The net current in the conductors and the value of the line integral

- The resultant remains same 3.2 *10^4 Tm
This is further explained below.
<h3>What is the net current in the conductors?</h3>
Generally,
To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).


B)
In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.
The resultant remains the same at 3.2 *10^4 Tm
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