problem set 6 - University of Toronto

UNIVERSITY OF TORONTO
FACULTY OF APPLIED SCIENCE AND ENGINEERING
The Edward S. Rogers Sr. Department of
Electrical and Computer Engineering
ECE357H1S – ELECTROMAGNETIC FIELDS
PROBLEM SET 6
Topics:
Review of vector calculus and operations.
Reading:
2.1-2.9
1.
Cheng P. 2-21
2.
Cheng P. 2-34
3.
Problem
3.44
Each of thevector
following
vector
fields is displayed
in Fig.
in the
Each of
the following
fields
is displayed
in the form
of aP3.44
vector
form
of
a
vector
representation.
Determine
∇
·
A
analytically
and
then
compare
thewith
representation. Determine div A analytically and then compare the result
result
with
your expectations
the basis
the displayed
pattern.
your
expectations
on theon
basis
of theofdisplayed
pattern.
(a) A = −x̂ cos x sin y + ŷ sin x cos y, for −π ≤ x, y ≤ π
a.
Figure P3.44(a)
Solution:
A = −x̂ cos x sin y + ŷ sin x cos y
∂ Ax ∂ Ay
∇·A =
+
∂x
∂y
∂
∂
(− cos x sin y) + (− sin x cos y)
=
∂x
∂y
= sin x sin y − sin x sin y = 0
Yes, A is divergenceless everywhere.
(b) A = −x̂ sin 2y + ŷ cos 2x, for −π ≤ x, y ≤ π
b.
Figure P3.44(b)
(c) A =
Solution:
−x̂ xy + ŷ y2 ,
for −10 ≤ x, y ≤ 10
A = −x̂ sin 2y + ŷ cos 2x
∂ Ax ∂ Ay
∇·A =
+
∂x
∂y
∂
∂
(− sin 2y) + (cos 2x) = 0
=
∂x
∂y
Yes, A is divergenceless everywhere.
c.
Figure P3.44(c)
(d) A = −x̂ cos x + ŷ sin y, for −π ≤ x, y ≤ π
Solution:
A = −x̂ xy + ŷ y2
∂ Ax ∂ Ay
∇·A =
+
∂x
∂y
∂
∂
(−xy) + (y2 ) = −y + 2y = y
=
∂x
∂y
NO, A is not divergenceless everywhere. It is divergenceless only at y = 0.
d.
Figure P3.44(d)
Solution:
A = −x̂ cos x + ŷ sin y
∂ Ax ∂ Ay
∇·A =
+
∂x
∂y
∂
∂
(− cos x) + (sin y) = sin x + cos y
=
∂x
∂y
(e) A = x̂ x, for −10 ≤ x ≤ 10
e.
(f) A =
Solution:
Figure P3.44(e)
x̂ xy2 ,
for −10 ≤ x, y ≤ 10
A = x̂ x
∂ Ax ∂ Ay ∂ Az
+
+
∂x
∂y
∂z
=1
∇·A =
This indicates that the divergence of A is the same at all points in the defined space.
In other words, every small volume is a source of flux (more flux leaving the volume
than entering it), and the net generated flux is the same at all locations.
f.
(g) A =
Solution:
Figure P3.44(f)
x̂ xy2 + ŷ x2 y,
for −10 ≤ x, y ≤ 10
A = x̂ xy2
∂ Ax ∂ Ay ∂ Az
+
+
∂x
∂y
∂z
2
=y
∇·A =
g.
Figure P3.44(g)
Solution:
A = x̂ xy2 + ŷ x2 y
∂ Ax ∂ Ay ∂ Az
∇·A =
+
+
∂x
∂y
∂z
2
2
= y +x
(h)
A = x̂ sin
! πx "
10
+ ŷ sin
! πy "
10 , for −10 ≤ x, y ≤ 10
h.
Figure P3.44(h)
(i) A = r̂ r + φ̂φ r cos φ , for
Solution:
!
0 ≤ r ≤ 10
0 ≤ φ ≤ 2π .
A = x̂ sin(π x/10) + ŷ sin(π y/10)
∂ Ax ∂ Ay ∂ Az
∇·A =
+
+
∂x
∂y
∂z
π
= [cos(π x/10) + cos(π y/10)]
10
i.
(j) A =
Solution:
Figure P3.44(i)
r̂ r2 + φ̂φ r2 sin φ ,
for
!
0 ≤ r ≤ 10
0 ≤ φ ≤ 2π .
A = r̂ r + φ̂φ r cos φ
1 ∂
1 ∂ Aφ ∂ Az
+
(rAr ) +
r ∂r
r ∂φ
∂z
= 2 − sin φ
∇·A =
j.
Figure P3.44(j)
Solution:
A = r̂ r2 + φ̂φ r2 sin φ
∇·A =
1 ∂
1 ∂ Aφ ∂ Az
+
(rAr ) +
r ∂r
r ∂φ
∂z
4.
Repeat problem 3 for curl A.
5.
For the vector field
Problem 3.46 For the vector field E = x̂xz − ŷyz2 − ẑxy, verify the divergence
heorem by computing:
verify the divergence theorem by computing:
(a) the total outward flux flowing
the surface
of a cubethrough
centeredthe
at surface
the
(a) the through
total outward
flux flowing
of a cube centered at the
origin and with sides equal origin
to 2 units
parallel
andeach
withand
sides
equaltotothe2 Cartesian
units eachaxes,
and parallel to the Cartesian axes,
and
and
(b) the integral of ∇ · E over(b)
thethe
cube’s
volume.
integral
of div E over the cube’s volume.
Solution:
6.
Verifyintegral
Stokes’s
(a) For a cube, the closed
surface
has theorem
6 sides: for the vector field
Problem
Stokes’s! theorem for the vector field B = (r̂r
cos φ + φ̂φ3.52
sin φ )Verify Stokes’s theorem for the vector field B = (r̂r cos φ + φ̂φ sin φ )
by evaluating:
!
by+evaluating:
E · ds = Ftop + Fbottom
Fright + Fleft + Ffront
+ Fback ,
B · dl over the semicircular contour shown in Fig. P3.52(a), and
(a)
(a)"Fig.
theP3.52(a),
closed contour
semicircular contour
and#$! C integral of B over the semicircular contour shown below,
! 1 shown
! 1 in
2
$ · (ẑ dy dx)
and,
Ftop =
x̂xz
− ŷyz − ẑxy
× B) · ds over the surface of the semicircle.
(b) z=1(∇×
y=−1
er the surface of thex=−1
semicircle.
S of curl B over the surface of the semicircle.
(b) the surface%integral
($1
& 2 2 '$1
! 1 ! 1
$
x y $$
$
= 0,y
=−
xy dy dx =
$
$
y
4
x=−1 y y=−1
y
y=−1 $
♥
♥
x=−1
2
L3
Fbottom
L2 =
0 L1
(a)
2
! 1
! 1
2
x=−1 y=−1
L3
! 1
=x
! 11
! 1
! 1
Fright =
0
L4
x=−1 y=−1
"
#$
2
$
x̂xz
L − ŷyz − ẑxy z=−1 · (−ẑ dy dx)
2
2
L2
2
L1
xy dy dx =x
1
" (b)
2
%&
($1
'$1
$
$
$
$
$ -2 =L0,3 0
$
4
y=−1 $
1
x2 y2
x=−1
#$
x̂xz − ŷyz2 − ẑxy $
(a)
L1
2
x
0
L2
L3
L4
L1
1
2
x
(b)
· (ŷ dz dx)
Figure P3.52: Contour paths for (a) Problem 3.52 and
($
(b) Problem 3.53.
'$
$1
3 $1
xz
−4
$
$
,
=−
z2 dz dx = −
=
$
3 $z=−1 $
3
x=−1 z=−1
Solution:
x=−1
! 1 ! 1 "
#$
(a)
Fleft =
x̂xz − ŷyz2 − ẑxy $y=−1 · (−ŷ dz dx)
!
!
!
!
x=−1 z=−1
!
!
($
%
1
&
'$
B · dl,
B · dl1=
!
! 1
$B · dl + −4B · dl +
xz3 $$
L2
L3
L1 $
B · dl,1
l+
B · dl +
2
=
=
−
z
dz
dx
=
−
,
$
L2
L3
3B · dl$z=−1
3 φ ) · (r̂ dr + φ̂
x=−1 z=−1
φr d φ + ẑ dz) = r cos φ dr + r sin φ d φ ,
= (r̂r $cos φ + φ̂φ sin
"! 0
#$
#$
+ φ̂φ sin φ ) · (r̂ dr +!φ̂φr1 d φ!+1ẑ dz) = r cos φ dr + r!sin φ d φ , "! x=−1
2
$
$
"
#$
"! 0
#$ 2
#$
$
$
B ··dl
+
r sin φ d φ $$
$Ffront =
x̂xz − ŷyz$ − ẑxy x=1
(x̂=
dz dy) r cos φ dr $
L1
r=0
φ =0
y=−1 z=−1
cos φ dr $$
+
r sin φ d φ $$
φ =0, z=0
z=0
($1 % &$2
%
φ
=0
φ =0, z=0!
z=0
'$
&
!
1
$ 1 r2 $ + 0 = 2,
1
1
=
yz2 $$
$ 2
r=0
0,
=
z dz dy =
$ "! =
#$
#$
"! π
+
0
=
2,
!
$
2
=0
$
$
2
z=−1
z=−1 $y=−1
#$
#$
"!y=−1
$
π
r sin φ d φ $$
B · dl =
r cos φ dr $ +
$
$
L2
r=2
φ =0
r sin φ d φ $$
cos φ dr $$ +
z=0
r=2, z=0
φ =0
π
z=0
r=2, z=0
= 0 + (−2 cos φ )|φ =0 = 4,
#$
"! 0
"! π
#$
cos φ )|πφ =0 = 4,
!
$
$
#$
"! π
#$
B · dl =
r cos φ dr $$
+
r sin φ d φ $$
$
$
L3
r=2
φ =π
φ =π ,z=0
z=0
cos φ dr $$
+
r sin φ d φ $$
%
&$
=
π
φ
φ =π ,z=0
z=0
1 2 $0
= − 2 r r=2 + 0 = 2,
$0
!
$ + 0 = 2,
r=2
B · dl = 2 + 4 + 2 = 8.
2 = 8.
z=−1
e P3.52: Contour pathsx=−1
for (a)
Problem 3.52 and %
&
! 1
1
(b) Problem! 3.53.
y=1
♥
♥

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