Question

For the magnetic field around a long, straight wire carrying a steady current, the line integral of the circular B-field, at some distance,r,from the wire, is given as (integral)B.dL, where dL is an infinitesimaly small displacement vector that points along the same direction as the B-field. This evaluates to B.2(pi) r, where 2(pi)r is the total path length. But, if dL is an infinitesimal displacement vector,and a journey around the loop will bring you back to the start point, why isn't the total displacement equal to zero, instead of 2(pi)r?

Answer 1

probably its not a conservative field

Answer 2

\[\int_0^r\int_0^{2\pi}\vec B\cdot d\vec L\]and as anonymous said, \(\vec B\) is not conservative around a current carrying wire

Answer 3

really more of a physics question

Answer 4

The magnetic field produced by a current carrying wire wraps around the wire in a circle.
So, as you say, B and DL are always parallel, which makes B dot dL a positive number at every step as you go around the loop. In fact, according to Ampere, you always get the same answer, regardless of the path you use to do the line integral, as long as you go all the way around the wire.

\[\int\limits_{?}^{?} B\cdot dL = \mu _{0}I\]