- #1
yungman
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Homework Statement
My question is mainly on the set up of the limits of the integral.
The original question is:
Find vector magnetic potential A distance s from a infinite long straight wire carrying DC current I.
The Attempt at a Solution
Let wire on z-axis and [itex] \vec I = \hat z I [/itex]
[tex]\nabla \times \vec A = \vec B = \hat {\phi} \frac {\mu_0 I}{2\pi r} [/tex]
We know A is same direction as I and is function of r in this case.
[tex] \nabla \times \vec A = \nabla \times \hat z A_{(r)}} = \hat {\phi} \frac {\partial A_{(r)}}{\partial r} = \hat {\phi} \frac {\mu_0 I}{2\pi r} [/tex]
[tex]\Rightarrow\; \vec A_{(s)} = \frac {\mu_0 I}{2\pi} \int \frac 1 r dr [/tex]
The book use
[tex] \vec A_{(s)} = \frac {\mu_0 I}{2\pi} \int^s_b \frac 1 r dr [/tex]
Where b is a constant.
My question is how do you justify using b as the lower limit. I understand b cannot be zero. But how do you justify using b constant?
My main question is how do you set the lower limit in this case.
Thanks
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