Magnetic Field inside plate

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SUMMARY

The magnetic field inside and outside a large, flat conducting plate with a linearly decreasing current density, described by J = J0(1-az), has been analyzed. Outside the plate, the magnetic field is calculated as B = μ0 J0 / 4a, which aligns with the solution provided in the textbook. However, the calculation for the magnetic field inside the plate is more complex, resulting in B = μ0 J0 (az²/2 - z + 1/4a), which includes an additional term of 1/4a that requires further investigation. The discrepancy in the internal magnetic field calculation highlights the need to consider contributions from both above and below the specified point within the plate.

PREREQUISITES
  • Understanding of Ampere's Law and its application in electromagnetism.
  • Familiarity with current density concepts and their mathematical representation.
  • Knowledge of magnetic field calculations in conductive materials.
  • Basic calculus for evaluating integrals related to magnetic fields.
NEXT STEPS
  • Review the derivation of magnetic fields using Ampere's Law in different geometries.
  • Study the effects of varying current density on magnetic field distributions.
  • Explore the concept of boundary conditions in electromagnetic theory.
  • Investigate the physical significance of the additional term in the magnetic field equation inside the plate.
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Students and professionals in physics and electrical engineering, particularly those focused on electromagnetism and magnetic field analysis in conductive materials.

deedsy
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Homework Statement


What is the magnetic field outside and inside of a large, flat conducting plate if the current density decreases linearly with depth z inside the plate. J = J0(1-az). The plate thickness is 1/a.

Homework Equations


*see below

The Attempt at a Solution


[itex]\oint \vec B \cdot d \vec l = \mu_0 I[/itex]

So outside the plate,
[itex]2Bl = \mu_0 \int J_0(1-az)ldz[/itex] *The integral is taken from from z=0 to z=1/a (the whole plate)
[itex]B=\frac{\mu_0 J_0}{4a}[/itex]

Which agrees with the back of the book solution.

Now inside is where I am having trouble...
I tried applying Ampere's Law again but I just got
[itex]\oint \vec B \cdot d \vec l = \mu_0 I[/itex]
[itex]2Bl = \mu_0 \int J_0(1-az)ldz[/itex] *taking the integral from 0 to z this time
[itex]B=\frac{\mu_o J_0}{2} (z-\frac{az^2}{2})[/itex]

but the back of the book says the answer is [itex]B = \mu_0 J_0 (\frac{az^2}{2} - z +\frac{1}{4a})[/itex]

So I'm not sure where the [itex]\frac{1}{4a}[/itex] came from... Does anyone know how i should go about finding the magnetic field inside the plate?
 
Last edited:
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At some point z, you have a field above and a field below. You should account for both of them.
 
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Got it - thanks for the help
 

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