Solutions Manual For Engineering Electromagnetics, 9th Edition by Hayt and Buck. All Chapters 1-14.

Solution Manual For
Engineering Electromagnetics
By; Hayt , Buck
9th Edition




1

,Chapter 1

1.1. Given The Vectors M = −10ax + 4ay − 8az And N = 8ax + 7ay − 2az, Find:
a) A Unit Vector In The Direction Of −M + 2n.
−M + 2n = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
(26, 10, 4)
A= = (0.92, 0.36, 0.14)
|(26, 10, 4)|

b) The Magnitude 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 Unit Vector In The Direction Of (A − B): First

A − B = (Ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−Ax − Ay + 5az)
√
W√H o s e Magnitude Is |A − B| = [(−Ax − Ay + 5az) · (−Ax − Ay + 5az)]1/2 = 1 + 1 + 25 =
3 3 = 5.20. The Unit Vector Is Therefore

Aab = (−Ax − Ay + 5az)/5.20

b) Find The Unit 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 Unit 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 Unit Vector Directed
From The Origin Toward Point—B Is (2, 2, 1)/3. If Points A And B Are Ten Units Apart, Find The
Coordinates Of Point B.
With A2= (6, −2, −4) 2And B = 1 B(2,
3
1
−2, 1), We Use The Fact That |B − A| = 10, Or
|(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
64−176
Or B2 − 8b − 44 = 0. Thus B = 8± 2
= 11.75 (Taking Positive Option) And So
2 2 1
B = (11.75)A − (11.75)A + (11.75)A = 7.83a − 7.83a + 3.92a
x y z x y z
3 3 3

2

,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 The Xy Plane, Is Tangent To The Circle — At ( 3,
1, 0), And Is In The General Direction Of Increasing Values Of Y:
A Unit 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 = −A √φ · 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 Unit 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 Unit Vector Directed From Q Toward
P:
(3, − 1, 4)
P−Q √ = (0.59, 0.20, −0.78)
AQP = = 26
|P − Q|

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 + 4x4 + 16x2 + 16 + 9z4


1.6. Find The Acute Angle Between The Two Vectors A = 2a x + Ay + 3az And B =—Ax 3ay + 2az
By Using The Definition Of:
√
a) The Dot Pro duct √ : First, A · B = √
2 − 3 + 6 = 5 = Ab Cos Θ, Where A = 22 + 12 + 32 =
√
14,
And Where B = 12 + 32 + 22 = 14. Therefore Cos Θ = 5/14, So That Θ = 69.1◦.
b) The Cross Product: Begin With

A B AyAx A A 7a
Ø Az Ø
× = 2 1 3 = 11 X − Y − Z
Ø 1 −3 2 Ø
√ √ √
And Then |A × ° B√| = 11
¢ + 1 ◦+ 7 = 171. So Now, With |A × B| = Ab Sin Θ = 171,
2 2 2

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 Surfaces On Which Ey = 0. With Ey = 2zy Sin 2x = 0, The Surfaces 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
3

, |Y| < 2,
|Z| < 2; 4) The Plane X = Π/2, With |Y| < 2, |Z| < 2.
b) The Region In Which Ey = Ez: This Occurs 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 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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