Solution:
Key concept: Ampere’s law gives another method to calculate the magnetic field due to a given current distribution.
Line integral of the magnetic field $\vec{\text{B}}$ around any closed curve is equal to $\mu_0$ times the net current i threading through the area enclosed by the curve, i.e.,
$\oint\vec{\text{B}}.\vec{\text{dI}}=\mu_0\sum\text{i}=\mu_0(\text{i}_1+\text{i}_3-\text{i}_2)$
Total current crossing the above area is (i1 + i3 - i2). Any current outside the area is not included in net current. $(\text{Outward}\ \circledcirc\rightarrow\ +\text{ve},\text{ Inward}\ \otimes\ \rightarrow\ -\text{ve})$
Applying the Ampere's circuital law, we have
$\oint\ \text{B.dl}=\text{i}_0(\text{I}-\text{I})=0$ (because current is in opposite sense.)
Also, there may be a point on C where B and dl are perpendiclular and hence,
$\oint\limits_\text{c}\text{B.dI}=0.$
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