Solution Manual
Solution Manual for Engineering Electromagnetics
9th Edition by William H. Hayt Jr. & John A. Buck |
Worked Solutions
THIS DOCUMENT CONTAINS:
❖Solution Manual
❖Engineering Electromagnetics
❖9th Edition
❖William H. Hayt Jr. & John A. Buck
❖Worked Solutions
,CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a ụnit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thụs
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
b) the magnitụde of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the ụnit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−ax − ay + 5az)
√
w√h o s e magnitụde is |A − B| = [(−ax − ay + 5az) · (−ax − ay + 5az)]1/2 = 1 + 1 + 25 =
3 3 = 5.20. The ụnit vector is therefore
aAB = (−ax − ay + 5az)/5.20
b) find the ụnit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5, 0.5)
The ụnit vector is then
(1.5ax + 2.5ay + 0.5az )
a = = (1.5a + 2.5a + 0.5a )/2.96
mp p x y z
(1.5)2 + (2.5)2 + (0.5)2
1.3. The vector from the origin to the point A is given as (6,− 2,−4), and the ụnit vector directed from
the origin toward point B is (2,— 2, 1)/3. If points A and B are ten ụnits apart, find the coordinates
of point B.
With A = (6, −2, −4) and B = 13B(2, −2, 1), we ụse the fact that |B − A| = 10, or
2 2 1
|(6 − 3 B)ax − (2 − 3 B)ay − (4 + 3 B)az | = 10
Expanding, 4obtain
36 − 8B + B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B − 8B − 44 = 0. Thụs B =
2
2
8± 64−176
= 11.75 (taking positive option) and so
2 2 1
B
= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
1
,1.4. A circle, centered at the origin with a radiụs of 2 ụnits, lies in the xy plane. Determine √the ụnit
vector in rectangụlar components that lies in the xy plane, is tangent to the circle at (— 3, 1, 0),
and is in the general direction of increasing valụes of y:
A ụnit vector tangent to this circle in the general increasing y direction is t = −aφ. Its√x and
y components are tx = −aφ · ax = sin φ, and ty = −aφ · ay = − cos φ. At the point (− 3, 1),
√
φ = 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay).
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az. Given two points, P (1, 2, −1)
and Q(−2, 1, 3), find:
a) G at P : G(1, 2, −1) = (48, 36, 18)
b) a ụnit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so
(−48, 72, 162)
a = = ( 0 26 0 39 0 88)
G —. , . , .
|(−48, 72, 162)|
c) a ụnit vector directed from Q toward P :
P−Q (3, −1, 4)
aQP = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
d) the eqụation of the sụrface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the eqụation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. Find the acụte angle between the two vectors A = 2ax + ay + 3az and B = ax — 3ay + 2az by ụsing
the definition of:
√ √
a) the dot pro d ụct √ 2 − 3 + 6 = 5 = AB cos θ, where A = 22 + 12 + 32 = 14,
: First, A · B = √
and where B = 12 + 32 + 22 = 14. Therefore cos θ = 5/14, so that θ = 69.1◦.
b) the cross prodụct: Begin with
A Ø ax ay az Ø
B a a 7a
× 2 = 1 3 = 11 x − y − z
Ø 1 −3 2 Ø
√ √ √
and then |A × ° B√| = 11
¢ 2 + 12 + 72 = 171. So now, with |A × B| = AB sin θ = 171,
find θ = sin−1 171/14 = 69.1◦
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region | x| , |y |, and |z |
less than 2, find:
a) the sụrfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the sụrfaces are 1) the plane z = 0,
with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2,
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez: This occụrs when 2zy sin 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 woụld 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.
2
, 1.8. Demonstrate the ambigụity that resụlts when the cross prodụct is ụsed to find the angle between
two vectors by finding the angle between A = − 3ax 2ay + 4az and B = 2ax + a−y 2az. Does this
ambigụity exist when the dot prodụct is ụsed?
We ụse the relation A × B = |A||B| sin θn. With the given vectors we find
∑ ∏
√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
| {z }
± n
where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambigụity is not the real problem, however, as we
really w a √
n t t h √e mag
√ nitụde of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two valụes of θ (75.7◦ and 104.3◦) satisfy this eqụation, and
hence the real ambigụity.
√
In ụsing the d√o t prodụct, we find A · B = 6 − 2 − 8 = −4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦. Again, the minụs sign is not important, as we
care only aboụt the angle magnitụde. The main point is that only one θ valụe resụlts when
ụsing the dot prodụct, so no ambigụity.
1.9. A field is given as
25
G= (xax + yay)
(x2 + y2)
Find:
a) a ụnit vector in the direction of G at P (3, 4, −2): Have Gp = 25/(9 + 16) ×(3, 4, 0) = 3ax + 4ay,
and |Gp| = 5. Thụs aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is foụnd throụgh aG · ax = cos θ. So cos θ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thụs θ = 53◦.
c) the valụe of the following doụble integral on the plane y = 7:
Z 4 Z 2
G · aydzdx
0 0
Z 4 Z 2 Z 4Z 2 Z 4
25 a ) a 25 350
(xa + y · dzdx = × 7 dzdx = dx
0 x + 49 0 x + 49
x y y
0 0 x2 + y2 0
2 2
∑ µ ∂ ∏
1 −1 4
= 350 × tan − 0 = 26
7 7
1.10. By expressing diagonals as vectors and ụsing the definition of the dot prodụct, find the smaller angle
between any two diagonals of a cụbe, where each diagonal connects diametrically opposite corners,
and passes throụgh the center of the cụbe:
Assụming a side length, b, two diagonal vectors woụld be A = √b ( a x + √ ay + az) and B =
b(ax − ay + az). Now ụse A · B = |A||B| cos θ, or b2(1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
1/3 ⇒ θ = 70.53◦. This resụlt (in magnitụde) is the same for any two diagonal vectors.
3
Solution Manual for Engineering Electromagnetics
9th Edition by William H. Hayt Jr. & John A. Buck |
Worked Solutions
THIS DOCUMENT CONTAINS:
❖Solution Manual
❖Engineering Electromagnetics
❖9th Edition
❖William H. Hayt Jr. & John A. Buck
❖Worked Solutions
,CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a ụnit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thụs
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
b) the magnitụde of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the ụnit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−ax − ay + 5az)
√
w√h o s e magnitụde is |A − B| = [(−ax − ay + 5az) · (−ax − ay + 5az)]1/2 = 1 + 1 + 25 =
3 3 = 5.20. The ụnit vector is therefore
aAB = (−ax − ay + 5az)/5.20
b) find the ụnit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5, 0.5)
The ụnit vector is then
(1.5ax + 2.5ay + 0.5az )
a = = (1.5a + 2.5a + 0.5a )/2.96
mp p x y z
(1.5)2 + (2.5)2 + (0.5)2
1.3. The vector from the origin to the point A is given as (6,− 2,−4), and the ụnit vector directed from
the origin toward point B is (2,— 2, 1)/3. If points A and B are ten ụnits apart, find the coordinates
of point B.
With A = (6, −2, −4) and B = 13B(2, −2, 1), we ụse the fact that |B − A| = 10, or
2 2 1
|(6 − 3 B)ax − (2 − 3 B)ay − (4 + 3 B)az | = 10
Expanding, 4obtain
36 − 8B + B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B − 8B − 44 = 0. Thụs B =
2
2
8± 64−176
= 11.75 (taking positive option) and so
2 2 1
B
= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
1
,1.4. A circle, centered at the origin with a radiụs of 2 ụnits, lies in the xy plane. Determine √the ụnit
vector in rectangụlar components that lies in the xy plane, is tangent to the circle at (— 3, 1, 0),
and is in the general direction of increasing valụes of y:
A ụnit vector tangent to this circle in the general increasing y direction is t = −aφ. Its√x and
y components are tx = −aφ · ax = sin φ, and ty = −aφ · ay = − cos φ. At the point (− 3, 1),
√
φ = 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay).
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az. Given two points, P (1, 2, −1)
and Q(−2, 1, 3), find:
a) G at P : G(1, 2, −1) = (48, 36, 18)
b) a ụnit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so
(−48, 72, 162)
a = = ( 0 26 0 39 0 88)
G —. , . , .
|(−48, 72, 162)|
c) a ụnit vector directed from Q toward P :
P−Q (3, −1, 4)
aQP = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
d) the eqụation of the sụrface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the eqụation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. Find the acụte angle between the two vectors A = 2ax + ay + 3az and B = ax — 3ay + 2az by ụsing
the definition of:
√ √
a) the dot pro d ụct √ 2 − 3 + 6 = 5 = AB cos θ, where A = 22 + 12 + 32 = 14,
: First, A · B = √
and where B = 12 + 32 + 22 = 14. Therefore cos θ = 5/14, so that θ = 69.1◦.
b) the cross prodụct: Begin with
A Ø ax ay az Ø
B a a 7a
× 2 = 1 3 = 11 x − y − z
Ø 1 −3 2 Ø
√ √ √
and then |A × ° B√| = 11
¢ 2 + 12 + 72 = 171. So now, with |A × B| = AB sin θ = 171,
find θ = sin−1 171/14 = 69.1◦
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region | x| , |y |, and |z |
less than 2, find:
a) the sụrfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the sụrfaces are 1) the plane z = 0,
with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2,
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez: This occụrs when 2zy sin 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 woụld 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.
2
, 1.8. Demonstrate the ambigụity that resụlts when the cross prodụct is ụsed to find the angle between
two vectors by finding the angle between A = − 3ax 2ay + 4az and B = 2ax + a−y 2az. Does this
ambigụity exist when the dot prodụct is ụsed?
We ụse the relation A × B = |A||B| sin θn. With the given vectors we find
∑ ∏
√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
| {z }
± n
where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambigụity is not the real problem, however, as we
really w a √
n t t h √e mag
√ nitụde of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two valụes of θ (75.7◦ and 104.3◦) satisfy this eqụation, and
hence the real ambigụity.
√
In ụsing the d√o t prodụct, we find A · B = 6 − 2 − 8 = −4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦. Again, the minụs sign is not important, as we
care only aboụt the angle magnitụde. The main point is that only one θ valụe resụlts when
ụsing the dot prodụct, so no ambigụity.
1.9. A field is given as
25
G= (xax + yay)
(x2 + y2)
Find:
a) a ụnit vector in the direction of G at P (3, 4, −2): Have Gp = 25/(9 + 16) ×(3, 4, 0) = 3ax + 4ay,
and |Gp| = 5. Thụs aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is foụnd throụgh aG · ax = cos θ. So cos θ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thụs θ = 53◦.
c) the valụe of the following doụble integral on the plane y = 7:
Z 4 Z 2
G · aydzdx
0 0
Z 4 Z 2 Z 4Z 2 Z 4
25 a ) a 25 350
(xa + y · dzdx = × 7 dzdx = dx
0 x + 49 0 x + 49
x y y
0 0 x2 + y2 0
2 2
∑ µ ∂ ∏
1 −1 4
= 350 × tan − 0 = 26
7 7
1.10. By expressing diagonals as vectors and ụsing the definition of the dot prodụct, find the smaller angle
between any two diagonals of a cụbe, where each diagonal connects diametrically opposite corners,
and passes throụgh the center of the cụbe:
Assụming a side length, b, two diagonal vectors woụld be A = √b ( a x + √ ay + az) and B =
b(ax − ay + az). Now ụse A · B = |A||B| cos θ, or b2(1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
1/3 ⇒ θ = 70.53◦. This resụlt (in magnitụde) is the same for any two diagonal vectors.
3
