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I have found the vector potential for a circuit $C$ carrying current $I$ such that $C$ is a segment of the z-axis $-L$ and $L$ to be

$A(r)=\frac{\mu_0k}{4\pi}I\ln\frac{L+\sqrt{\rho^2+L^2}}{-L+\sqrt{\rho^2+L^2}}$ where $\rho$ is the distance from an arbitrary point $P$ to the z-axis

which the textbook agrees with so that's fine.

It then asks to consider the case where $L$ is much larger than $\rho$ (an infinite straight wire) and to use the binomial expansion $\sqrt{\rho^2+L^2}=L(1+\frac{\rho^2}{L^2})$ as a hint. But when I do this binomial expansion (excluding terms bigger order than $\frac{\rho^2}{L^2}$) I get $L+\frac{\rho^2}{2L}$ and substituting I get:

$A(r)=\frac{\mu_0k}{4\pi}I\ln\frac{\rho^2+L^2}{\rho^2}$

But the text book gives the answer ( and also mentions I can drop an inessential constant??):

$A(r)=\frac{-\mu_0k}{2\pi}I\ln\rho$

Any help or hints would be appreciated

dahaka5
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1 Answers1

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Substituting the binomial expansion we have:

$$ A=\frac{\mu_0kI}{4\pi}\;\ln\left( \frac{2L^2+\rho^2}{\rho^2}\right) $$ now, since $\rho << L$, the argument of the log becomes

$ \sim 2\left(\frac{L}{\rho}\right)^2 $

and we have

$$ A=\frac{\mu_0kI}{4\pi}\; \left(\ln 2 -2 \ln \rho +2\ln L \right) $$ that, dropping the constant $\frac{\mu_0kI}{4\pi}\; \left(\ln 2 +2\ln L \right) $, is the result in your book.

Emilio Novati
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