Answer
$5.2\times10^{-8}\,T\cdot m$
Work Step by Step
Ampere's law states that
$\int \vec{B}\cdot d\vec{s}=\mu_{0}i_{enc}$
But $i_{enc}=Total\,current\,i\times\frac{encircled\,area\,HW}{full\,plate\,area\,L^{2}}$
Therefore, we get
$\int \vec{B}\cdot d\vec{s}=\mu_{0}\times i\times\frac{WH}{L^{2}}$
Plug the given values in the equation to get
$\int \vec{B}\cdot d\vec{s}=(4\pi\times10^{-7}\,T\cdot m/A)\,0.75\,A\times\frac{4.0\,cm\times2.0\,cm}{(12\,cm)^{2}}$
$=5.2\times10^{-8}\,T\cdot m$
