Urgent: Understanding Vector Potential from a Straight Wire

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SUMMARY

The vector potential A at a distance s from an infinite straight wire carrying current I is given by A(s) = (μ₀I/2π)ln(s/a)z, where a is a constant. This formulation confirms that the divergence of A is zero (∇·A = 0) and that the curl of A yields the magnetic field B. The vector potential A points in the same direction as the current I due to the fixed direction of the current density J in the wire.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and curl operations.
  • Familiarity with electromagnetic theory, particularly the concept of vector potential.
  • Knowledge of cylindrical coordinates and their application in physics.
  • Basic understanding of current density and its representation in electromagnetic fields.
NEXT STEPS
  • Study the derivation of the vector potential in cylindrical coordinates.
  • Learn about the implications of ∇·A = 0 in electromagnetic theory.
  • Explore the relationship between vector potential and magnetic fields using Ampère's Law.
  • Investigate the physical significance of current density J in electromagnetic fields.
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to vector potentials and magnetic fields.

fredrick08
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urgent EMM question

Homework Statement


find the vector potential a distance s from an infinite straight wire carrying current I, check that dell dot A=0 and dellxA=B...

ok i know the answer is A=A(s) z, in cylindricals. A(s)=(mu*I/2*pi)ln(s/a)z when a is a constant.

can someone please explain to me why A points in the same direction as I?

please, i need this asap, because i have no idea.
 
Last edited:
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[tex]\vec{A}=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J}}{|\vec{r}-\vec{r}\:'|}d^{3}r'[/tex]

For an infinite straight wire, the direction of the current density J is fixed. Therefore ...
 


Thankyou! = )
 

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