# Question on derivation of potential of a small current loop

1. Jun 29, 2013

### yungman

The vector magnetic potential is given
$$\vec A=\frac{\mu I}{4\pi}\oint\frac{e^{-j\beta R_1}}{R_1}dl'$$
After a few steps, the equation becomes:
$$\vec A=\frac{\mu I}{4\pi}e^{-j \beta R}\left[ (1+j\betaR)\oint\frac{dl'}{R_1}-j\beta\oint dl'\right]$$
The Book claim the second integral obviously vanishes!!!

I don't understand this. For a small loop, $dl'=rd\phi$
$$\oint dl'=\int_0^{2\pi}rd\phi=2\pi r$$

That is not zero by any stretch. Can anyone explain this?

thanks

2. Jun 29, 2013

### tiny-tim

hi yungman!

A is a vector, and so is dl'

(and think "polygon" )

3. Jun 29, 2013

### yungman

So I have to integrate the vector in a closed loop and is zero?

Thanks

4. Jun 29, 2013

### tiny-tim

yes, "the vector" is just a bit of the loop

add them end-to-end (as in a polygon, like a vector triangle), and they obviously make zero

5. Jun 29, 2013

### yungman

Thanks for your help Tim. I want to verify:
$$\oint d\vec l'=\int_0^{2\pi}\hat \phi rd\phi=2\pi r=\int_0^{2\pi}(-\hat x \sin\phi+\hat y \cos \phi)r d\phi$$

Integrate sine and cosine from 0 to $2\pi$ are both zero. So the integral is zero.

Last edited: Jun 29, 2013
6. Jun 29, 2013

### tiny-tim

(i'm ignoring the = 2πr = )

yeees

but you're assuming the loop is a circle, and my method works for any loop!

7. Jun 29, 2013

### yungman

Thanks, I understand that it works for non circle. My original question is a circle, that's why.