- #1

Bassalisk

- 947

- 2

I posted an actual problem in advanced physics but no answer so i will try to get an math part answer from it.

Suppose I have to solve this integral:

[itex] I=\int {\vec{dl} × \vec A } [/itex]

Where [itex] \vec A = -\frac {1}{x} \vec a_{z}[/itex]

So it has only a z component and I have to find the vector cross of the field with the contour depicted in the picture

http://pokit.org/get/img/7ea5523fdd63cfc5c9374a07f6ab8bb6.jpg [Broken]

First question:

For the part 3:

If I take that the [itex] \vec dl = - dy \vec {a_y} [/itex] will I integrate from b to c or from c to b?

In my intuition if I took the minus into the account in the differential there is no need for flipping the c and b, so we would integrate from c to b?

When you vector cross this all, there shouldn't be net y component am i right?

Link to thread in physics:

https://www.physicsforums.com/showthread.php?t=630013

Suppose I have to solve this integral:

[itex] I=\int {\vec{dl} × \vec A } [/itex]

Where [itex] \vec A = -\frac {1}{x} \vec a_{z}[/itex]

So it has only a z component and I have to find the vector cross of the field with the contour depicted in the picture

http://pokit.org/get/img/7ea5523fdd63cfc5c9374a07f6ab8bb6.jpg [Broken]

First question:

For the part 3:

If I take that the [itex] \vec dl = - dy \vec {a_y} [/itex] will I integrate from b to c or from c to b?

In my intuition if I took the minus into the account in the differential there is no need for flipping the c and b, so we would integrate from c to b?

When you vector cross this all, there shouldn't be net y component am i right?

Link to thread in physics:

https://www.physicsforums.com/showthread.php?t=630013

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