerielight thinks a varying B field inside will cause a varying E field outside... Maybe using the integral form of Maxwell's equations, it looks possible. But using the differential form, if there is no B field outside, then there will not be a varying E field outside.