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Electromagnetism Problem (conductors, dielectrics)

  • Thread starter roam
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  • #1
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Homework Statement



http://img209.imageshack.us/img209/1508/problemhd.jpg [Broken]

Homework Equations



The Ampere law: [itex]\oint_C \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}[/itex]

The field and flux is proportional to the current: [itex]\Phi = LI[/itex]

The Attempt at a Solution



(a)

[itex]\oint_C \vec{B} \cdot d\vec{l} = \vec{B} \oint_C d\hat{l}[/itex]

[itex]= \vec{B} (2 \pi s)= \mu_0 I_{enc}[/itex]

[itex]\therefore \ \vec{B}= \frac{\mu_0 I_{enc}}{2 \pi s}[/itex]

The magnetic field is 0 for s<a. So for the space between the conductors do I have to evaluate this between a<s<b?

[itex]\therefore \ \vec{B}= \frac{\mu_0 I_{enc}}{2 \pi b-2 \pi a}[/itex]

Is this correct?

(b) The magnetic flux produced by the current in the conductor in the middle is

[itex]\Phi = \int B.dA = B(2 \pi a h)[/itex]

Here 2πah is the surface area. So how can we evaluate this if we do not know h (height)? :confused:

(c) [itex]\Phi = LI \implies L = \frac{\Phi}{I} = \frac{B(2 \pi r h)}{I}[/itex]

Is this the correct approach?

Any help is greatly appreciated.
 
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Answers and Replies

  • #2
TSny
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The Attempt at a Solution



(a)

[itex]\oint_C \vec{B} \cdot d\vec{l} = \vec{B} \oint_C d\hat{l}[/itex]

[itex]= \vec{B} (2 \pi s)= \mu_0 I_{enc}[/itex]

[itex]\therefore \ \vec{B}= \frac{\mu_0 I_{enc}}{2 \pi s}[/itex]
Hi, roam. That looks fine.

The magnetic field is 0 for s<a.
The magnetic field for s<a would depend on how the current is distributed in the inner conductor. But, apparently they don't want you to worry about that in this problem.

So for the space between the conductors do I have to evaluate this between a<s<b?

[itex]\therefore \ \vec{B}= \frac{\mu_0 I_{enc}}{2 \pi b-2 \pi a}[/itex]

Is this correct?
I don't understand this at all. The symbol B stands for the value of the magnetic field at a point of space.

(b) The magnetic flux produced by the current in the conductor in the middle is

[itex]\Phi = \int B.dA = B(2 \pi a h)[/itex]

Here 2πah is the surface area. So how can we evaluate this if we do not know h (height)? :confused:
You'll need to think about the direction of B and construct an area perpendicular to B and find the flux through that area. When you integrate over the area you won't be able to pull B out of the integral since it is not a constant over the area. Also, note that the question asks for the flux per unit length of the cable.

(c) [itex]\Phi = LI \implies L = \frac{\Phi}{I} = \frac{B(2 \pi r h)}{I}[/itex]

Is this the correct approach?
Yes, it's the correct approach. But you'll need to find the correct expression for the flux.
 
  • #3
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You'll need to think about the direction of B and construct an area perpendicular to B and find the flux through that area. When you integrate over the area you won't be able to pull B out of the integral since it is not a constant over the area. Also, note that the question asks for the flux per unit length of the cable.
Hi,

Thank you for your input. So, the magnetic field is due to the inner conductor in the center carrying I. The direction of B is that it circles around the central wire anti-clockwise (using the right hand rule for instance). We would have:

[itex]\int B.da =\int \frac{\mu I}{2 \pi s}.da[/itex]

So I'm a bit confused, do I now pull everything out of the integral except s? Also, why is B not constant? Isn't the central wire carrying a steady current?
 
  • #4
TSny
Homework Helper
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Thank you for your input. So, the magnetic field is due to the inner conductor in the center carrying I. The direction of B is that it circles around the central wire anti-clockwise (using the right hand rule for instance). We would have:

[itex]\int B.da =\int \frac{\mu I}{2 \pi s}.da[/itex]
That looks good.
So I'm a bit confused, do I now pull everything out of the integral except s? Also, why is B not constant? Isn't the central wire carrying a steady current?
Are you clear on the meaning of the symbol s? Have you selected the surface over which you are integrating to find the flux? You'll need to choose a flat rectangular area that extends from the inner to outer conductor and that's oriented perpendicular to B as shown in the attached figure. s is not constant over this surface. Since B contains s, B is not constant over the surface.

Since you want the flux per unit length, you can take the length of the surface that is parallel to the current to be 1 unit long.
 

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