Magnetic field of a Coaxial Cable

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic field of a coaxial cable with inner and outer conductors carrying equal and opposite currents of 5.00 A. The dimensions of the conductors are specified as a = 1.20 cm, b = 5.40 cm, and c = 7.30 cm. The magnetic field at a distance of 0.36 cm from the center axis is derived using Ampère's Law, specifically the equation \(\oint B \cdot dl = \mu_0 I_{\text{enclosed}}\). The correct approach involves recognizing that the magnetic field inside the inner conductor is proportional to the enclosed current, leading to the conclusion that the magnetic field strength must be recalculated considering the fraction of the current enclosed.

PREREQUISITES
  • Understanding of Ampère's Law and its application in magnetostatics
  • Familiarity with coaxial cable geometry and current distribution
  • Knowledge of magnetic field calculations in cylindrical coordinates
  • Basic principles of electromagnetism, particularly magnetic fields generated by current-carrying conductors
NEXT STEPS
  • Study the application of Ampère's Law in different geometries, particularly cylindrical systems
  • Learn about the magnetic field calculations for coaxial cables in various configurations
  • Explore the concept of current density and its implications in magnetic field calculations
  • Investigate the effects of varying current magnitudes and directions on magnetic field strength
USEFUL FOR

Physics students, electrical engineers, and anyone studying electromagnetism or working with coaxial cables in practical applications.

evanclear
Messages
4
Reaction score
0

Homework Statement



A long coaxial cable consists of two concentric conductors with dimensions a = 1.20 cm, b = 5.40 cm and c = 7.30 cm. There are equal and opposite currents of 5.00 A in the conductors. Find the magnitude of the magnetic field at a distance of 0.36 cm from the center axis.

Homework Equations



\ointB*dl=μ_0*I_enclosed


The Attempt at a Solution



Far as i can tell because the outer portion of the wire is not enclosed I enclosed for that portion of the problem is 0 so:

B=I*r/2∏a2

But I'm not getting the right answer.

Any help would be greatly appreciated
 

Attachments

  • CoaxialCable.gif
    CoaxialCable.gif
    3.6 KB · Views: 868
Last edited:
Physics news on Phys.org
0.36cm is within the inner cable - just some fraction of the 5A current flow in the inner wire is enclosed in your integral.
You can assume that (electric) current density is constant in the wire.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
Replies
10
Views
2K
  • · Replies 23 ·
Replies
23
Views
6K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 26 ·
Replies
26
Views
4K
Replies
10
Views
5K