Ampere's Law with a coaxial cable

In summary: Thank you!In summary, the problem involves a current flowing through a long, cylindrical coaxial cable. The magnetic field is to be determined at different regions, including inside and outside the conductor. Using Ampere's law, the magnetic field is found to be directly proportional to the enclosed current and inversely proportional to the distance from the current. The enclosed current is calculated by considering the current passing through an area that has the loop as its boundary. In the case of (a), there is a current in both the inner and outer conductors, resulting in a non-zero magnetic field. For (b) and (c), the enclosed current is only from the inner conductor, resulting in a smaller magnetic field. Finally, for (d), there
  • #1
Ignitia
21
5

Homework Statement


A portion of a long, cylindrical coaxial cable is shown
in the accompanying figure. A current I flows down the
center conductor, and this current is returned in the outer
conductor. Determine the magnetic field in the regions (a)
R ≤ r1, (b) r2 ≥ R ≥ r1, (c) r3 ≥ R ≥ r2, and (d)
R ≥ r3. Assume that the current is distributed uniformly
over the cross sections of the two parts of the cable.

media%2F088%2F088be225-7b5c-44c5-98d6-f936f309dda3%2FphpxpG6y9.png


Homework Equations


B∫dl = μI

Inner current = I1
Outer Current = I2

I = J*A
J = Current Density
A = Area

The Attempt at a Solution


(a) R ≤ r1
B∫dl = μI
B*2πR = μ*J*A
B*2πR = μ*(I/πr12)*πR2
B = (μ*I*R)/2πr12
The answer given states it's only B = (μ*I*R)/r12 Why is that?

(b) r2 ≥ R ≥ r1

My thinking was that since this is outside the conductor, B = 0T since I = 0. The answer says the enclosed charge would be I, not 0, but doesn't go into any further detail. Can someone please explain why?

(c) r3 ≥ R ≥ r2
My answer was the same as a and was wrong, and likely related to B. the book stated that I = I1 - I2. Why is this?

(d) R ≥ r3
Since it's outside the conductor, then I = 0 and B = 0T This was correct.

Thanks for the help.
 

Attachments

  • media%2F088%2F088be225-7b5c-44c5-98d6-f936f309dda3%2FphpxpG6y9.png
    media%2F088%2F088be225-7b5c-44c5-98d6-f936f309dda3%2FphpxpG6y9.png
    16.7 KB · Views: 2,196
Physics news on Phys.org
  • #2
It seems you have some problems understanding what the enclosed current in Ampere's law is. That there is no current where the loop is does not matter. The only thing that matters is how much current passes through an area that has the loop as its boundary. So for example, in the case of ##r_1 < r < r_2##, the area could be a disk with radius ##r##, what would be the current through this disk? This problem affects both (b) and (c).

For (a), there definitely should be a ##2\pi## in the denominator. The field is ##B = \mu_0 I_{\rm enc}/(2\pi r)## and ##I_{\rm enc} = I r^2/r_1^2##.
 
  • Like
Likes Ignitia
  • #3
Okay, I see what my confusion is now. I was thinking the area was shaped like a washer, and only integrated from r1 to r2, not from 0 to r2. I got it now!
 

Related to Ampere's Law with a coaxial cable

1. What is Ampere's Law with a coaxial cable?

Ampere's Law with a coaxial cable is a physics principle that describes the relationship between the magnetic field and the electric current flowing through a coaxial cable. It states that the magnetic field surrounding a long, straight coaxial cable is directly proportional to the electric current passing through the cable.

2. How do you calculate the magnetic field using Ampere's Law with a coaxial cable?

To calculate the magnetic field using Ampere's Law with a coaxial cable, you will need to know the electric current passing through the cable, the distance between the cable's inner and outer conductors, and the permeability of the material surrounding the cable. The formula for calculating the magnetic field is B = (μ0 * I) / (2πr), where B is the magnetic field, μ0 is the permeability of free space, I is the electric current, and r is the distance between the cable's conductors.

3. What is the significance of the permeability of the material surrounding a coaxial cable in Ampere's Law?

The permeability of the material surrounding a coaxial cable is a crucial factor in Ampere's Law as it determines how much the magnetic field will be affected by the current passing through the cable. Materials with higher permeability will have a stronger magnetic field, while materials with lower permeability will have a weaker magnetic field.

4. Can Ampere's Law be applied to other types of cables besides coaxial cables?

Yes, Ampere's Law can be applied to any type of long, straight cable as long as the magnetic field is perpendicular to the cable. However, it is most commonly used for coaxial cables due to their cylindrical shape and symmetrical design.

5. How is Ampere's Law with a coaxial cable used in practical applications?

Ampere's Law with a coaxial cable is used in various practical applications, such as in the design of electromagnets, transformers, and other electrical devices. It is also used in the telecommunications industry for signal transmission and in medical imaging technologies such as MRI machines.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
246
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
955
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top