Determining the direction of magnetic vector potential

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SUMMARY

The discussion focuses on determining the magnetic vector potential A for a cylindrical wire with a uniform current density J. The professor asserts that due to symmetry, the vector potential can be expressed as \(\vec{A} = A\hat{z}\), leading to the conclusion that \(\nabla \times A = \frac{-dA(s)}{ds}\hat{\phi}\). The reasoning behind this symmetry is that when the current density \(\vec{j}\) flows uniformly in one direction, it establishes a preferred axis, thus aligning \(\vec{A}\) along that axis. This conclusion is valid as long as the current density is consistent throughout the wire.

PREREQUISITES
  • Understanding of vector calculus, particularly curl and divergence operations.
  • Familiarity with electromagnetic theory, specifically magnetic vector potential.
  • Knowledge of cylindrical coordinate systems and their applications in physics.
  • Basic principles of symmetry in physical systems.
NEXT STEPS
  • Study the derivation of magnetic vector potential in cylindrical coordinates.
  • Explore the implications of symmetry in electromagnetic fields.
  • Learn about the relationship between current density and magnetic fields in different geometries.
  • Investigate the use of Ampère's Law in calculating magnetic fields and potentials.
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Students and professionals in physics, particularly those focusing on electromagnetism, as well as educators teaching concepts related to magnetic vector potentials and current distributions.

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Homework Statement


Im doing this practice question and I am to determine the magnetic vector potential A for a cylindrical wire with a uniform current density J. i have already determined B both inside and outside the wire no problem.My issue is in the solution given my professor states that due to symmetry \vec{A}=A\hat{z} then \nabla \times A=\frac{-dA(s)}{ds}\hat{\phi}

The part I don't understand is the statement "due to symmetry \vec{A}=A\hat{z}", how does he know that just by looking?

Is it because A is determined by the integral of the current, therefore A points in the same direction as the current?
 
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In general, the direction of ##\vec{A}## will not necessarily be aligned with that of ##\vec{j}## (e.g. if the current density field points in different directions at different points, that is if the current doesn't all flow in one direction, then you can't immediately be certain that the direction of ##\vec{A}## aligns with that of ##\vec{j}## from point to point) but if ##\vec{j}## flows in one direction everywhere then what you said is basically true up to a sign.

Think about it: if ##\vec{j}## points in a single direction everywhere in space then this picks out a preferred axis in space right? If ##\vec{A}## didn't point along the same axis, how will it even know which way to point when there is no directional preference in the system off of that axis?
 

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