Answer
$
\mu_0 i_d\left(\frac{\text { enclosed area }}{\text { total area }}\right)\\=52 \mathrm{nT} \cdot \mathrm{m} .
$
Work Step by Step
6. The integral of the field along the indicated path is, equal to
$
\mu_0 i_d\left(\frac{\text { enclosed area }}{\text { total area }}\right)\\=\mu_0(0.75 \mathrm{~A}) \frac{(4.0 \mathrm{~cm})(2.0 \mathrm{~cm})}{12 \mathrm{~cm}^2}\\=52 \mathrm{nT} \cdot \mathrm{m} .
$
