Answer
The flux:
$\Phi \left( t \right) \approx - 1.56 \times {10^{ - 5}}{{\rm{e}}^{ - 0.1t}}$ ${\rm{T - }}{m^2}$
The voltage drop:
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - 1.56 \times {10^{ - 6}}{{\rm{e}}^{ - 0.1t}}$ volts
Work Step by Step
Referring to Figure 17, and also see the figure attached, we can consider the triangular region ${\cal R}$ be defined by
${\cal R} = \left\{ {\left( {x,y} \right):3 \le y \le 9,y - 9 \le y \le 9 - y} \right\}$
The parametrization of ${\cal R}$ is given by $G\left( {x,y} \right) = \left( {x,y} \right)$.
Referring to Example 5 and Figure 13(B), we have ${\bf{N}} = {\bf{k}}$. Since ${\bf{B}}$ points into the page, so
${\bf{B}} = - \dfrac{{{\mu _0}i}}{{2\pi y}}{\bf{k}}$
By Eq. (3), the flux $\Phi \left( t \right)$ of ${\bf{B}}$ through ${\cal R}$ at time t is
$\Phi \left( t \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\bf{B}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{y = 3}^9 \mathop \smallint \limits_{x = y - 9}^{9 - y} {\bf{B}}\cdot{\bf{N}}{\rm{d}}x{\rm{d}}y$
$\Phi \left( t \right) = - \mathop \smallint \limits_{y = 3}^9 \mathop \smallint \limits_{x = y - 9}^{9 - y} \dfrac{{{\mu _0}i}}{{2\pi y}}{\rm{d}}x{\rm{d}}y$
$\Phi \left( t \right) = - \dfrac{{{\mu _0}i}}{{2\pi }}\mathop \smallint \limits_{y = 3}^9 \dfrac{1}{y}\left( {18 - 2y} \right){\rm{d}}y = - \dfrac{{{\mu _0}i}}{{2\pi }}\left( {\mathop \smallint \limits_{y = 3}^9 \dfrac{{18}}{y}{\rm{d}}y - \mathop \smallint \limits_{y = 3}^9 2{\rm{d}}y} \right)$
$\Phi \left( t \right) = - \dfrac{{{\mu _0}i}}{{2\pi }}\left( {18\ln y|_3^9 - 12} \right) = - \dfrac{{{\mu _0}i}}{{2\pi }}\left( {18\ln 3 - 12} \right)$
Since $i\left( t \right) = 10{{\rm{e}}^{ - 0.1t}}$ and ${\mu _0} = 4\pi \cdot{10^{ - 7}}$, so
$\Phi \left( t \right) = - \dfrac{{4\pi \cdot {{10}^{ - 7}}}}{{2\pi }}\left( {10{{\rm{e}}^{ - 0.1t}}} \right)\left( {18\ln 3 - 12} \right) \approx - 1.56 \times {10^{ - 5}}{{\rm{e}}^{ - 0.1t}}$
So, the flux is $\Phi \left( t \right) \approx - 1.56 \times {10^{ - 5}}{{\rm{e}}^{ - 0.1t}}$ ${\rm{T - }}{m^2}$.
Next, we use Faraday's Law to determine the voltage drop around the rectangular loop:
By Eq. (5):
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - \dfrac{d}{{dt}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\bf{B}}\cdot{\rm{d}}{\bf{S}}$
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - \dfrac{d}{{dt}}\left( { - 1.56 \times {{10}^{ - 5}}{{\rm{e}}^{ - 0.1t}}} \right)$
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - 1.56 \times {10^{ - 6}}{{\rm{e}}^{ - 0.1t}}$
So, the voltage drop is $\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - 1.56 \times {10^{ - 6}}{{\rm{e}}^{ - 0.1t}}$ volts.
