Magnetic field for coaxial cable

[NYK]

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 = I_encμ_o

I_enc=∫J⋅da

The Attempt at a Solution

For r < a:

I_enc = I

∫B⋅nda = 2πr = I_enc μ_o

B = (I_enc μ_o)/(2πr)

For r > b:

I_enc = I - I

B =0

For a < r < b:

dI = J⋅nda = IJIda da = 2πrdr

dI = J_o(b/r)(2πrdr)

I_enc = ∫J_o(b/r)(2πrdr) = J_o2πb∫dr (integrating from b to a)

I_enc = J_o2πb(b-a)

∫B⋅nda = B2πr = I_enc μ_o

B2πr = μ_o( J_o2πb(b-a))

B = (μ_o/r)( J_ob(b-a))
I think I did this correctly just looking for a double check

 

[TSny]

The Attempt at a Solution

For r < a:

∫B⋅nda = 2πr = I_enc μ_o

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 = (I_enc μ_o)/(2πr)

This looks correct for r < a.

For r > b:

I_enc = I - I

B =0

Correct.

For a < r < b:

dI = J⋅nda = IJIda da = 2πrdr

dI = J_o(b/r)(2πrdr)

OK

I_enc = ∫J_o(b/r)(2πrdr) = J_o2πb∫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 I_enc?

 

[NYK]

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 = I_encμ_o

B(2πr) = I_encμ_o

B = (I_encμ_o)/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 I_enc?

That makes sense the current in the middle wire is for sure having an effect on the total current so;

I_enc = I_o - I_meat

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

I_meat = J_o(2πb)(r-a)

Then; I_enc = I_o - J_o(2πb)(r-a)

So, B = (μ_o/r)(I_o - J_o(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,

 

[TSny]

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.

 

[NYK]

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 J_o = I_o

So for a < r < b

the B field would be simplified to:B = (μ_o/r)(I_o(1-b(r-a)))

 

[TSny]

Could I define J_o = I_o

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?

 

[NYK]

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:

I_meat = ∫ J_o(b/r)(2πrdr) = J_o2πb∫dr (integrating from a to b)

I_meat = J_o2πb(b-a)

Not sure I am understanding what you are saying though

 

[TSny]

I integrated over that area to begin with and came up with:

I_meat = ∫ J_o(b/r)(2πrdr) = J_o2πb∫dr (integrating from a to b)

I_meat = J_o2πb(b-a)

Not sure I am understanding what you are saying though

OK.

Now, how is I_meat related to I_o?

 

[NYK]

OK.

Now, how is I_meat related to I_o?

Well their reciprocals of one another but the shaded portion has a current density while the middle portion does not.

So,

I_enc = I_o - I_meat = I_o - J_o2πb(b-a)

am I on the right track?

 

[TSny]

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 J_o2πb(b-a) is equal to. Hence, you can solve for J_o. Then substitute into your final expression for B in post #3.

 

[NYK]

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 J_o2πb(b-a) is equal to. Hence, you can solve for J_o. Then substitute into your final expression for B in post #3.

Pretty sure I got it:

If I_meat = J_o2πb(b-a)

then by the same method of integrating

I_o = J_o2πb(r-a)

putting htis back into the equation:B = (μ₀/r)(J_o2πb((b-a)(r-a))

How does that look?

 

[TSny]

Pretty sure I got it:

If I_meat = J_o2πb(b-a)

then by the same method of integrating

I_o = J_o2πb(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)(J_o2πb((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.

 

[NYK]

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) = (μ_o/2πr)I_o(1-(b/r)((r²-a²)/(b²-a²)))

so now when r meets b, B = 0

 

[TSny]

Your right,

I re worked the problem and came up with:

for a < r < b:

B(r) = (μ_o/2πr)I_o(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 $I_o$?

 

[NYK]

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 J_o = I_o/π(b²-a²)

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

So then I_meat = I_o/π(r²-a²)(b/r)(π(r²-a²) = I_o(b/r)

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

so, B(r) = (μ_o/2πr)(I(1-(b/r))

 

[TSny]

I said in the equation the J_o = I_o/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) = (μ_o/2πr)(I(1-(b/r))

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

 

[NYK]

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) = J_obln(r/a)

Then for a < r < b

I_enc = I_o - J_ob(ln(r/a))

So, B(r) =(μ_o/2πr)(I_o - J_ob(ln(r/a))

 

[TSny]

In your first post you had

I_enc = ∫J_o(b/r)(2πrdr) = J_o2πb∫dr (integrating from b to a)

I_enc = J_o2πb(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 $I_o$.

 

[NYK]

Pretty sure I got it:

If I_meat = J_o2πb(b-a)

then by the same method of integrating

I_o = J_o2πb(r-a)

putting htis back into the equation:B = (μ₀/r)(J_o2πb((b-a)(r-a))

How does that look?

So using I_o = J_o2πb(b-a)

and solving for J_o

I come up with J_o = (I_o/2πb(b-a))

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

B = (μ_o/r)(I_o(r-a))

 

[TSny]

So using I_o = J_o2πb(b-a)

and solving for J_o

I come up with J_o = I_o/(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 = (μ_o/r)(I_o(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.

 

[NYK]

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 = (μ_o/r)I_o(1-((r-a)/(b-a))

 

[TSny]

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

for a < r < b;

B = (μ_o/r)I_o(1-((r-a)/(b-a))

That looks correct. You can simplify (1-((r-a)/(b-a)) somewhat further, if you wish.

 

[NYK]

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 = (μ_o/(2πr))I_o(1-((r-a)/(b-a))

the 2π is canceled it in the I_enc so there is still the 2π left over in the B2πr = I_encμ_o equation i started with

I was cancelling out the 2π because i hadn't made the substitution for J_o yet, which ends up having a 2π in its denominator.

 

[TSny]

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.

 

[NYK]

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 = (μ_o/2πr)I_o(1-((a-a)/(b-a))

B = (μ_o/2πr)I_o(1 - 0)

So B = (μ_o/2πr)I_o as expected :)

Thank you for all your help TSny!

 

[TSny]

Great. Good work!