Urgent: Understanding Vector Potential from a Straight Wire

AI Thread Summary
The discussion focuses on finding the vector potential A at a distance s from an infinite straight wire carrying current I. The vector potential is expressed as A(s) = (μI/2π)ln(s/a)z, indicating it points in the z-direction. The reason A points in the same direction as the current I is due to the fixed direction of the current density J along the wire. The relationship between the vector potential and the current density is confirmed through the integral formulation provided. Understanding this relationship is crucial for solving electromagnetic problems related to current-carrying wires.
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.
 
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\vec{A}=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J}}{|\vec{r}-\vec{r}\:'|}d^{3}r'

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


Thankyou! = )
 
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