How Do Magnetic Fields Behave Around Infinite Current Sheets?

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SUMMARY

The discussion focuses on calculating the magnetic fields generated by infinite current sheets using Ampere's Law. For part (a), the magnetic field above and below a single infinite current sheet carrying current I along the positive z-axis is determined to be B = ±(μ₀/2)I in the x direction, depending on the position relative to the sheet. In part (b), the challenge is to find the magnetic field configuration between two parallel sheets separated by distance d, both carrying the same current I. The solution requires careful application of Ampere's Law and consideration of the enclosed current from multiple wires.

PREREQUISITES
  • Understanding of Ampere's Law and its mathematical formulation
  • Familiarity with magnetic fields generated by current-carrying conductors
  • Knowledge of the concept of permeability of free space (μ₀)
  • Ability to visualize and analyze magnetic field lines in three-dimensional space
NEXT STEPS
  • Study the application of Ampere's Law in different geometries, including cylindrical and planar symmetries
  • Learn about the superposition principle for magnetic fields from multiple current sources
  • Explore the effects of varying current densities on magnetic field distributions
  • Investigate the implications of magnetic fields in electromagnetic theory and applications
USEFUL FOR

Physics students, electrical engineers, and anyone interested in electromagnetism and magnetic field analysis will benefit from this discussion.

christinaaa.g
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Hi everyone,

1.
Part (a)
Find the magnetic field above and below an infinite current sheet (infinite straight wires with current I along the positive z-axis). The sheet is on the xz plane. There are N wires per meter counting along the axis.

Part (b)
There are two parallel sheets with the same givens as part (a) except that the second sheet is separated by a distance d above the first. The sheet above carries current I along the positive z-axis. Find the magnetic field above, between, and below the sheets.

2.
Ampere's Law:
Integral of (gradient cross B) times dA=Integral of B times dl
with B=magentic field
A=area
l=length

3. I'm having more trouble with part (b) than I did with part (a).
This is how I went about solving part (a). I'm not sure if it's completely right or not though.

I drew a rectangular loop parallel to the xy plane extended an equal distance above and below the surface. I then used ampere's law...
Integral (B times dl) = 2Bl
It is 2Bl because one is from the top segment and one from the bottom.
2Bl=(mu sub 0)(I enclosed)(l)
where I is the current and l is the length and mu is the permeability of free space

Setting 2Bl = 0(I enclosed l)
B=+ (mu sub 0/2)I in the x direction (when y<0)
B= - (mu sub 0/2)I in the x direction (when y>0)




If anybody has any ideas to solve part (b) or any corrections to my part (a) solution, I would really appreciate it! Thanks!
 
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Welcome to PF.

(a)

2Bl=(mu sub 0)(I enclosed)(l)

This should be
2BL=(mu0)(I enclosed)

That is, no "L" at the far right. And I've switched to uppercase L to avoid confusing it with current I.

Then it's a matter of figuring out (I enclosed):
Each wire has a current I.
There are N wires per unit length along x.
The loop spans a length L along x.
 

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