Magnetic and Electric Field outside Solenoid

In summary, the conversation discusses the fields of solenoids and how they can vary with a sinusoidal current. The participants address the misconception that the magnetic field outside a solenoid is always zero and explain that it only approaches zero as the length of the solenoid becomes infinite. They also discuss how the varying electric field outside the solenoid can induce a magnetic field, creating an electromagnetic wave. Finally, they clarify that Ampere's law can still be applied to a solenoid, but the geometry must be taken into consideration.
  • #1
*FaerieLight*
43
0
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!
 
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  • #2
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
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:
  • #4
*FaerieLight* said:
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!

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
marcusl said:
B=0 outside of a solenoid occurs in the limit as the length L of the solenoid becomes infinite.
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
Redbelly98 said:
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?
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
Redbelly98 said:
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?
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
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
marcusl said:
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.
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.

BruceW said:
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.
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.

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'.
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
Redbelly98 said:
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.
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
@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:
*FaerieLight* said:
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.
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.
 

1. What is a solenoid and how does it create a magnetic field?

A solenoid is a coil of wire that produces a magnetic field when an electric current passes through it. This is due to the flow of electrons in the wire, which creates a magnetic field around the wire. The strength of the magnetic field can be increased by increasing the number of wire turns in the solenoid.

2. How does the magnetic field outside of a solenoid differ from the magnetic field inside?

The magnetic field inside a solenoid is uniform and strong, while the magnetic field outside the solenoid is weaker and more spread out. This is because the magnetic field lines inside the solenoid are tightly packed due to the close proximity of the wire turns, while the field outside is more spread out and less concentrated.

3. How is the direction of the magnetic field outside a solenoid determined?

The direction of the magnetic field outside a solenoid is determined by the right-hand rule. If you point your thumb in the direction of the current flow in the wire, the fingers of your hand will curl in the direction of the magnetic field lines outside the solenoid.

4. Can the direction of the magnetic field outside a solenoid be changed?

Yes, the direction of the magnetic field outside a solenoid can be changed by changing the direction of the current flow in the wire. This can be done by reversing the direction of the current or by changing the polarity of the power source.

5. How does the electric field outside a solenoid compare to the magnetic field?

The electric field outside a solenoid is very weak compared to the magnetic field. This is because the electric field is created by the changing magnetic field inside the solenoid, and the magnetic field inside is much stronger than the changing field outside.

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