Apr 29, 2018

See below

#### Explanation:

Sorting out the units in the current density formulation:

• $- {10}^{4} m A \equiv - 10 A$

Surface $S$ is on the plane $y = 1$, so has unit normal vector, in the direction in which flux $I$ is to be calculated, as follows:

• $m a t h b f \hat{n} = \left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right)$

Following the definition of flux:

$I = {\int}_{S} \setminus m a t h b f J \cdot m a t h b f \hat{n} \setminus \mathrm{dS}$

$= {\int}_{S} \setminus \textcolor{red}{- 10} \left(\begin{matrix}\sin \left(2 x\right) \setminus {e}^{- 2 y} \\ \cos \left(2 x\right) \setminus {e}^{- 2 y} \\ 0\end{matrix}\right) \cdot \left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right) \setminus \mathrm{dS}$

$= - 10 {\int}_{S} \setminus \cos \left(2 x\right) \setminus {e}^{- 2 y} \setminus \mathrm{dS}$

On surface $S$:

• $y = 1 \implies {e}^{- 2 y} = \frac{1}{e} ^ 2$.

• $\mathrm{dS} = \mathrm{dx} \setminus \mathrm{dz}$.

$\implies I = - \frac{10}{e} ^ 2 {\int}_{0}^{2} {\int}_{0}^{1} \setminus \cos \left(2 x\right) \setminus \mathrm{dx} \setminus \mathrm{dz}$

$= - \frac{10}{e} ^ 2 {\int}_{0}^{2} \setminus {\left(\frac{\sin \left(2 x\right)}{2}\right)}_{0}^{1} \setminus \mathrm{dz}$

$= - \frac{10}{e} ^ 2 \setminus \frac{\sin \left(2\right)}{2} \setminus {\int}_{0}^{2} \setminus \setminus \mathrm{dz}$

$= - \frac{10}{e} ^ 2 \setminus \frac{\sin \left(2\right)}{2} \cdot 2$

$\approx - 1.23 \text{ A}$

Apr 30, 2018

As explained by @ultrilliam we need to take argument of $\sin$ term as in radians to get the posted answer.

$I = - {10}^{4} {e}^{- 2} \sin 2$
$\implies I = - {10}^{4} \times 0.1353 \times 0.9093$
$\implies I = - 1230 \setminus k A$
$\implies I = - 1.230 \setminus M A$

Only problem is how on earth the student will know this !

Link deleted as it is not longer required.