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Magnetic field of a cylinder with an offset hole

  • Thread starter georgia
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  • #1
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Homework Statement



A current of 1A is flowing down a long cylindrical conductor of radius 8mm. A long cylindrical hole of radius 2mm runs along the conductor with the axis of the hole offset 3mm from the centre of the conductor. What is the magnetic field
i) on the axis of the conductor, and
ii) on the axis of the hole?

Ans: 3.54A/m, 7.96A/m

Homework Equations



B field in a cylinder radius a with a hole through the centre = μIr/2pi(a^2)

The Attempt at a Solution



I found that if you have a cylinder with a hole bored off-centre so that the axis of the hole is displaced from the axis of the cylinder then the field in the hole is perfectly uniform and the flux lines are parallel straight lines along the length of the cylinder.

How do I know what the strength of the magnetic field in the hole is? And then how do I use this to answer the question?
 

Answers and Replies

  • #2
marcusl
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This problem was first solved by none other than Nobel prize winner I.I. Rabi, back in the 20's or 30's. Tell us how you found that the field is uniform--you may have the answer to the field strength there...
 
  • #3
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I don't know how you find that the field is uniform...
 
  • #4
marcusl
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I don't know how you find that the field is uniform...
Hmm, I'm confused because you wrote
I found that if you have a cylinder with a hole bored off-centre so that the axis of the hole is displaced from the axis of the cylinder then the field in the hole is perfectly uniform and the flux lines are parallel straight lines along the length of the cylinder.
Did you find that the field was perfectly uniform?

It might be helpful to know what class are you taking (intro E&M, sophomore/junior E&M?). Are you using vector calculus (cross product, e.g.), and have you seen Ampere's Law, Law of Biot and Savart, etc.?
 
  • #5
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I didn't find that the field was uniform, I was just told it. I have no idea how to work that out. I know what Ampere's Law and Biot Savart Law is...
 
  • #6
marcusl
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Good, I'll try to walk you through this. It's a cool problem, too.

The first step is to calculate the B field inside of a long solid cylindrical conductor of radius R carrying a uniform current density J. Imagine a circle inside, centered on the cylinder axis, of radius r<R. What direction is the B field at any point on the circle? Use Ampere's Law to find the line integral of B around the circle, then realize that B itself is just 2*pi*r times the value of line integral (do you see why?). Let us know what you get.

The next step will be to take a second smaller cylinder carrying current density J in the opposite direction, and superposing that into the first. That creates the hole, and the sum of fields from the two cylinders gives the net field in the hole and outside the hole (ie., on the axis of the first).
 
  • #7
Isnt the net magnetic field inside a conductor 0, cause the current flows on the skin of the condutor?
 
  • #8
marcusl
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At DC the skin depth becomes infinite, so the conductor carries a uniform current density.
 
  • #9
Ok. If A is the area of the circle under consideration, then the current flowing is I=J.A.

So, [tex]\int B.dl=\mu I_{enclosed}[/tex]

[tex]B\int dl=\mu J.A[/tex]
[tex]B(2\pi r)=\mu JA[/tex]
[tex]B=\frac{\mu JA}{2\pi r}[/tex]

What do you do next?
 
  • #10
marcusl
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Right, what you wrote in terms of radius in the first cylinder is
[tex]B_1=\frac{\mu Jr_1}{2}[/tex]
or in terms of vectors
[tex]\vec{B_1}=\frac{\mu}{2} \vec{J}\times\vec{r_1} .[/tex]

Georgia, do you follow so far?

The next step is to add a second cylinder with equal and opposite J. Write down the total field, which is the sum of the fields inside each cylinder, at an arbitrary point inside the resulting hole. You'll see it's in terms of a distance vector that's constant, establishing that B is constant and uniform.
 
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  • #11
This would be because the current density is constant throughout the cylinder, right?

So that would be:
[tex]\vec{B_2}=-\frac{\mu}{2} \vec{J}\times\vec{r_2} .[/tex]
The net field would be:
[tex]\vec{B_net}=\frac{\mu}{2} \vec{J}\times(\vec{r_1}-\vec{r_2}) .[/tex]
 
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  • #12
Im sorry. Im slightly confused now. r2 is the radius of the second cylinder. This cylinder is at a distance a from the center of the first cylinder. I cant figure out the vector part of this. I get what youre saying, but Im messing up tha math.
 
  • #13
marcusl
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Im sorry. Im slightly confused now. r2 is the radius of the second cylinder. This cylinder is at a distance a from the center of the first cylinder. I cant figure out the vector part of this. I get what youre saying, but Im messing up tha math.
Well r2 isn't the radius of cylinder 2, it's the distance from axis of cylinder 2 to a point of interest, and r1 is distance from axis of cylinder 1 to that same point. The vector r1-r2 is a thus a constant vector pointing from the axis of cylinder 1 to 2, having magnitude a.

Your expression for Bnet is therefore a constant, establishing the fact that B is uniform within the hole and oriented perpendicular to both r1-r2 and J.

Pretty cool derivation!
 
  • #14
Ok. Thanks. Wow. This one always escaped me.
 
  • #15
marcusl
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ur welcome :rolleyes:
 
  • #16
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Well r2 isn't the radius of cylinder 2, it's the distance from axis of cylinder 2 to a point of interest, and r1 is distance from axis of cylinder 1 to that same point. The vector r1-r2 is a thus a constant vector pointing from the axis of cylinder 1 to 2, having magnitude a.

Your expression for Bnet is therefore a constant, establishing the fact that B is uniform within the hole and oriented perpendicular to both r1-r2 and J.

Pretty cool derivation!
Hey, I don't know if anyone still checks this forum, but I still don't understand how to find the magnetic field inside the hole. I followed the derivation of why B is uniform, but I don't know where to go from there.
 
  • #17
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Nevermind, I got it.
 

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