Answer
$\int~B\cdot ds = -2.5\times 10^{-6}~T\cdot m$
Work Step by Step
By the right hand rule for the direction of integration, current that is directed upward is positive.
We can find $\int~B\cdot ds$:
$\int~B\cdot ds = \mu_0~i_{enc}$
$\int~B\cdot ds = (\mu_0)~(i_2-i_1)$
$\int~B\cdot ds = (4\pi\times 10^{-7}~H/m)~(3.0~A-5.0~A)$
$\int~B\cdot ds = (4\pi\times 10^{-7}~H/m)~(-2.0~A)$
$\int~B\cdot ds = -2.5\times 10^{-6}~T\cdot m$
