Magnetic field for coaxial cable

In summary, for a given current flowing through a thin wire, the magnetic field inside and outside a coaxial cable with a metal coating can be calculated using the equations ∫B⋅nda = Iencμo and Ienc=∫J⋅da. For r < a, the magnetic field can be expressed as B = (Ienc μo)/(2πr), for r > b, the magnetic field is 0, and for a < r < b, the magnetic field can be found using the equation B = (μo/r)( Io - Jo(2πb)(r-a)), where Io is the current in the wire and Jo is the current in the metal coating.
  • #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
 
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  • #2
NYK said:

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
TSny said:
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
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
TSny said:
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
NYK said:
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
TSny said:
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
NYK said:
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
TSny said:
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
NYK said:
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
TSny said:
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
NYK said:
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
TSny said:
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
NYK said:
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
TSny said:
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
NYK said:
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
TSny said:
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
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 said:
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
NYK said:
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
TSny said:
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
NYK said:
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
TSny said:
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 canceled 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.
 
Last edited:
  • #24
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
TSny said:
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!
 
  • #26
Great. Good work!
 
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1. What is a coaxial cable and how does it work?

A coaxial cable is a type of cable that is commonly used for transmitting electrical signals. It consists of two concentric conductors - a central wire surrounded by a cylindrical conducting shield - separated by an insulating material. This design allows for the efficient transmission of high frequency signals without interference.

2. How is a magnetic field generated in a coaxial cable?

A magnetic field is generated in a coaxial cable when current flows through the central conductor. The flow of electrical current creates a circular magnetic field around the central conductor, which is then confined and strengthened by the conducting shield surrounding it.

3. What is the purpose of the magnetic field in a coaxial cable?

The magnetic field in a coaxial cable serves two main purposes. First, it helps to maintain the integrity of the electrical signal being transmitted by reducing interference from external sources. Second, it helps to contain the electromagnetic radiation within the cable, preventing it from interfering with nearby electronic devices.

4. How does the strength of the magnetic field in a coaxial cable affect its performance?

The strength of the magnetic field in a coaxial cable directly affects its performance. A stronger magnetic field can help to reduce interference and improve the quality of the transmitted signal. However, if the magnetic field becomes too strong, it can cause distortion of the signal and affect the overall performance of the cable.

5. What factors can affect the strength of the magnetic field in a coaxial cable?

The strength of the magnetic field in a coaxial cable can be affected by several factors, including the amount of current flowing through the central conductor, the distance between the central conductor and the shield, and the type and quality of the materials used in the cable. External factors such as nearby sources of electromagnetic interference can also impact the strength of the magnetic field.

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