Question

(0.0, 0.0, 1.5)m S2 (top) S. (back) (1.0, 0.0, 0.0)m - Sg (right side) S5 (front) S. (bottom) (0.0, 2.0, 0.0)m Figure 3: 3. The current density flowing through the rectangular box shown in Figure 3 is given by J = (2.ri-2y) +ryk) A/m² = (2x, -2y, xy) A/m2. And he current through any surface is I = j.d. (a) The area vector for S3(right side) and Si(left side) arerespectively: (i) + (2.5m²)k, +(1.5m²)ì (ii) +(1.5m²)ì,-(1.5m²)ì (iii) + (2.0m²)ì, –(2.0m²)ì (iv) none of the above (b) On the surfaces S3(right side) and and Si(left side the expression J .dS reads[respectively [(2«, – 2y, xy)A/m². (dxdzī)] = -2.0ydxdz\y=2.0m A/m2 = – 4.Odydz +1 = z=1.5 S2=2.0(-4dydz) = -4.5A [(2.4, –2y, xy)A/m² • (-dzdzi) . = +2.0ydrdz|y=0.0m A/m² = 0.0 — I = 0.0A . (True, False) g=2.0m ly=0.0m (c) The current flowing through all 6 faces of the cube is (i)2.04 (ii)0.04 (iii)4.04 (iv) none of the above

Answer 1

height of the rectangle ( along Z-axis ) , and,

h = 1.5 m

width of the rectangle ( along X-axis ) ,

w = 1 m

length of the rectangle ( along Y-axis ) ,

L = 2.0 m

and current through any surface is given as,

, _______________________________relation 1

=I

where,

$\vec{J}.\vec{ds}$ is infinitesimal current flowing through the infinetesimal area $\vec{ds}$ .

a)

Area vector $\vec{A}$ of a plane surface of area   is given as,

___________________________________________relation 2

A = An

Where,

$\hat{n}=$ unit vector normal to the plane containing area .

we can choose vector $\hat{n}$ pointing on either side of the surface,

but in our case we need to choose the normal unit vectors $\hat{n}$ of surfaces of rectangle which are pointing outward the cube or inward the cube .

looking at options of this question the former convention ( unit vectors pointing outward the rectangle ) has been fallowed,

Otherwise if we choose unit vectors of surfaces otherwise, we need to be extra careful and aware about signs eachtime while doing calculations which becomes an extra and unessesory task.

we have,

length of surface $S_{3}=$   

h = 1.5 m

width of surface  $S_{3}=$  

w = 1 m

area   of surface   $S_{3}=$
h.w = 1.5 x1 = 1.5 m2

Unit vector normal to the surface $S_{3}=$   $\hat{n}_{3}$  =  $\hat{j}$

so as per relation 2 the surface vector $\vec{S}_{3}$ of surface $S_{3}$   will be given as,

S3 = Az n3 = 1.5

Similarly,

area of surface $S_{1}=$ area of surface $S_{3}$ ,

and,

A1 = 43 = 1.5 m

unit vector perpendicular to surface $S_{1}$ =

- =

so as per relation 2 the surface vector $\vec{S}_{1}$ of surface $S_{1}$   will be given as,

S3 = Aj ñ = -1.5

So the answer is,

ii) + (1.5 m²) j, -(1.5 m2)

b)

Answer - False

the calculations after the fallowing step are wrong,

-2.0 y dx dz|y=20mA/m

correct calculations are,

-2.0 y do dzy=2.0 mA/m2 = -4.0 do dz +

p=1.5 pe=1 p=1.5 I = (-4.0 dc dz) = / -4.0[2]. dz = T=0 SO

p=1.5 -4.0 dz = -4.0[211.5 = -4.0 x 1.5 = -6.0 A A=0

everything else is correct and true.

c)

from part b we have currents through surfaces $S_{3}$ and $S_{1}$ of the rectangle,

current $I_{1}$ flowing through surface $S_{1}$

=0.

current $I_{3}$ flowing through surface $S_{3}$   

= -6

now we need to calculate currents through remaining surfaces,

Using relation 1 current $I_{2}$ flowing through surface $S_{2}$ will be given as,

$I_{2}=\int_{S_{2}}\vec{J}.\vec{ds}|_{z=1.5\:m}$

$I_{2}=\int_{S_{2}}(2x\:\hat{i}-2y\:\hat{j}+xy\:\hat{k}).(dx\:dy)\:\hat{k}|_{z=1.5\:m}$

12 = 1 | ry de dyl:=1.5 m Jo Jo

$I_{2}=\int_{0}^{2} y\left[\frac{x^{2}}{2} \right ]_{0}^{1}\:dy=\int_{0}^{2} y\left[\frac{1}{2} \right ]\:dy$

$I_{2}=\frac{1}{2} \left[ \frac{y^{2}}{2}\right ]_{0}^{2}\:dy$

$I_{2}=\frac{1}{2} \left[ \frac{4}{2}\right ]\:dy$

$I_{2}=1\:A$

Similarly,

Using relation 1 current $I_{4}$ flowing through surface $S_{4}$ will be given as,

$I_{4}=\int_{S_{4}}\vec{J}.\vec{ds}|_{z=0\:m}$

$I_{4}=\int_{S_{4}}(2x\:\hat{i}-2y\:\hat{j}+xy\:\hat{k}).(-dx\:dy)\:\hat{k}|_{z=0\:m}$

14 = - 1 my dr dyl:=0 m Jo Jo

$I_{4}=-\int_{0}^{2} y\left[\frac{x^{2}}{2} \right ]_{0}^{1}\:dy=-\int_{0}^{2} y\left[\frac{1}{2} \right ]\:dy$

$I_{4}=-\frac{1}{2} \left[ \frac{y^{2}}{2}\right ]_{0}^{2}\:dy$

$I_{4}=-\frac{1}{2} \left[ \frac{4}{2}\right ]\:dy$

14 = -14

Similarly,

Using relation 1 current $I_{5}$ flowing through surface $S_{5}$ will be given as,

$I_{5}=\int_{S_{5}}\vec{J}.\vec{ds}|_{x=1\:m}$

$I_{5}=\int_{S_{5}}(2x\:\hat{i}-2y\:\hat{j}+xy\:\hat{k}).(dy\:dz)\:\hat{i}|_{x=1\:m}$

$I_{5}=\int_{0}^{1.5}\int_{0}^{2} 2x\:dy\:dz|_{x=1\:m}$

$I_{5}=\int_{0}^{1.5}\int_{0}^{2} 2(1)\:dy\:dz$

$I_{5}=\int_{0}^{1.5} 2\left[y \right]_{0}^{2}\:dz=\int_{0}^{1.5}4\:dz$

15 = 4 [211.5

$I_{5}=4(1.5)$

$I_{5}=6\:A$

Similarly,

Using relation 1 current $I_{6}$ flowing through surface $S_{6}$ will be given as,

$I_{6}=\int_{S_{6}}\vec{J}.\vec{ds}|_{x=0\:m}$

$I_{6}=\int_{S_{6}}(2x\:\hat{i}-2y\:\hat{j}+xy\:\hat{k}).(dy\:dz)\:\hat{i}|_{x=0\:m}$

$I_{6}=\int_{0}^{1.5}\int_{0}^{2} 2x\:dy\:dz|_{x=0\:m}$

$I_{6}=\int_{0}^{1.5}\int_{0}^{2} 2(0)\:dy\:dz$

$I_{5}=0\:A$

So,

net current flowing through all the surfaces is given as,

$I=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}$

$I=(0)+(1)+(-6)+(-1)+(6)+(0)$

$I=0\:A$

so answer is,