Simple line integral:confused with directions

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Homework Help Overview

The discussion revolves around evaluating a line integral involving a vector along a loop, specifically the integral ∫(1/Ω)dl, where Ω represents the distance from the origin to points on the loop. The problem is divided into four regions, with participants attempting to set up the integral for each region and questioning the signs and directions of the vectors involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the integral in different regions, particularly focusing on the expressions for dl and the limits of integration. There is confusion about the cancellation of results from regions 1 and 3, as well as the signs in regions 2 and 4. Questions arise about the correctness of the vector representations and the implications of directionality.

Discussion Status

Some participants are offering guidance on the setup of the integrals and questioning the assumptions made about the directions of the vectors. There is an ongoing exploration of the mathematical representations and their physical meanings, with no clear consensus yet on the correct approach.

Contextual Notes

Participants are grappling with the implications of vector directions and the positivity of the integral, as well as the need for formal mathematical reasoning in the context of calculating a magnetic vector potential.

Aziza
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I am supposed to do the integral ∫(1/Ω)dl, where dl is the vector pointing along the loop and Ω is the distance from the origin to a point on the loop.

Let region 1 be from x=a to x=b
region 2: outer semicircle of radius b
region 3: x=-b to x=-a
region 4: inner semicircle of radius a.

Let's just look at regions 1 and 3:
in these regions, dl=dx x, x is the unit vector. dx x is a positive quantity, since it is parallel (not antiparallel) to the arrows in the above figure drawn in these regions (my path of integration). Also in regions 1 and 3, Ω=x.
Thus i set up the integral in region 1 as:

x∫(1/x)dx , with limits from a->b

and in region 3:

x∫(1/x)dx , with limits from -b->-a.

However, when I add these two results, they cancel, but the solution says they should add. I am having same problem with regions 2 and 4 (they should cancel but in my case they add). Thus I must probably be confused with signs/directions, but i do not see where my error is.

Help is much appreciated!
 
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Hi Aziza! :smile:
Aziza said:
… and in region 3:

x∫(1/x)dx

nooo :wink:
 
tiny-tim said:
Hi Aziza! :smile:nooo :wink:

huh?

if you agree with my region 1 set up, then region 3 is same...except for the limits
 
|x| :wink:
 
tiny-tim said:
|x| :wink:

ohhh ;DDDD yes you are right, thank you!

However i am still having trouble with regions 2 and 4:

region 2 i set up as:
(1/b)∫dl2 from 0->pi

and region 4:
(1/a)∫dl4 from pi->0

Now,
dl2 = bθdθ, where θ is unit vector and positive if pointing in direction of increasing θ (ie counterclockwise)

So since in region 4, θ is pointing clockwise:
dl4 = -aθ

The sum of regions 2 and 4 however then gives me 2∫dθ from 0->pi, but it should give zero...
 
Last edited:
Aziza said:
dl2 = b dθ

that (dθ) doesn't make any sense :redface:

the RHS needs to be a vector, but dθ isn't a vector
 
tiny-tim said:
that (dθ) doesn't make any sense :redface:

the RHS needs to be a vector, but dθ isn't a vector

oops you are right, that is not wat i meant. i edited the post above,
 
Aziza said:
So since in region 4, θ is pointing clockwise:
dl4 = -aθ

why the minus? :confused:

i think you're being too analytical about this whole problem …

just use common-sense: look at the diagram, and remember that ∫dl is always the length (and is always positive) :wink:
 
tiny-tim said:
why the minus? :confused:

i think you're being too analytical about this whole problem …

just use common-sense: look at the diagram, and remember that ∫dl is always the length (and is always positive) :wink:

but in the end what i am calculating is the magnetic vector potential, so the result of that integral must be a vector.

lol i have no common sense, thus i really need the formal mathematical way of doing this
 
  • #10
sorry, i got carried away, i meant to leave out the "∫" :redface:

dl is always positive (ie if you use l instead of θ, you shouldn't get confused)
 

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