# Math Help - Calculate the line integral over a vector field

1. ## Calculate the line integral over a vector field

Hello!
I encountered the following problem in an exercise book for Electro magnethism. It is in a section that is used to refresh ones mathematical knowledge obtained in Calculus.

The vector field H = ( I / ( 2 pi * R) ) * phi is given in cylindrical coordinates.
The curve C is the circumference to the circle: R = a, z = 0, 0 <= pi <= 2 pi.

Calculate the line integral over the closed integral C of H* dl

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Anyone that can give me some hints of the methodology to use? I have tried to solve it, but I do not succeed.

Thanks!

Anders Branderud
bloganders.blogspot.com

2. Originally Posted by andersbranderud
Hello!
I encountered the following problem in an exercise book for Electro magnethism. It is in a section that is used to refresh ones mathematical knowledge obtained in Calculus.

The vector field H = ( I / ( 2 pi * R) ) * phi is given in cylindrical coordinates.
Do you mean $\vec{H}= \frac{\phi}{2\pi R}\vec{i}$?

The curve C is the circumference to the circle: R = a, z = 0, 0 <= pi <= 2 pi.
Parametric equations for the circle are $x= a cos(\phi)$, $y= a sin(\phi)$, z= 0 so that $d\vec{s}= (-a sin(\phi)\vec{i}+ a cos(\phi)\vec{j})d\phi$ (Yes, $0\le \pi\le 2\pi$ but that doesn't make sense here. I presume you meant $0\le \phi\le 2\pi$.)

Calculate the line integral over the closed integral C of H* dl
Take the dot product of $\vec{H}$ and $d\vec{s}$ and integrate with respect ot $\phi$ from 0 to $2\pi$.[/quote]
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Anyone that can give me some hints of the methodology to use? I have tried to solve it, but I do not succeed.

Thanks!

Anders Branderud
bloganders.blogspot.com[/QUOTE]

3. HallsofIvy, Thanks alot for your help!

Originally Posted by HallsofIvy
Do you mean $\vec{H}= \frac{\phi}{2\pi R}\vec{i}$?
Yes, that is correct.
I saw now that they write: "Intensity is often denoted with the scalar <I>I</I>, but one should remember that it actually is a vector <B>I</B>.

Anders Branderud
bloganders.blogspot.com