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Magnetic field for coaxial cable

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  • #1
NYK
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Homework Statement


https://scontent-b.xx.fbcdn.net/hphotos-xpa1/v/t1.0-9/10846336_1517298108546407_4211362504714893270_n.jpg?oh=2f11c165933554a0e4003b398c877004&oe=5542789C

Homework Equations


∫B⋅nda = Iencμo

Ienc=∫J⋅da

The Attempt at a Solution


For r < a:

Ienc = I

∫B⋅nda = 2πr = Ienc μo

B = (Ienc μo)/(2πr)

For r > b:

Ienc = I - I

B =0

For a < r < b:

dI = J⋅nda = IJIda da = 2πrdr

dI = Jo(b/r)(2πrdr)

Ienc = ∫Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from b to a)

Ienc = Jo2πb(b-a)

∫B⋅nda = B2πr = Ienc μo

B2πr = μo( Jo2πb(b-a))

B = (μo/r)( Job(b-a))



I think I did this correctly just looking for a double check
 

Answers and Replies

  • #2
TSny
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The Attempt at a Solution


For r < a:

∫B⋅nda = 2πr = Ienc μo
The integral on the left should be an integral along a path rather than an integral over an area. Also, your middle expression is incomplete.

B = (Ienc μo)/(2πr)
This looks correct for r < a.

For r > b:

Ienc = I - I

B =0
Correct.

For a < r < b:

dI = J⋅nda = IJIda da = 2πrdr

dI = Jo(b/r)(2πrdr)
OK

Ienc = ∫Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from b to a)
You need to find B at some arbitrary value of r that lies between a and b. To find B at r, what will be the choice of the path of integration? How does this affect the integration limits?

Also, will the path of integration enclose the thin wire at the center of the cable? If so, should the current in the wire also be included as part of Ienc?
 
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  • #3
NYK
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The integral on the left should be an integral along a path rather than an integral over an area. Also, your middle expression is incomplete.



So I could say ∫Bdl = Iencμo

B(2πr) = Iencμo

B = (Iencμo)/2πr

You need to find B at some arbitrary value of r that lies between a and b. To find B at r, what will be the choice of the path of integration? How does this affect the integration limits?

Limits on integration would be from a to r, not a to b.

Also, will the path of integration enclose the thin wire at the center of the cable? If so, should the current in the wire also be included as part of Ienc?
That makes sense the current in the middle wire is for sure having an effect on the total current so;

Ienc = Io - Imeat

and Imeat = ∫Jo(b/r)(2πrdr) = Jo(2πb)∫dr (integrating from a to r)

Imeat = Jo(2πb)(r-a)

Then; Ienc = Io - Jo(2πb)(r-a)

So, B = (μo/r)(Io - Jo(2πb)(r-a))

That makes more sense to me because that is saying that that current is the current running through the meat subtracted from the current running through the wires. Which it should be since the currents are going in opposite directions,
 
  • #4
TSny
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OK, that all looks good. There is one more thing you should probably do. The problem states that the metal carries the same current ##I_o## as the wire. This means that you can express ##J_o## in terms of ##I_o## and simplify your result.
 
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  • #5
NYK
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OK, that all looks good. There is one more thing you should probably do. The problem states that the metal carries the same current ##I_o## as the wire. This means that you can express ##J_o## in terms of ##I_o## and simplify your result.
Could I define Jo = Io

So for a < r < b

the B field would be simplified to:


B = (μo/r)(Io(1-b(r-a)))
 
  • #6
TSny
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Could I define Jo = Io
No. ##J_o## has the units of current density while ##I_o## has the units of current. So, they cannot be equal.

What would you get if you integrated ##J## over the area between r = a and r = b?
 
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  • #7
NYK
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No. ##J_o## has the units of current density while ##I_o## has the units of current. So, they cannot be equal.

What would you get if you integrated ##J## over the area between r = a and r = b?
I integrated over that area to begin with and came up with:

Imeat = ∫ Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from a to b)

Imeat = Jo2πb(b-a)

Not sure I am understanding what you are saying though
 
  • #8
TSny
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I integrated over that area to begin with and came up with:

Imeat = ∫ Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from a to b)

Imeat = Jo2πb(b-a)

Not sure I am understanding what you are saying though
OK.

Now, how is Imeat related to Io?
 
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  • #9
NYK
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OK.

Now, how is Imeat related to Io?
Well their reciprocals of one another but the shaded portion has a current density while the middle portion does not.

So,

Ienc = Io - Imeat = Io - Jo2πb(b-a)

am I on the right track?
 
  • #10
TSny
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Well their reciprocals of one another ....
No. Note that you got ##I_{meat}## by integrating the current density ##J## over the entire cross-section of the metal. Therefore, ##I_{meat}## is the total current carried by the metal between r = a and r = b. The problem statement tells you what this total current is equal to.

Therefore you know what Jo2πb(b-a) is equal to. Hence, you can solve for Jo. Then substitute into your final expression for B in post #3.
 
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  • #11
NYK
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No. Note that you got ##I_{meat}## by integrating the current density ##J## over the entire cross-section of the metal. Therefore, ##I_{meat}## is the total current carried by the metal between r = a and r = b. The problem statement tells you what this total current is equal to.

Therefore you know what Jo2πb(b-a) is equal to. Hence, you can solve for Jo. Then substitute into your final expression for B in post #3.

Pretty sure I got it:

If Imeat = Jo2πb(b-a)

then by the same method of integrating

Io = Jo2πb(r-a)

putting htis back into the equation:


B = (μ0/r)(Jo2πb((b-a)(r-a))

How does that look?
 
  • #12
TSny
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Pretty sure I got it:

If Imeat = Jo2πb(b-a)

then by the same method of integrating

Io = Jo2πb(r-a)
OK. What I would do here is solve for ##J_o## in terms of ##I_o## and then substitute for ##J_o## in the result of post #3. This way, you will get B expressed in terms of ##I_o##.

putting htis back into the equation:


B = (μ0/r)(Jo2πb((b-a)(r-a))

How does that look?
This doesn't look correct. Note that B should go to zero when r = b since you know that B = 0 for r>b. Also you can check that the units on the right side of your result are not the correct units for B. Check the part that is in red.
 
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  • #13
NYK
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OK. What I would do here is solve for ##J_o## in terms of ##I_o## and then substitute for ##J_o## in the result of post #3. This way, you will get B expressed in terms of ##I_o##.



This doesn't look correct. Note that B should go to zero when r = b since you know that B = 0 for r>b. Also you can check that the units on the right side of your result are not the correct units for B. Check the part that is in red.

Your right,

I re worked the problem and came up with:

for a < r < b:

B(r) = (μo/2πr)Io(1-(b/r)((r2-a2)/(b2-a2)))

so now when r meets b, B = 0
 
  • #14
TSny
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Your right,

I re worked the problem and came up with:

for a < r < b:

B(r) = (μo/2πr)Io(1-(b/r)((r2-a2)/(b2-a2)))

so now when r meets b, B = 0
But this expression also implies that B = 0 when r = a. So something's wrong.

What did you get for the relationship between ##J_o## and ##I_o##?
 
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  • #15
NYK
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But this expression also implies that B = 0 when r = a. So something's wrong.

What did you get for the relationship between ##J_o## and ##I_o##?

I said in the equation the Jo = Io/π(b2-a2)

But i just tried redefining that to: Jo = Io/π(r2-a2)

So then Imeat = Io/π(r2-a2)(b/r)(π(r2-a2) = Io(b/r)

so then my Ienc = I(1-(b/r))

so, B(r) = (μo/2πr)(I(1-(b/r))
 
  • #16
TSny
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I said in the equation the Jo = Io/π(b2-a2)
The current density ##J(r)## is not uniform. It is given by the equation ##J(r) = J_ob/r##. To find the relation between the constant ##J_o## and the total current in the outer conductor, ##I_o##, integrate ##J(r)## over the cross-section of the outer conductor.

B(r) = (μo/2πr)(I(1-(b/r))
This goes to zero for r = b, but does it give the correct answer for r = a?
 
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  • #17
NYK
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The current density ##J(r)## is not uniform. It is given by the equation ##J(r) = J_ob/r##. To find the relation between the constant ##J_o## and the total current in the outer conductor, ##I_o##, integrate ##J(r)## over the cross-section of the outer conductor.



This goes to zero for r = b, but does it give the correct answer for r = a?

I integrated J(r) from a to r and came up with:

J(r) = Jobln(r/a)

Then for a < r < b

Ienc = Io - Job(ln(r/a))

So, B(r) =(μo/2πr)(Io - Job(ln(r/a))
 
  • #18
TSny
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In your first post you had

Ienc = ∫Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from b to a)

Ienc = Jo2πb(b-a)
Use this to relate ##J_o## to ##I_o##.

Then use this in your result for B in post #11 to express B in terms of ##I_o##.
 
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  • #19
NYK
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Pretty sure I got it:

If Imeat = Jo2πb(b-a)

then by the same method of integrating

Io = Jo2πb(r-a)

putting htis back into the equation:


B = (μ0/r)(Jo2πb((b-a)(r-a))

How does that look?

So using Io = Jo2πb(b-a)

and solving for Jo

I come up with Jo = (Io/2πb(b-a))

Then plugging this into the B field equation quoted above I come up with:

B = (μo/r)(Io(r-a))
 
  • #20
TSny
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So using Io = Jo2πb(b-a)

and solving for Jo

I come up with Jo = Io/(2πb(b-a))
OK. (I modified the parentheses a bit to make sure the (b-a) factor is in the denominator.)

Then plugging this into the B field equation quoted above I come up with:

B = (μo/r)(Io(r-a))
Oops! My fault. You should substitute for ##J_o## in your expression that you obtained in post #3. Your expression in post #11 was incorrect. Sorry.
 
  • #21
NYK
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OK. (I modified the parentheses a bit to make sure the (b-a) factor is in the denominator.)



Oops! My fault. You should substitute for ##J_o## in your expression that you obtained in post #3. Your expression in post #11 was incorrect. Sorry.

No problem, I just really appreciate all your help.

So after making the corrctions to the B field in post 3 I come up with,

for a < r < b;

B = (μo/r)Io(1-((r-a)/(b-a))
 
  • #22
TSny
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So after making the corrctions to the B field in post 3 I come up with,

for a < r < b;

B = (μo/r)Io(1-((r-a)/(b-a))
That looks correct. You can simplify (1-((r-a)/(b-a)) somewhat further, if you wish.
 
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  • #23
NYK
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That looks correct. You can simplify (1-((r-a)/(b-a)) somewhat further, if you wish.

I just looked at it and would it actually be:

B = (μo/(2πr))Io(1-((r-a)/(b-a))

the 2π is cancelled it in the Ienc so there is still the 2π left over in the B2πr = Iencμo equation i started with

I was cancelling out the 2π because i hadn't made the substitution for Jo yet, which ends up having a 2π in its denominator.
 
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  • #24
TSny
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Yes. I overlooked the absence of the ##2 \pi##. As you noted, there should be a factor of ##2 \pi## in the denominator. Check to make sure that your answer gives the expected result for r = a.
 
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  • #25
NYK
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Yes. I overlooked the absence of the ##2 \pi##. As you noted, there should be a factor of ##2 \pi## in the denominator. Check to make sure that your answer gives the expected result for r = a.

when r = a, B = (μo/2πr)Io(1-((a-a)/(b-a))

B = (μo/2πr)Io(1 - 0)

So B = (μo/2πr)Io as expected :)

Thank you for all your help TSny!!
 

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