Why is the position of the wire important for Ampere's Law to hold?

In summary, a long straight cylindrical shell with inner radius Ri and outer radius Ro carries a current i that is uniformly distributed over its cross section. A wire placed parallel to the cylinder axis in the hollow region (r<Ri) must carry a current in the opposite direction of the current in the shell in order to cancel out the magnetic field outside the shell. The only position the wire can be in to achieve this is on the cylinder axis. This can be explained using the superposition of magnetic fields, where the magnetic fields of the shell and wire must cancel out everywhere outside the shell for the magnetic field to be zero. The current carried by the wire must be equal to the current carried by the shell for this to occur.
  • #1
mbrmbrg
496
2

Homework Statement



A long straight cylindrical shell has an inner radius Ri and an outer radius Ro. It carries a current i, uniformly distributed over its cross section. A wire is parallel to the cylinder axis, in the hollow region (r < Ri). The magnetic field is zero everywhere outside the shell (r > Ro). We conclude that the wire:
  1. may be anywhere in the hollow region but must be carrying current i in the same direction as the current in the shell
  2. may be anywhere in the hollow region but must be carrying current i in the direction opposite to that of the current in the shell
  3. does not carry any current
  4. is on the cylinder axis and carries current i in the same direction as the current in the shell
  5. is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell

(each option is a radio button)

Homework Equations



[tex]\oint \vec{B} \cdot d\vec{s} = \mu_0 i_{enc}[/tex]

The Attempt at a Solution



Total enclosed current is the algebraic sum of each portion of enclosed current. Therefore, to get [tex]\oint B \cdot ds = 0[/tex] any current within the amperian loop (in this case, loop is the outer surface of the shell) must be compensated for by current flowing in the opposite direction. I don't see why the relative positions of the current-carrying bodies are of any relevence. So I said that the answer was 3. But the book says the answer is 5: the wire must be on the cylinder's axis. Can I have a pointer as to why the position of the wire matters, as long as the wire is within the amperian loop?

edit: stupid mistake. I said that the answer was 2 not 3.
 
Last edited:
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  • #2
Just guessing again, but I also picked the answer 5.

- It cannot be 3, because the the magnetic field outside the wire would not be 0 due to the current carried by the conductor
-And, the ll wire cannot carry current in the same direction because, that would amplify the magnetic field created by the outer cylinder (since B is dir. proportional to the current)
-and now, that leaves only 5 and 2.

If, it is 2, then that means one side of the cylinder would be more closer to the ll wire than the other, so there would be different B fields on the two opposite sides of the cylinder.

So, that means the answer is 5.

I would also add that the current the ll wire carries is more than the current carried by the cylinder. Because the magnitude of the B field created by the ll wire must be equal, and opposite in the direction to the B field created by the cylinder.

Use right hand rule, to find the directions of the B fields created by both things, and so perhaps that would surely help
 
  • #3
Lets look at the shell. Outside the shell the magnetic field is as if all the current is concentrated at the center. Now firstly inorder to cancel the magnetic field of the present shell we'll need a wire carying current in the opposite direction, using the right hand rule you will see that the fields of the shell will cancel with the fields of the wire everywhere outside the shell if the wire was placed in the center of the shell.

This is similar to electric fields, if you have a uniformly distributed shell of charge q and measure its E outside the shell then it will apear as if all of q is concentrated in the center of the sphere. Similarly can be drawn for gravatational bodies.

Magnetic fields are vectors and you can add them using super position. If you try placing the wire in other places the magnetic fields will not cancel everywhere outside the shell, you can use the right hand rule to see this.

Hope this helps!
 
  • #4
rootX said:
- It cannot be 3, because the the magnetic field outside the wire would not be 0 due to the current carried by the conductor

erm... if I told you that I actually answered 2, and the 3 is a typo, would you believe me? Because it's true.:redface:

-And, the ll wire cannot carry current in the same direction because, that would amplify the magnetic field created by the outer cylinder (since B is dir. proportional to the current)

My thoughts exactly.

-and now, that leaves only 5 and 2.

If, it is 2, then that means one side of the cylinder would be more closer to the ll wire than the other, so there would be different B fields on the two opposite sides of the cylinder.

So, that means the answer is 5.

That makes sense, but how does the equation reflect that?

I would also add that the current the ll wire carries is more than the current carried by the cylinder.

I don't think so. The currents must be equal in order to cacel each other; it is the current density of the wire that must be greater than the current density of the cylinder.
 
  • #5
Four said:
Lets look at the shell. Outside the shell the magnetic field is as if all the current is concentrated at the center. Now firstly inorder to cancel the magnetic field of the present shell we'll need a wire carying current in the opposite direction, using the right hand rule you will see that the fields of the shell will cancel with the fields of the wire everywhere outside the shell if the wire was placed in the center of the shell.

This is similar to electric fields, if you have a uniformly distributed shell of charge q and measure its E outside the shell then it will apear as if all of q is concentrated in the center of the sphere. Similarly can be drawn for gravatational bodies.

So far, so good.

Magnetic fields are vectors and you can add them using super position. If you try placing the wire in other places the magnetic fields will not cancel everywhere outside the shell, you can use the right hand rule to see this.

Hope this helps!

But why won't it cancel everywhere? Assume that the cylinder's magentic field acts like a long wire along the cylinder's axis. If I put a wire with equal and opposite current anywhere in the same amperian loop as this hypothetical wire, the integral will equal 0. The wires don't need to be on top of each other!
(Clearly, I'm wrong. I just don't see why.)

edit: And thanks, rootX and Four, for your help!
 
  • #6
I think they wudn't cancel cuz
B is inv. proportional to r,

so, if the ll wire is towards the left of the central axis, then it's B field wud be out of phase with the B field of the cylinder
(I really cannot describe my visualization, but try to visualize it may help.)
 
  • #7
rootX said:
I think they wudn't cancel cuz
B is inv. proportional to r,

so, if the ll wire is towards the left of the central axis, then it's B field wud be out of phase with the B field of the cylinder
(I really cannot describe my visualization, but try to visualize it may help.)

My visualization is of two sets of circular ripples... Oh, forget it. See attachment for rough sketch. Scenario A (wire and cylinder not co-axial) won't cancel out. Scenario B (wire on cylinder's axis) does cancel. (B is purple because how else can I draw red and blue circles on top of each other?)
So when I visualize it, it makes sense.
But why why why does the equation seem to neglect the positioning of the enclosed current-carrying objects?
 

Attachments

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1. What is Ampere's Law?

Ampere's Law is a fundamental law in electromagnetism, named after the French physicist André-Marie Ampère. It states that the magnetic field created by an electric current is proportional to the magnitude of the current and the distance from the current.

2. How is Ampere's Law used in practical applications?

Ampere's Law is used in many practical applications, such as in the design of electrical motors and generators, in electronic circuit design, and in the development of magnetic levitation technology. It also plays a crucial role in understanding the behavior of electromagnetic waves.

3. Can Ampere's Law be derived from other laws?

Yes, Ampere's Law can be derived from other fundamental laws in electromagnetism, such as Gauss's Law and Faraday's Law. It is also a consequence of the principle of conservation of charge.

4. What is the mathematical equation for Ampere's Law?

The mathematical equation for Ampere's Law is ∮B · dl = μ0I, where B is the magnetic field, dl is an element of length along the closed loop path, μ0 is the permeability of free space, and I is the electric current passing through the loop.

5. How does Ampere's Law relate to Maxwell's equations?

Ampere's Law is one of the four Maxwell's equations, which are a set of fundamental equations that describe the behavior of electromagnetic fields. Specifically, it is the integral form of the fourth Maxwell's equation, also known as the Maxwell-Ampere Law.

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