Answer
The flux:
$\Phi \left( t \right) \approx - 9.66 \times {10^{ - 7}}t\left( {12 - t} \right)$ ${\rm{T - }}{m^2}$
The voltage drop:
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = 9.66 \times {10^{ - 7}}\left( {12 - 2t} \right)$ volts.
Work Step by Step
Similar to Example 5, we have the rectangle region ${\cal R}$ given by
${\cal R} = \left\{ {\left( {x,y} \right):0 \le x \le 3,d \le y \le 2 + d} \right\}$
Since $d = 0.5$, so
${\cal R} = \left\{ {\left( {x,y} \right):0 \le x \le 3,0.5 \le y \le 2.5} \right\}$.
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
$\Phi \left( t \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\bf{B}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{x = 0}^L \mathop \smallint \limits_{y = d}^{2 + d} {\bf{B}}\cdot{\bf{N}}{\rm{d}}y{\rm{d}}x$
$\Phi \left( t \right) = - \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0.5}^{2.5} \dfrac{{{\mu _0}i}}{{2\pi y}}{\rm{d}}y{\rm{d}}x$
$\Phi \left( t \right) = - \dfrac{{{\mu _0}i}}{{2\pi }}\mathop \smallint \limits_{x = 0}^3 {\rm{d}}x\mathop \smallint \limits_{y = 0.5}^{2.5} \dfrac{1}{y}{\rm{d}}y = - \dfrac{{3{\mu _0}i}}{{2\pi }}\left( {\ln y} \right)|_{0.5}^{2.5} = - \dfrac{{3{\mu _0}i}}{{2\pi }}\ln 5$
Since $i\left( t \right) = t\left( {12 - t} \right)$ and ${\mu _0} = 4\pi \cdot{10^{ - 7}}$, so
$\Phi \left( t \right) = - \dfrac{{3\left( {4\pi \cdot {{10}^{ - 7}}} \right)\ln 5}}{{2\pi }}t\left( {12 - t} \right) \approx - 9.66 \times {10^{ - 7}}t\left( {12 - t} \right)$
So, the flux is $\Phi \left( t \right) \approx - 9.66 \times {10^{ - 7}}t\left( {12 - t} \right)$ ${\rm{T - }}{m^2}$.
Next, we use Faraday's Law to determine the voltage drop around the rectangular loop:
By Eq. (5):
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - \dfrac{d}{{dt}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\bf{B}}\cdot{\rm{d}}{\bf{S}}$
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = - \dfrac{d}{{dt}}\left( { - 9.66 \times {{10}^{ - 7}}t\left( {12 - t} \right)} \right)$
$\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = 9.66 \times {10^{ - 7}}\left( {12 - 2t} \right)$
So, the voltage drop is $\mathop \smallint \limits_C^{} {\bf{E}}\cdot{\rm{d}}{\bf{r}} = 9.66 \times {10^{ - 7}}\left( {12 - 2t} \right)$ volts.
