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Old Apr 27th 2011, 09:05 PM   #1
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Please help in vector potential involve line integral with log.

This is a new question, I am just using the old thread because the tittle applys.

I want to find the vector potential A at origin due to a current segment I(t)=kt flowing along x axis from -b<x<-a where b>a. This mean from -b to -a on the left of the origin.





I did not write the second part because that is not part of the question. But the book said it is:



Please help me on this, thanks

Alan
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Old Apr 28th 2011, 03:42 AM   #2
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Originally Posted by yungman View Post
This is a new question, I am just using the old thread because the tittle applys.

I want to find the vector potential A at origin due to a current segment I(t)=kt flowing along x axis from -b<x<-a where b>a. This mean from -b to -a on the left of the origin.



I did not write the second part because that is not part of the question. But the book said it is:



Please help me on this, thanks

Alan
(I had to edit your quote because the system is a bit feral when it comes to how many images a post can have.)

Your integral is slightly off.
<--For negative x.
(Plus the arbitrary constant, of course.) This is the source of the extra negative sign outside the ln() function.

-Dan
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Old Apr 28th 2011, 07:41 AM   #3
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Originally Posted by topsquark View Post
(I had to edit your quote because the system is a bit feral when it comes to how many images a post can have.)

Your integral is slightly off.
<--For negative x.
(Plus the arbitrary constant, of course.) This is the source of the extra negative sign outside the ln() function.

-Dan
Is sgn(x) the sign of x?

Is this because if x>0 then |x|=x and d|x|/dx=1. But if x<0, then |x|=-x and d|x|/dx=-1. So if we do substitution of U=|x|, dU=sgn(x)dx?

So .

Thanks

Alan

Last edited by yungman; Apr 28th 2011 at 07:45 AM.
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Old Apr 28th 2011, 08:42 AM   #4
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Originally Posted by yungman View Post
Is sgn(x) the sign of x?

Is this because if x>0 then |x|=x and d|x|/dx=1. But if x<0, then |x|=-x and d|x|/dx=-1. So if we do substitution of U=|x|, dU=sgn(x)dx?

So .

Thanks

Alan
Exactly. Note that (integral)dx/|x| will involve ln|x|. This might have been another problem with your result, but since both -a and -b are negative and it is (-b)/(-a) that enters the logarithm this does not matter.

-Dan
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Old Apr 28th 2011, 05:00 PM   #5
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Originally Posted by topsquark View Post
Exactly. Note that (integral)dx/|x| will involve ln|x|. This might have been another problem with your result, but since both -a and -b are negative and it is (-b)/(-a) that enters the logarithm this does not matter.

-Dan

Can you give an example of this? I don't quite get what you saying here.

Thanks
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Old Apr 28th 2011, 05:49 PM   #6
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Originally Posted by yungman View Post
Can you give an example of this? I don't quite get what you saying here.

Thanks
Sure. Let a and b be positive. Then


In this case we need to beware of that negative sign on the -a. The absolute value gets rid of it so we have


Similar to your problem, if we have


the negative signs cancel out anyway, so there is nothing to worry about.

-Dan
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Old Apr 28th 2011, 06:04 PM   #7
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I am confused again, my original question is like your last example:



How can it be your original answer that I thought I understand that



Thanks

Alan

Last edited by yungman; Apr 28th 2011 at 06:07 PM.
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Old Apr 28th 2011, 06:07 PM   #8
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Originally Posted by yungman View Post
How do you post the equation again? The method you showed don't work anymore!!!

I am confused again, my original question is like the last example

Right. I don't mean to make too much out of this. All I'm saying is that the two negative signs cancel so you don't need to worry about the absolute value. Nothing more than that.

To post an equation using LaTeX copy the following line:
[img]http://latex.codecogs.com/png.latex?*LaTeX commands go here*[/img]

-Dan
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Old Apr 28th 2011, 06:09 PM   #9
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Originally Posted by topsquark View Post
Right. I don't mean to make too much out of this. All I'm saying is that the two negative signs cancel so you don't need to worry about the absolute value. Nothing more than that.

To post an equation using LaTeX copy the following line:
[img]http://latex.codecogs.com/png.latex?*LaTeX commands go here*[/img]

-Dan
I modified the last post, please read it again. Somehow the Latex work now!!!
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Old Apr 29th 2011, 07:33 AM   #10
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Originally Posted by yungman View Post
Note: I have removed your quote. Apparently there is a limit to the number of tex statements we can put in the posts and the ones in the quote count.
Oh, I'm sorry. First we have to change this back to the original integral:


This last step follows from both limits on the integral are negative, x is negative so sgn(x) = -1. And the negative signs in the ln cancel. Also, we can remove the absolute value bars since both a and b are positive.

Now recall that


So your integral becomes:




-Dan
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