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Hardik Batra
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I have attached a problem.

∫dl = A1B1 = 1

Here dl is a length of wire but in problem it takes distance between the point A1 and B1

Why does this happen ?
 

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its important to notice that the integral contains vector dl and not just the magnitude of dl.

You can view that integral as a sum of vectors where the end of each vector is the start of the next one. We know from vector addition that the sum is the vector that has as start the start of the first vector and end the end of the last vector (for example for 3 vectors AB, BC and CD its easy to se that the vector sum AB+BC+CD=AD). The start of the "first dl" vector is A1 and the end of the "last dl" is B1. Hence that integral is the vector A1B1.
 
Delta² said:
its important to notice that the integral contains vector dl and not just the magnitude of dl.

You can view that integral as a sum of vectors where the end of each vector is the start of the next one. We know from vector addition that the sum is the vector that has as start the start of the first vector and end the end of the last vector (for example for 3 vectors AB, BC and CD its easy to se that the vector sum AB+BC+CD=AD). The start of the "first dl" vector is A1 and the end of the "last dl" is B1. Hence that integral is the vector A1B1.

From figure you can see than...
The length of wire is greater than 1m.
because A1B1 = 1m. and wire is not straight.
 
Hardik Batra said:
From figure you can see than...
The length of wire is greater than 1m.
because A1B1 = 1m. and wire is not straight.
All true, but if you view things as Delta² suggests, you can view that integral as a sum of vectors aligned head to tail. The sum of those vectors, the resultant, is just the line A1B1.

You can also realize that the components of the vectors perpendicular to the line A1B1 cancel out.
 
If it was [tex]\int dl[/tex] then it would be equal to length of the wire. But it is

[tex]\int \vec{dl}[/tex] so it is equal to the vector [tex]\vec{A_1B_1}[/tex]. The reason is as i said in my first post , you can view Doc Al's reason as well. I think the critical thing you "filtering" in your mind is that we have an integral of infinitesimal vectors [tex]\vec{dl}[/tex] and not an integral of infinitesimal line segments [tex]dl[/tex]
 
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