Question

How about Answer ?

Answer 1

See below

Explanation:

Sorting out the units in the current density formulation:

• `-10^4 mA 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:

• `mathbf hat n = ((0),(1),(0))`

Following the definition of flux:

`I = int_S \ mathbf J * mathbf hat n \ dS`

`= int_S \ color(red)(-10) ((sin (2x) \ e^(-2y)),(cos (2x) \ e^(-2y)),(0)) * ((0),(1),(0)) \ dS`

`= - 10 int_S \ cos (2x) \ e^(-2y) \ dS`

On surface `S`:

• `y =1 implies e^(- 2y) = 1/e^2`.

• `dS = dx \ dz`.

`implies I = - 10/e^2 int_0^2 int_0^1 \ cos (2x) \ dx \ dz`

`= - 10/e^2 int_0^2 \ (( sin (2x))/2)_0^1 \ dz`

`= - 10/e^2 \ ( sin (2))/2 \ int_0^2 \ \ dz`

`= - 10/e^2 \ ( sin (2))/2 * 2`

`approx - 1.23 " A"`

Answer 2

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) sin2`
`=>I=-10^4 xx0.1353xx 0.9093`
`=>I=-1230\ kA`
`=>I=-1.230\ MA`

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

Link deleted as it is not longer required.