S O L UT I O NS MA N U A L
to accompany
E L E C T R I C I T Y A N D MA G N E T I S M
MUNIR H. NAYFEH
MORTON K. BRUSSEL
University of Illinois
JOHN WILEY &SONS
New York Chichester Brisbane Toronto Singapore
,Copyright@ 1986 by John Wiley & Sons, Inc.
This ma.terial ma.y be reproduced for testing or
instructional purposes by people using the text.
ISBN 0 471 80692 7
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
, CHAPTER 1
+ + + +
1.1 The cross product Ax Bis perpendicular to both A and B.
Since we want a unit vector, we just divide by the
+ + +
magnitud e: n -- +A X B/IA X Bl
X y z
+ +
A X B = 2 -6 -3 l 5x - lOy + 3oz
4 3 -1
IA X s1 = 2
/1s + 10 + 30
2 2
35
Hence,
A
TI
s
= 1 x - lOy
+ 30; 3x - 2v + 6z
3'i 7
1 .2 The position vectors of these three points are
+
rl 2x - A A
+
y + z, r2 = 3x + 2 y
A
- A
z, and
+ -x + 3y + 2 z.
r3 The position vector of an arbitrary point is
+ A A A
r = xx + yy + zz. If all vectors lay in the plane, then the
following triple cross product vanishes.
which gives llx +Sy+ 13z - 30 O.
1 .3 The position vectors of these points and an arbitrary point
are rl
+ A
+
Jx + y + 2z, r2 = X
A
,, -
- ,._y A
4z, and
t = xx - yy+ zz. The equation of the plane is governed by the
, condition (t t2)•(r2 - r1) = 0 which gives the equation
2x + 3y +6z +28 = 0 for the plane.
r dtnltl/dr = r/r
V(l/r) = r d(l/r)/dr = -;/r 2
1.5 Consider the surface defined by f where
f(x,y,z) = 2xz
2
- 3xy - 4x - 7 = O. Recall that Vf is normal to
surface f(x,y,z) = O, then
af
- af af
Vf x+-y+-z
A A A
ax ay az
At (1,-1,2) we have: Vf = (8 + 3 - 4)x - 3 y+8z
7x - 3 y+ 8z. The unit vector normal to the surface at this
A A
point is:
n = Vf =
7i - 3i+ s:£ 7;_ - 3i+ 8:i
"fvTT
/2 2 + 3 2+8 2 ✓1 22
1.6 By definition we have H/ds = I Vt 1- Now Vt = 2xyz
3
;+
max
x 2 z 3 y+ 3x2yz2 ;, thus at X 2, y = l, z = -1 we have
Vt -4x - 4y + 1 2 z. Hence, dt/ds is maximum along
A A A
=
(-4i - 4y +12 �)//4 +4 + 1 22 direction and
2 2
/4 +4 +1 2
2
IVtl = ✓176
2 2
2
to accompany
E L E C T R I C I T Y A N D MA G N E T I S M
MUNIR H. NAYFEH
MORTON K. BRUSSEL
University of Illinois
JOHN WILEY &SONS
New York Chichester Brisbane Toronto Singapore
,Copyright@ 1986 by John Wiley & Sons, Inc.
This ma.terial ma.y be reproduced for testing or
instructional purposes by people using the text.
ISBN 0 471 80692 7
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
, CHAPTER 1
+ + + +
1.1 The cross product Ax Bis perpendicular to both A and B.
Since we want a unit vector, we just divide by the
+ + +
magnitud e: n -- +A X B/IA X Bl
X y z
+ +
A X B = 2 -6 -3 l 5x - lOy + 3oz
4 3 -1
IA X s1 = 2
/1s + 10 + 30
2 2
35
Hence,
A
TI
s
= 1 x - lOy
+ 30; 3x - 2v + 6z
3'i 7
1 .2 The position vectors of these three points are
+
rl 2x - A A
+
y + z, r2 = 3x + 2 y
A
- A
z, and
+ -x + 3y + 2 z.
r3 The position vector of an arbitrary point is
+ A A A
r = xx + yy + zz. If all vectors lay in the plane, then the
following triple cross product vanishes.
which gives llx +Sy+ 13z - 30 O.
1 .3 The position vectors of these points and an arbitrary point
are rl
+ A
+
Jx + y + 2z, r2 = X
A
,, -
- ,._y A
4z, and
t = xx - yy+ zz. The equation of the plane is governed by the
, condition (t t2)•(r2 - r1) = 0 which gives the equation
2x + 3y +6z +28 = 0 for the plane.
r dtnltl/dr = r/r
V(l/r) = r d(l/r)/dr = -;/r 2
1.5 Consider the surface defined by f where
f(x,y,z) = 2xz
2
- 3xy - 4x - 7 = O. Recall that Vf is normal to
surface f(x,y,z) = O, then
af
- af af
Vf x+-y+-z
A A A
ax ay az
At (1,-1,2) we have: Vf = (8 + 3 - 4)x - 3 y+8z
7x - 3 y+ 8z. The unit vector normal to the surface at this
A A
point is:
n = Vf =
7i - 3i+ s:£ 7;_ - 3i+ 8:i
"fvTT
/2 2 + 3 2+8 2 ✓1 22
1.6 By definition we have H/ds = I Vt 1- Now Vt = 2xyz
3
;+
max
x 2 z 3 y+ 3x2yz2 ;, thus at X 2, y = l, z = -1 we have
Vt -4x - 4y + 1 2 z. Hence, dt/ds is maximum along
A A A
=
(-4i - 4y +12 �)//4 +4 + 1 22 direction and
2 2
/4 +4 +1 2
2
IVtl = ✓176
2 2
2
