Find `oint B. dl` for the paths shown in (a) and (b)

Explanation

a) Going around the path in the anticlockwise direction, `I_(1)` is taken as positive while `I_(3)` is negative. Currents `I_(2) and I_(4)` do not matter as they are NOT enclosed by the path.
`oint B. dl=mu_(0) (I_(1)-I_(3))`
Note: Currents `I_(2) and I_(4)` create magnetic fields fall around them and B due to them on any element of the path would be non-zero. However, the sum B. dl due to them would be zero.
b) Calculation of B. dl for the entire path can broken up into two separate calculations, one covering all contributions from an anti-clockwise traversal around `I_(1)` and the other covering all contributions from a clockwise traversal around `I_(3)`. Thus
`underset(1) int B. dl = mu_(0) I_(1)` for all elements around `I_(1)` traversed in an anti-clockwise direction
`underset(2) int B. dl =mu_(0)I_(3)` for all elements around `I_(3)` traversed in a clockwise direction direction, `I_(3)` taken as positive because it is flowing into the plane. Thus the total `oint B. dl =mu_(0) (I_(1)-I_(3))`