ENGINEERING ELECTOMAGNETICS 7TH
EDATION WILLIAM H HAYT SOLUTION
MANUAL (1) COMPATIBILITY MODE
1
, 6 2 / 8 7 , 2 1 0 $ 1 8 $ / ( 1 * , 1 ( ( 5 , 1
+ + $ < 7 : , 7 + & 2 0 3 / ( 7 ( & +
CHAPTER 1
1.1. Given the vectors M = Š10ax +4 ay Š 8az and N = 8 ax +7 ay Š 2az , “nd:
a) a unit vector in the direction of ŠM +2 N.
ŠM + 2 N = 10ax Š 4ay +8 az + 16ax + 14ay Š 4az = (26, 10,4)
Thus
(26, 10,4)
a= = (0 .92,0.36,0.14)
|(26, 10,4)|
b) the magnitude of 5ax + N Š 3M:
(5, 0, 0)+ (8 , 7, Š2) Š (Š30, 12,Š24) = (43 , Š5, 22), and |(43, Š5, 22)| = 48.6.
c) |M|| 2N|(M + N):
|(Š10, 4, Š8)|| (16, 14,Š4)|(Š2, 11,Š10) = (13.4)(21.6)(Š2, 11,Š10)
= ( Š580.5,3193,Š2902)
1.2. The three vertices of a triangle are located atA(Š1, 2, 5), B (Š4, Š2, Š3), and C(1, 3, Š2).
a) Find the length of the perimeter of the triangle: Begin with AB = ( Š3, Š4, Š8), BC = (5 , 5, 1),
and CA = ( Š2, Š1, 7). Then the perimeter will be = |AB| + |BC| + |CA| = 9 + 16+ 64 +
25+ 2 5 + 1 + 4 + 1 + 49 = 23 .9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC : The vector from the origin to the midpoint of AB is M AB 1= 1 (A + B) =1 1 (Š5a x +2 az ).
The vector from the origin to the midpoint of BC is M 2 2
BC = 2 (B + C) = 2 (Š3ax + ay Š 5az ).
1
The vector from midpoint to midpoint is now M AB Š M BC = 2
(Š2ax Š ay + 7 az ). The unit
vector is therefore
M AB Š M BC (Š2ax Š ay +7 az )
aMM = = = Š0.27ax Š 0.14ay +0 .95az
|M AB Š M BC | 7.35
where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC . First we “nd AC = 2 ax + ay Š 7az , which we recognize
as Š7.35 aMM . The vectors are thus parallel (but oppositely-directed).
1.3. The vector from the origin to the point A is given as (6,Š2, Š4), and the unit vector directed from
the origin toward point B is (2, Š2, 1)/3. If points A and B are ten units apart, “nd the coordinatesof
point B .
With A = (6 , Š2, Š4) and B = 13B (2, Š2, 1), we use the fact that |B Š A| = 10, or
|(6 Š 2 B )ax Š (2 Š 2 B )ay Š (4 + 1 B )az | = 10
3 3 3
Expanding,4 obtain
36 Š 8B + B 2 + 4 Š 8
B+ 4
B 2 + 16+ 8
B+ 1
B 2 = 100
9 3 9 3 9
or B 2 Š 8B Š 44 = 0. Thus B = 8± 64Š176
2
= 11.75 (taking positive option) and so
2 2 1
B =
2
, (11.75)ax Š (11.75)ay + (11.75)az = 7 .83ax Š 7.83ay +3 .92az
3 3 3
3
, 1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector
in rectangular components that lies in thexy plane, is tangent to the circle at ( 3, 1, 0), andis in the
general direction of increasing values oyf:
A unit vector tangent to this circle in the general increasing y direction is t = a . Its x and y
components aret x = a · ax = Š sin , andt y = a · ay = cos . At the point ( 3, 1), = 30 , and
sot = Š sin 30 ax + cos 30 ay = 0 .5(Šax + 3ay ).
1.5. A vector “eld is speci“ed as G = 24xy ax + 12(x 2 + 2) ay + 18z2az . Given two points, P(1, 2, Š1)
and Q(Š2, 1, 3), “nd:
a) G at P: G(1, 2, Š1) = (48 , 36,18)
b) a unit vector in the direction of G at Q: G(Š2, 1, 3) = ( Š48, 72,162), so
(Š48, 72,162)
aG = = ( Š0.26, 0.39,0.88)
|(Š48, 72,162)|
c) a unit vector directed from Q toward P:
P Š Q = (3, Š1, 4) = (0 .59,0.20,Š0.78)
aQP =
|P Š Q| 26
d) the equation of the surface on which|G| = 60: We write 60 = |(24xy, 12(x2 + 2) , 18z2)|, or
10 = |(4xy, 2x2 + 4 , 3z2)|, so the equation is
100 = 16x2y2 + 4 x4 + 16x2 + 1 6+ 9 z4
1.6. If a is a unit vector in a given direction, B is a scalar constant, andr = xax + yay + zaz , describe
the surfacer · a = B . What is the relation between the the unit vector a and the scalar B to this
surface? (HINT: Consider “rst a simple example with a = ax and B = 1, and then consider anya
and B .):
We could consider a general unit vector,a = A1ax + A2ay + A3az , where A2 + A2 + A2 = 1.
1 2 3
Then r · a = A 1 x + A 2 y + A 3 z = f (x, y, z ) = B . This is the equation of a planar surface, where
f = B . The relation of a to the surface becomes clear in the special case in whicah= ax . We
obtain r · a = f (x ) = x = B , where it is evident that a is a unit normal vector to the surface
(as a look ahead (Chapter 4), note that taking the gradient of f gives a).
1.7. Given the vector “eld E = 4 zy2 cos 2xax + 2 zy sin 2xay + y2 sin 2xaz for the region |x|, |y|, and |z|
less than 2, “nd:
a) the surfaces on whichEy = 0. With Ey = 2 zy sin 2x = 0, the surfaces are 1) the planez = 0 ,with
|x| < 2, |y| < 2; 2) the planey = 0 , with |x| < 2, |z| < 2; 3) the planex = 0 , with |y| < 2,
|z| < 2; 4) the plane x = /2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez : This occurs when 2zysin 2x = y2 sin 2x, or on the plane 2z = y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0 : We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0 , with |x| < 2, |z| < 2.
4
EDATION WILLIAM H HAYT SOLUTION
MANUAL (1) COMPATIBILITY MODE
1
, 6 2 / 8 7 , 2 1 0 $ 1 8 $ / ( 1 * , 1 ( ( 5 , 1
+ + $ < 7 : , 7 + & 2 0 3 / ( 7 ( & +
CHAPTER 1
1.1. Given the vectors M = Š10ax +4 ay Š 8az and N = 8 ax +7 ay Š 2az , “nd:
a) a unit vector in the direction of ŠM +2 N.
ŠM + 2 N = 10ax Š 4ay +8 az + 16ax + 14ay Š 4az = (26, 10,4)
Thus
(26, 10,4)
a= = (0 .92,0.36,0.14)
|(26, 10,4)|
b) the magnitude of 5ax + N Š 3M:
(5, 0, 0)+ (8 , 7, Š2) Š (Š30, 12,Š24) = (43 , Š5, 22), and |(43, Š5, 22)| = 48.6.
c) |M|| 2N|(M + N):
|(Š10, 4, Š8)|| (16, 14,Š4)|(Š2, 11,Š10) = (13.4)(21.6)(Š2, 11,Š10)
= ( Š580.5,3193,Š2902)
1.2. The three vertices of a triangle are located atA(Š1, 2, 5), B (Š4, Š2, Š3), and C(1, 3, Š2).
a) Find the length of the perimeter of the triangle: Begin with AB = ( Š3, Š4, Š8), BC = (5 , 5, 1),
and CA = ( Š2, Š1, 7). Then the perimeter will be = |AB| + |BC| + |CA| = 9 + 16+ 64 +
25+ 2 5 + 1 + 4 + 1 + 49 = 23 .9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC : The vector from the origin to the midpoint of AB is M AB 1= 1 (A + B) =1 1 (Š5a x +2 az ).
The vector from the origin to the midpoint of BC is M 2 2
BC = 2 (B + C) = 2 (Š3ax + ay Š 5az ).
1
The vector from midpoint to midpoint is now M AB Š M BC = 2
(Š2ax Š ay + 7 az ). The unit
vector is therefore
M AB Š M BC (Š2ax Š ay +7 az )
aMM = = = Š0.27ax Š 0.14ay +0 .95az
|M AB Š M BC | 7.35
where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC . First we “nd AC = 2 ax + ay Š 7az , which we recognize
as Š7.35 aMM . The vectors are thus parallel (but oppositely-directed).
1.3. The vector from the origin to the point A is given as (6,Š2, Š4), and the unit vector directed from
the origin toward point B is (2, Š2, 1)/3. If points A and B are ten units apart, “nd the coordinatesof
point B .
With A = (6 , Š2, Š4) and B = 13B (2, Š2, 1), we use the fact that |B Š A| = 10, or
|(6 Š 2 B )ax Š (2 Š 2 B )ay Š (4 + 1 B )az | = 10
3 3 3
Expanding,4 obtain
36 Š 8B + B 2 + 4 Š 8
B+ 4
B 2 + 16+ 8
B+ 1
B 2 = 100
9 3 9 3 9
or B 2 Š 8B Š 44 = 0. Thus B = 8± 64Š176
2
= 11.75 (taking positive option) and so
2 2 1
B =
2
, (11.75)ax Š (11.75)ay + (11.75)az = 7 .83ax Š 7.83ay +3 .92az
3 3 3
3
, 1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector
in rectangular components that lies in thexy plane, is tangent to the circle at ( 3, 1, 0), andis in the
general direction of increasing values oyf:
A unit vector tangent to this circle in the general increasing y direction is t = a . Its x and y
components aret x = a · ax = Š sin , andt y = a · ay = cos . At the point ( 3, 1), = 30 , and
sot = Š sin 30 ax + cos 30 ay = 0 .5(Šax + 3ay ).
1.5. A vector “eld is speci“ed as G = 24xy ax + 12(x 2 + 2) ay + 18z2az . Given two points, P(1, 2, Š1)
and Q(Š2, 1, 3), “nd:
a) G at P: G(1, 2, Š1) = (48 , 36,18)
b) a unit vector in the direction of G at Q: G(Š2, 1, 3) = ( Š48, 72,162), so
(Š48, 72,162)
aG = = ( Š0.26, 0.39,0.88)
|(Š48, 72,162)|
c) a unit vector directed from Q toward P:
P Š Q = (3, Š1, 4) = (0 .59,0.20,Š0.78)
aQP =
|P Š Q| 26
d) the equation of the surface on which|G| = 60: We write 60 = |(24xy, 12(x2 + 2) , 18z2)|, or
10 = |(4xy, 2x2 + 4 , 3z2)|, so the equation is
100 = 16x2y2 + 4 x4 + 16x2 + 1 6+ 9 z4
1.6. If a is a unit vector in a given direction, B is a scalar constant, andr = xax + yay + zaz , describe
the surfacer · a = B . What is the relation between the the unit vector a and the scalar B to this
surface? (HINT: Consider “rst a simple example with a = ax and B = 1, and then consider anya
and B .):
We could consider a general unit vector,a = A1ax + A2ay + A3az , where A2 + A2 + A2 = 1.
1 2 3
Then r · a = A 1 x + A 2 y + A 3 z = f (x, y, z ) = B . This is the equation of a planar surface, where
f = B . The relation of a to the surface becomes clear in the special case in whicah= ax . We
obtain r · a = f (x ) = x = B , where it is evident that a is a unit normal vector to the surface
(as a look ahead (Chapter 4), note that taking the gradient of f gives a).
1.7. Given the vector “eld E = 4 zy2 cos 2xax + 2 zy sin 2xay + y2 sin 2xaz for the region |x|, |y|, and |z|
less than 2, “nd:
a) the surfaces on whichEy = 0. With Ey = 2 zy sin 2x = 0, the surfaces are 1) the planez = 0 ,with
|x| < 2, |y| < 2; 2) the planey = 0 , with |x| < 2, |z| < 2; 3) the planex = 0 , with |y| < 2,
|z| < 2; 4) the plane x = /2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez : This occurs when 2zysin 2x = y2 sin 2x, or on the plane 2z = y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0 : We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0 , with |x| < 2, |z| < 2.
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