Problem 1-7 Wangsness Electromagnetic Fields Show that A~ (B~ C~ ) equals the volume of a parallelepiped whose sides at a common corner are parallel to the vectors A~ , B~ and C~ . Solution:
The vector product B~ C~ is numerically equal to the product of the lengths B and C and the sine of the angle between them. This is equal to the base B times height A sin AB of the wall which they bound. This is just the area of this wall which we shall refer as the \base" of the parallelepiped. Dotting this with A~ gives this area times the length of A times the cosine of the angle between A~ and the normal to this base. So the overall product is the base height of the parallelepiped which in turn is just the volume.
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Problem 1-8 Wangsness Electromagnetic Fields A family of hyperbolae in the xy plane is given by u = xy. a) Find r~ u. b) Find the component of the vector A~ = 3 x^ + 2 y^ + 4 z^ in the direction of r~ u at the point on the curve for which u = 3 and x = 2. Solution: a)
@u x^ + @u y^ + @u z^ r~ u = @x @y @z where u = xy. Therefore
r~ u = y x^ + x y^ b) The component of the vector A~ = 3 x^ + 2 y^ + 4 z^ in the direction of r~ u is ~ u= r ~ u = p3y 2+ 2x2 A~ r x +y When u = 3 and x = 2 then xy = 2y = 3 so that x = 2 and y = 3=2. Therefore ~ u = 17=5: A~ r
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Problem 1-9 Wangsness Electromagnetic Fields The equation giving a family of ellipsoids is: 2 2 2 u = xa2 + yb2 + zc2
Find the unit vector normal to each point of the surface of these ellipsoids. Solution:
The unit normal to a surface u(~r) =constant is
~ n^ = r~ u ru In the present case:
r~ u = 2 ax2 x^ + by2 y^ + cz2 z^ Therefore the normal at point (x; y; z ) is
n^ =
x x^ + y y^ + z z^ x2 + y2 + z2 a2 b2 c2 a4 b4 c4
4
!
1 2
Problem 1-13 Wangsness Electromagnetic Fields A vector eld is given by A~ (~r) = xy x^ + yz y^ + zx z^
a) Evaluate the ux through the rectangular parallelepiped sides a; b; c respectively in the x; y; z dimensions and with corner at the origin and in the rst octant. ~ over the volume of this same parallelepiped b) Evaluate r~ Ad and compare your result with that from part a). R
Solution: a) The ux through the rectangular parallelepiped is Z
S
A~ d~a
where S is the whole surface. For the side I at x = 0 the vector d~a = dydzx^, for the side at y = 0 then d~a = dxdzy^ and for the side at z = 0 then d~a = dxdyz^. The sides opposite these have these vectors reversed in sign. Therefore the surface integral is Z
S
A~ d~a =
b
Z
Z
Z
c a
dy dz dx
c
Z
Z
a
Z
b
dz xy
Z
x=a
dx yz
y =b
dy zx
z=c
dy
Z
c
dz xy
+
x=0 Z c Z a dz dx yz + y=0 Z a Z b dx dz zx z=0
= 21 ab2 c + abc2 + a2 bc
5
b
b) The divergence of A~ is y + z + x. Therefore the integral over the volume of the parallelepiped V is Z
V
~ = r~ Ad
a
Z
dx
b
Z
dy
Z
c
dz(y + z + x)
= 12 a2 bc + ab2 c + abc2 This is the same result as in a) as we would expect from the divergence theorem.
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Problem 1-15 Wangsness Electromagnetic Fields A vector eld is given by
A~ (~r) = x2y x^ + xy2 y^ + a3 e y cos x z^ where a; and are constants (independent of x; y; z).
a) Evaluate the line integral over the closed loop C in the plane z = 0 along the curve y2 = kx from x = 2 to x = 0, then along the y axis back to the y at which x = 2 and then at constant y back to the starting point. b) Evaluate S (r~ A~ ) d~a over the at surface enclosed by the loop C and compare your result with that from part a). R
Solution: a) The line integral along the curved part (y2 = kx) is rst. We have s
y = kx ; 2ydy = kdx ; dy = 12 dx xk Therefore this integral is 2
x=
Z
x=2
(Ax dx + Ay dy) = =
x=
Z
x=2Z
p
k
2
x2 y dx + xy2 dy
dx x5=2 + 12 kx3=2
p 16 4 7 5 = k 27 x 2 + 15 kx 2 = 2k 7 + 5 k 2 R Along the y axis the integral is Ay dy at x = 0 which ispzero. The part of the integral parallel to the x axis is at y = 2k and is Z 2 Z 2 p Axdx = x2ydx = 2k 83 The path integral around C is therefore p 8 4 2k 21 5k
1 2
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b) The curl of A~ is
r~ A~ = a3e y cos x x^ + a3 e y sin x y^ + (y2 x2) z^ The vector d~a is dxdyz^ so that the integral of the curl over the surface is Z
S
Z
~ A~ ) d~a = (r
2
dx
Z
p 2k p dy (y2 kx
x2 ) Z 2 p p k 2 = dx 2k kx 3 (2 x) x Z p 2 r x 2 k 2k = 2k dx 1 x + x
2 = 2k 31 x3 + k6 x2 23k x p 8 2k 4k 16 = 2k 3 + 3 3 7 p 8 4k = 2k 21 5
p
which agrees with part a) as expected.
8
3
3
x 2 x3 + 2k x2 4k x2 2 7 15 9 8k + 8k
r
15
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Problem 1-19 Wangsness Electromagnetic Fields A vector eld is given by A~ (~r) = a ^ + b ^ + c z^
where a; b and c are constants (independent of ; and z). Is A~ a constant vector? Evaluate r~ A~ . Evaluate r~ A~ . Find A~ in Cartesian coordinates with components in terms of x; y and z. e) Find A~ in spherical polar coordinates with components in terms of r; and .
a) b) c) d)
Solution: a) The vector changes direction as the position is changed because ^ and ^ change their direction. The magnitude of A~ however does remain constnt. b) The divergence of A~ is r~ A~ = 1 @ (A ) + 1 @A + @Az = a
@
NOTE - this is not zero as it would have been were A~ a constant
vector. c) The curl of A~ is ~r A~ = ^ 1 @Az = z^ b
@
@
@z
@A + ^ @A @Az + z^ 1 @ (A ) 1 @A @z @z @ @ @
NOTE - this also is not zero as it would have been were A~ a constant vector. 9
d) The unit vectors are related by:
^ = cos x^ + sin y^ ; ^ = sin x^ + cos y^ ; z^ = z^ Therefore our vector can be written as
A~ = (a cos b sin ) x^ + (a sin + b cos ) y^ + c z^ Eliminate using
cos = p 2x 2 ; sin = p 2y 2 x +y x +y
to obtain:
A~ = (ax byp) x^2+ (ay2 + bx) y^ + c z^ x +y
e) The unit vectors in spherical coordinates are related to those for cylindrical coordinates by: ^ = sin r^ + cos ^ ^ = ^ z^ = cos r^ sin ^ Therefore our vector can be written as
A~ = (a sin + c cos ) r^ + (a cos c sin ) ^ + b ^
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Problem 1-21 Wangsness Electromagnetic Fields Find the divergence r~ r for the position vector ~r in rectangular, cylindrical and spherical polar coordinates and show that it is the same in all these. Solution:
In rectangular coordinates
~r = x x^ + y y^ + z z^ x + @ry + @rz = 3 r~ ~r = @r @x @y @z In cylindrical coordinates
~r = r r^ @ r2r + 1 @ (sin r ) + 1 @r = 3 r~ ~r = r12 @r r r sin @ r sin @ Clearly all three coordinates give the same result.
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Problem 1-23 Wangsness Electromagnetic Fields a) For a vector eld A~ = 4 r^+3 ^ 2 ^ nd the line integral counter clockwise around a plane loop in the xy plane consisting of the rst quadrant of a circle radius r centred at the origin and the portions of the x and y axes inside this quarter circle. b) Find the surface integral of r~ A~ over the xy plane enclosed by the loop. Solution: a) The path around the circular arc has path d~r = rd ^. So the line integral for this arc is Z
A~ d~r =
=2
Z
( 2)r d =
r
For the path along the y axis heading towards the origin the path is d~r = dr r^ So the line integral for this arc is Z Z ~A d~r = 4dr = 4r r
For the path along the x axis heading away from the origin the path is d~r = dr r^ So the line integral for this arc is Z
A~ d~r =
Z
r
4dr = 4r
Therefore the path integral around the whole loop is Z
C
A~ d~r =
r
b) The curl of A~ is
r~ A~
^ 1 @Ar @ @A = r sin @ (sin A ) @ + r sin @ 2 3 ^ ^ = r cot r^ + + 2
r^
12
@ (rA ) + ^ @ (rA ) @Ar @r r @r @
The surface di erential is
d~a = r2 sin dd r^ r sin drd ^ rdrd ^ Then the product 3 r cos dd + sin drd + 2 drd Since we work in the xy plane, = =2 and only the ^ component of the surface di erential is non zero. Given the counter clockwise circulation around the loop, this is the negative value rdrd . So
r~ A~ d~a = 2
Z
S
~ A~ ) d~a = (r
2
Z
r
dr
Z
=2
d =
r
which is the same as the line integral as expected from Stokes' theorem.
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Problem 1-25 Wangsness Electromagnetic Fields a) Apply the divergence theorem to a constant, though arbitrary, vector eld A~ (~r) to show that the total vector area d~a around a closed surface is zero. b) Using such a eld show also that for any closed path the vector sum d~r = 0. R
R
Solution:
a) Let A~ (~r) = a x^ + b y^ + c z^ (where a; b and c are independent of x; y and z ) be the constant but arbitrary eld. Since all derivatives of a; b and c with respect to x; y and z are zero, the divergence of A~ is also zero everywhere. Apply the divergence theorem to a volume V enclosed by a closed surface S to obtain Z Z ~A d~a = (r~ A~ )d = 0 so S
Z
V
Z
Z
a dax + b day + c daz = 0 S S S for all a; b and c. This can only be so if each of the integrals is inde-
pendently zero. Therefore Z
S
d~a =
Z
S
Z
Z
dax x^ + day y^ + daz z^ = 0 Q.E.D. S
S
b) Since all derivatives of a; b and c with respect to x; y and z are zero, the curl of A~ is also zero everywhere. Applying Stokes' theorem to a closed loop C anywhere and an arbitrary surface S bounded by this loop, we see that the path integral Z
Z
Z
C
A~ d~r = (r~ A~ ) d~a = 0 so S
Z
Z
a drx + b dry + a drz = 0 C C C for arbitrary a; b or c. Therefore the integrals are again separately zero and therefore Z
