How to Convert Ampere's Law to Laplace Equations and Solve Numerically?

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SUMMARY

This discussion focuses on converting Ampere's Law into Laplace and Poisson equations for numerical solutions using MATLAB. The differential form of Ampere's Law, represented as ∇ × H = J, is transformed into the Poisson equation through vector identities. The user seeks assistance in extending their MATLAB code from a single wire scenario to a two-wire configuration, leveraging the principle of superposition to combine magnetic fields based on distance.

PREREQUISITES
  • Understanding of Ampere's Law and its differential form
  • Familiarity with vector calculus identities
  • Proficiency in MATLAB programming for numerical simulations
  • Knowledge of electromagnetic theory, specifically magnetic fields and potentials
NEXT STEPS
  • Research how to implement the Poisson equation in MATLAB
  • Learn about the principle of superposition in electromagnetic fields
  • Explore numerical methods for solving partial differential equations
  • Investigate the use of MATLAB's built-in functions for vector calculus
USEFUL FOR

Electromagnetic engineers, MATLAB programmers, and students working on numerical simulations of electromagnetic fields will benefit from this discussion.

yoyobarn
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Homework Statement


I need to write ampere's law (differential form) in the form of Laplace equations / Poisson equations and then solve them numerically using Matlab.


Homework Equations


Del x H = J


The Attempt at a Solution


pls see my attachment


I need help from someone who is good at electromagnetics, and also have some numerical programming background.

Thank you very much for help!
 

Attachments

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Ok now I have come up with code for one wire..

Now i need to code for two wires!

Any idea how to do that?
 

Attachments

  • wire_square.png
    wire_square.png
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<br /> \nabla \times B=\mu_0J<br />
<br /> B=\nabla \times A<br />
so then we substitute this into amperes law.
<br /> \nabla \times(\nabla \times A)<br />
then we use the vector identity to re-write the curl of the curl.
so we get
<br /> \nabla(\nabla \cdot A) - (\nabla)^2A=\mu_0J<br />
<br /> \nabla \cdot A = 0
I did this in terms of the B field but
<br /> B=\mu H<br />
So now we have reduced this to Poisson equation.
On your second post you said you need code for two wires . well the B fields obey superposition so you can just add the field to the other based on distance.
 
Last edited:

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