Solucionário - Eletromagnetismo para engenheiros PDF

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Solucionario

Eletromagnetismo para Engenheiros

Clayton R. Paul

Chapter 1 Problem Solutions

111 .. (a) 25 × 10 4 Ω = 250 × 10 3 Ω = 250 k Ω 2 (b) 0.035 × 10 4 Ω = 35 . × 10 Ω = 350 Ω

(c) 0 .00045F = 450 ×10 −6 F = 450 µ F (d) 0.003 × 10 − 7 F = 0.3 × 10 − 9 F = 0.3nF (e) 0 .005 ×10 − 2 H = 50 ×10 − 6 H = 50 µ H 11 . .2 5280ft 12in 2.54cm 1m 1 km × × × × = 48.28 km 1mile 1ft 1in 100cm 1000m 12in 1000mils (b) 1ft × × = 12, 000mils 1ft 1in 3ft 12in 2.54cm 1m . m (c) 100yds × × × × = 9144 100cm 1yd 1ft 1in 100cm 1m 1in 1000 mils (d) 5mm × × × × = 196.85mils 1in 1000mm 1m 2.54cm 1in 1000 mils 100cm 1m × × × = 0.7874 mils (e) 20 µ m × 6 1in 2.54cm 1m 10 µ m

(a) 30miles ×

(f) 880yds ×

3ft 12in 2.54cm 1m × × × = 804.67m 100cm 1yd 1ft 1in

1.2.1 3 × 108 m s = 33333 . km = 2,071.2 miles 90 Hz 3 × 10 8 m s λ= = 300 km = 186.41 miles 103 Hz 3 × 10 8 m s = 85714 λ= . m = 0.533 miles 350 × 103 Hz 3 × 10 8 m s = 250 m = 820.2 ft λ= 1.2 × 106 Hz 3 ×10 8 m s λ= = 8.57 m = 28.12 ft 35 × 106 Hz

(a) λ = (b) (c) (d) (e)

1-1

(f) λ = (g) λ =

3 × 10 8 m s 110 × 10 6 Hz 3 × 10 8 m s

= 2. 73m = 8.95 ft

= 89.55 cm = 2.94 ft 335 × 106 Hz 3 × 10 8 m s (h) λ = = 5 cm = 1.97 in 6 × 109 Hz 3 ×10 8 m s (i) λ = = 6.67 mm = 262.5 mils 45 × 109 Hz

1.2.2 (a)

λ=

3 × 108 m s = 5 × 106 m 50 miles = 80.46 km 60 Hz

(

)

1 λ 80.46 × 10 3 m = ? × 5 × 106 m ∴ ? = 0. 0161λ = 14 4244 3 62 14243 length

λ

(b)

λ=

3 × 10 8 m s

= 600 m 500 feet = 152.4 m 500 × 103 Hz m = ? × (600 m ) ∴ ? = 0 .254 λ 152 12 4 .44 3 12 4 4 3 λ

length

(c)

λ=

3 × 108 m s

= 2.73 m 4.5 feet = 1.37 m 110 × 106 Hz 137 . 23 m = ? × (2.73 m ) ∴? = 0.5 λ 1 12 4 4 3 length

λ

(d)

λ=

15. ×10 8 m s

= 7 5. cm 2 inches = 5.08 cm 2 × 109 Hz 5123 .084 cm = ? × (7.5 cm ) ∴ ? = 0.677 λ 12 4 4 3 length

λ

1-2

1.2.3 (a)

β = 2.2 × 10 −2 T=

m ω v rad , f = 1 MHz, v = = 2.856 × 108 , λ = = 285.6 m, s m f β

d = 10.5 µ s, φ = β d = 66 rad = 3781.5o v

(b) m ω v rad , f = 3 GHz, v = = 2.5 × 108 , λ = = 833 . mm, s m f β d 4 inches = 0.102 m, T = = 0.41 ns, φ = β d = 7.66 rad = 438.92 o v

β = 75.4

(c) rad m ω v , f = 150MHz, v = = 2.99 × 108 , λ = = 1995 . m, s m f β d 20 feet = 6.1m, T = = 20.4 ns, φ = β d = 19.2 rad = 1100.2o v

β = 315 .

(d) rad m ω v , f = 3 kHz, v = = 15 . × 108 , λ = = 50 km, s m f β d 50 miles = 80.5 km, T = = 0.54 ms, φ = β d = 1014 . rad = 580.9o v × 10 −3 β = 0126 .

1-3

Chapter 2 Problem Solutions

211 .. F1 = m ω 2d sin θ , F2 = mg

cos θ =

9.78

( 2π)

2

sin θ g 2π × rpm = 2π , , F1 = F2 , ∴ cos θ = 2 , ω = 60 cosθ ω d

= 0 .5 , ∴ θ = 60.3o

× 0.5 ␻

d ␪ F1

F2

m

␪ mg

21 . .2 W . o, cos 45o , ∴ θ = 813 200 mi o GS + Wsin45 = 200 cos θ , GS = 169.71 hr W × cos 45 o = 200 × sin θ , sin θ =

45° W GS 200 mi/hr ␪

2-1

21 . .3 4 sin θ = 3 , sin θ =

3 , ∴θ = 48.6 o 4

N 4 mi/hr 3 mi/hr ␪

21 . .4 2

2

v mv , −W = N cos θ = mg , ∴ tan θ = , r rg mi 5280 ft ft 1 hr v = 60 × × = 88 , ∴ tan θ = 016 . o . , ∴ θ = 916 hr 1mi 3600s s

F = N sin θ =

N

F —W

N ␪

W

2.31 . C = A +B, ∴C 2 = C • C = (A + B ) • ( A + B) = A • A + B • B + 2 A • B = A 2 + B 2 + 2AB cos α But α = 180o − θ AB and ∴ cos α = − cos θ A B ∴ C 2 = A 2 + B 2 − 2 AB cosθ AB which is the law of cosines.

C B ␪ AB

A

2-2

␣

2.41 . A2 ×3 A + A× B+ A× C= 0 A + B + C = 0 , A × (A + B + C ) = 0 = 1 0

(

)

o

A × B = − AB sin α C a n = − AB sin 180 − θC a n where a n is a unit normal into the

page. Also A × C = AC sin α Ba n = AC sin θ Ba n . ∴ AB sin θ C = AC sin θ B giving the B C law of sines: = . Similarly for B × ( A + B + C) = 0 and sin θ B sin θ C

C × ( A + B + C) = 0 .

␪B C

B ␪C

B

␪C

␪B A

A ␣C

C

␣B

A

2.4.2 (a) (A • B ) gives a scalar which cannot be “crossed with” a vector. (c) (A • B ) gives a scalar which cannot be “dotted with” a vector. (d) (B • C) gives a scalar which cannot be added to a vector.

2.51 . (a) A = (3 − 0 ) a x + ( −4 − 2) a y + (5 − ( −4) ) a z = 3 a x − 6 a y + 9 a z (b) A = (c) a A =

(3)2 + ( −6)2 + (9)2

= 1122 . m

A = 0.27a x − 0.53a y + 0.8a z A

2.5.2 (a) A + B = 3a x + 4 a y − 3a z , (b) B − C = −2a x + 2 a y − 3a z , (c)

A + 3B - 2C = − a x + 8a y − 9a z , (d) A = 22 + 32 + 12 = 3.74 , (e) aB =

B 1 = a x + a y − 2 a z =.41a x + 0.41a y − 0.82 a z , (f) A • B = 7 , (g) B • A = 7 , B 6

(

)

(h) B × C = − a x − 7a y − 4a z , (i) C × B = a x + 7a y + 4a z , (j) A • B × C = −19

2-3

2.5.3 (a) A • B = 2 − 2 − 3 = −3 = A B cos θ . Therefore B cos θ =

A•B 3 = = 0.8 , A 14

−3 A•B = ⇒ θ = 1091 . o , (c) AB 14 6 A × B −a x − 7a y − 5a z = −0115 unit vector = = . a x − 0.808a y −0.577 a z . A ×B 1 +49 +25

(b) cos θ =

2.5.4 A •B = α + 2 − 3 = 0 ∴α = 1

2.5.5 A × B = (−18 − 3 β)a x + (3α + 9 )a y + (β − 2α )a z = 0 . Hence α = −3 and β = −6 .

2.5.6 A × B = −14 a x − 9 a y + a z gives a vector that is perpendicular to the planes containing

both A and B and hence is perpendicular to both A and B. The length of this vector is

(−14 )2 + (−9 )2 + (1) 2 C=

= 16 .67 . Hence

10 −14 a x − 9a y + a z = −8.4a x − 54 . a y + 0.6 a z . Check that A • C = 0 and 16.67

(

)

B•C = 0. 2.5.7

(

)

(

)

B × C = B yC z − B zC y a x + ( B zC x − B xC z ) a y + B xC y − B yC x a z so that

) ( ( ) + ( Az ( By Cz − Bz C y ) − Ax ( Bx C y − B y C x )) a y + ( Ax ( Bz Cx − Bx Cz ) − Ay ( B y Cz − Bz C y )) a z

A × (B × C ) = A y B xC y − B y C x − A z ( B zC x − B xC z ) a x

(

)

(

)

(

)

B( A • C) = B x A xC x + A yC y + A zC z a x + B y A xC x + A yC y + A zC z a y

(

)

+ B z A xC x + A yC y + A zC z a z

(

)

C( A • B) = C x A x B x + A y B y + A zB z a x + C y A xB x + A yB y + A zB z a y

(

)

+ C z A xB x + A yB y + A zB z a z Matching components we find that A × ( B × C) = B( A • C) − C( A • B) .

2-4

2.5.8

(

)

( ) (A × B ) × C = (( A z B x − A xB z )C z − ( A xB y − A yB x )C y )a x + ( ( Ax B y − A y B x )C x − ( A y B z − A z B y )C z) a y + ( ( Ay Bz − Az By )C y − ( Az Bx − Ax B z ) C x ) a z

A × B = A y B z − A zB y a x + (A zB x − A xB z )a y + A xB y − A yB x a z so that

Comparing terms to the result in the previous problem we see that A × (B × C ) ≠ (A × B ) × C . 2.5.9

The distance is D =

(3 − (−1 )) 2 + (−1 − 2 )2 + (5 − (−4 ))2 = 10.296 .

By integration we

3 5 integrate D = ∫ PP2 dl . A straight line between the two points is governed by y = − x + 1 4 4 and z =

2 9 2 9 7  3 x − . Hence dl = dx 1 +  −  +   = 2.574dx and  4  4 4 4

x=3

∫ 2.574dx = 10.296 .

D=

x = −1

2.510 .

The surface is drawn below and lies in the yz plane at x=1. Hence the surface area is z= 2 y = 2z − 1

A= ∫ z= 1

z =2

∫ dydz = ∫ ( 2z − 2)dz = 1 . Directly it is the area of a triangle of height 1 and

y =1

base 2 or A =

z =1

1 ( 2)( 1) = 1. 2 z y = 2z – 1

(1, 3, 2)

y (1, 1, 1) x

2-5

(1, 3, 1)

2.61 .

Drawing the rectangular and cylindrical coordinate system axes as shown below we see that x = r cos φ , y = r sin φ , and z = z . From this we form

(

)

x 2 + y 2 = r 2 cos 2 φ + sin 2 φ and hence r = x 2 + y 2 . Similarly we form 1442443 1 y r sin φ = = tan φ . x r cos φ z z

r P(r, ␾, z)

r

y

␾ x

2.6.2

Draw the coordinate system and use the right-hand rule. 2.6.3

Drawing the vector in the xy plane as shown below shows that A x = A r cos φ − Aφ sin φ and A y = Ar sin φ + Aφ cos φ . Ay y

Ax ␾

Ar

A␾ ␾

x

2-6

2.6.4

At point P, φ =

π 3

= 60o . Hence Ax = 2 cos φ + 3 sin φ =3.598 and

A y = 2 sin φ − 3 cos φ =0.232 and Bx = 4 cos φ − 6 sin φ =-3.196 and B y = 4 sin φ + 6 cos φ =6.464. Directly in cylindrical coordinates A • B = 8 − 18 − 2 = −12 . In rectangular coordinates A • B = (3598 . × −3196 . ) + (0.232 × 6.464 ) − 2 = −12 . 2.6.5  π From problem 2.6.1 x = r cos φ , y = r sinφ , z=z. At P1 2, ,1 , x 1 = 0, y 1 = 2, z 1 = 1  2   π  . , y2 = 2.598, z2 = −2 . Hence the distance between the two and at P2 = 3, ,−2  , x2 = 15  3 

points is D =

( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + (z 2 − z 1) 2 =

3.407 .

2.6.6

The surface is 1/6 of the surface of a cylinder of length 4-1=3 and radius 2. Hence the (2π × 2 × 3) = 2π . By direct integration we have surface area is S = 6 4

π

3

r = 2 ) dφ dr = 2π . ∫ (1 4243 z =1 φ = 0 ds r

S = ∫

2.6.7

The volume is 1/6 of the volume of a cylinder of radius 2 minus the volume of a cylinder 1 π 2 2 of radius 1 or V = π (2 ) × 1 − π (1) × 1 = . By direct integration, 6 2

(

1

V= ∫

π

∫

)

3 2

π

dz = . φ4 ∫ rdrd 1 42 3 2 z= 0φ =0 r =1 dv

2.7.1

Drawing the rectangular and spherical coordinate system axes as shown below we see that

x = r sin θ cos φ , y = r sin θ sin φ , and z = r cosθ . From this we form

(

)

x 2 + y 2 + z 2 = r 2sin 2 θ cos 2 φ + sin 2 φ + r 2 cos 2 θ = r 2 and hence 1442443 1

2-7

r=

y r sin θ sin φ x 2 + y 2 + z 2 . Similarly we form = = tan φ and x r sin θ cos φ

x2 + y2 z

=

r sin θ = tan θ . r cos θ z z

r sin ␪

␪

r y

␾ x

2.7.2

Draw the coordinate system and use the right-hand rule. 2.7.3

Drawing the vector in the zx or zy plane as shown below shows that Az = Ar cos θ − Aθ sin θ . The components parallel to the xy plane are Ar sin θ + Aθ cos θ

and the φ component, Aφ . Hence the x and y components of these are

Ax = ( Ar sin θ + Aθ cos θ ) cos φ − Aφ sin φ and A y = ( Ar sin θ + Aθ cos θ ) sin φ + Aφ cos φ .

Ay

z Ar sin ␪ Ar cos ␪

Ax ␪

Ar

␾

A␾ Ar sin ␪ + A␪ cos ␪

A␪

␪ A␪ sin ␪

A␪ cos ␪

x x, y plane

2-8

y

2.7.4

2π π = 120o and φ = = 60o . Hence 3 3

At point P, θ =

Ax = 2 sin θ cos φ + 3 cos θ cos φ − sin φ =-0.75 and A y = 2 sin θ sin φ + 3 cos θ sin φ + cos φ =0.701 and Az = 2 cos θ − 3 sin θ =-3.598. . , B y = 0.634 , and B z = −3732 . . Directly in spherical coordinates Similarly, Bx = 383 A • B = 8 + 6 − 3 = 11 . In rectangular coordinates A • B = (−0.75 × 383 . ) + (0.701 × 0.634 ) + (−3598 . × −3732 . ) = 11 . 2.7.5  π 2π From problem 2.7.1 x = r sin θ cos φ , y = r sin θ sin φ , and z = r cosθ . At P1 2, ,  ,  2 3  π π . , z 2 = 15 . . . , z 1 = 0 and at P2 = 3, , −  , x 2 = 2.25, y 2 = −1299 x 1 = −1, y 1 = 1732  3 6

Hence the distance between the two points is

(x2 − x1 ) 2 + (y 2 − y1 )2 + (z 2 − z 1 )2 = 4.69 .

D=

2.7.6

The surface is 1/8 of the surface of a sphere of radius 4. Hence the surface area is

( 4π × (4) ) S= 2

π

∫

S =

π

∫

θ =π 2 φ = π 2

8

= 8π . By direct integration we have

2 r = 4) sin θ dθ dφ = 8π . (1 4442444 3

dsr

2.7.7 π

The volume is S =

3 2π

2

r = 2 ) sin θ dφ dθ = 5.21 . ∫ (1 4442444 3 π 0 = φ θ= 4 ds ∫

r

2-9

2.81 . 2

∫ F • dl =

4

1

∫ 2 xdx + ∫ 4 dy − x = −1

y =3

∫ ydz . Third integral with respect to z contains y. So we z = −2

1 11 z+ . Hence the line integral 3 3 2 4 1 1 11 7 becomes ∫ F • dl = ∫ 2xdx + ∫ 4dy − ∫  z +  dz = − .  3 2 x =−1 y= 3 z =− 2 3

need to determine the equation of the path as y =

2.8.2 P2

1

1

2

P1

x =0

y =0

z =0

W = ∫ F • dl = ∫ 2 xdx + ∫ 3 zdy + ∫ 4 dz . But along the path z = 2 y which when

substituted into the second integral gives W=1+3+8=12J. 2.8.3 P2

3

2

0

P1

x =0

y =0 P2

z =2

(a) ∫ F • dl = ∫ xdx + ∫ 2xydy − ∫ ydz . Along this path x = 3

2

0

x= 0

y= 0

z= 2

3 y and y = − z + 2 . 2

substituting these gives ∫ F • dl = ∫ xdx + ∫ 3 y 2dy − ∫ ( − z + 2) dz = P1

29 . (b) The 2

integral is the sum of the integrals along the two paths: P2

0 0 0 3 2 0 3   ∫ F • dl = ∫ xdx + ∫ 2 xydy − ∫ ( y = 0 )dz + ∫ xdx + ∫ 2 x = y  ydy − ∫ ydz =0+0+0+  2  P1 x=0 y =0 z= 2 x= 0 y= 0 z= 0

9/2+8+0=25/2. Along the first segment of this path neither x nor y change and y=0. Hence the integral along this first path is zero. Along the second segment of the path, 3 there is no change in z and we substitute the equation for the path, x = y , into the y 2 integration. 2.8.4 P2

∫ F • dl = ∫ 2rdr + ∫ zrdφ + ∫ 4dz . The two paths are sketched below.

P1 P2

0

P1 P2

r =0 P3

P4

P2

P1

P1

P3

P4

0

0

(a) ∫ F • d l = ∫ 2(r = 0)dr + ∫ z ( r = 0) d φ + ∫ 4 dz = 0 + 0 − 12 = − 12 (b)

φ =0

z =3

∫ F • dl = ∫ F •dl + ∫ F • dl + ∫ F • dl

2-10

π P3

8

4

P1

r =0

φ=

3

∫ F • d l = ∫ 2rdr + ∫ ( z = 3) rd φ + ∫ 4dz = 8 + 0 + 0 = 8 π

z =3

4

π P4

8

∫ F • dl =

P3

(

4

r= 8

)

φ=

(

0

)

z r = 8 dφ + ∫ 4 dz = 0 + 0 − 12 = − 12

∫ 2 r = 8 dr + ∫

π

z =3

4

π P2

0

∫ F • dl =

P4

r= 8

0

4

∫ ( z = 0 ) rdφ + ∫ 4 dz = −8 + 0 + 0 = −8

∫ 2 rdr +

φ=

π

z =0

4

P2

P3

P4

P2

P1

P1

P3

P4

∫ F • dl = ∫ F • dl + ∫ F • dl + ∫ F • dl = 8 − 12 − 8 = − 12

z P1(0, 0, 3)

P3(2, 2, 3)

P2(0, 0, 0) y P4(2, 2, 0) x

2.8.5 P2

∫ F • dl = ∫ rdr + ∫ 2 rdφ − ∫ zdz . The two paths are sketched below.

P1 P2

P0

P2

P1

P1

P0

(a) ∫ F • dl = ∫ F • dl + ∫ F • dl P0

0

P1 P2

r =0 3

0

0

P0

r =0

φ =0

z=0

0

0

∫ F • dl = ∫ ( r = 0) dr + ∫ 2( r = 0) d φ − ∫ zdz = 0 + 0 + 2 = 2 φ=0

z =2

∫ F • dl = ∫ rdr + ∫ 2rdφ − ∫ ( z = 0 ) dz =

9 9 + 0+ 0 = 2 2

2-11

P2

P0

P2

∫ F • dl = ∫ F • dl + ∫ F • dl = 2 +

P1

P1

P0

9 13 = 2 2

P2

P3

P4

P2

P1

P1

P3

P4

∫ F • dl = ∫ F •dl + ∫ F • dl + ∫ F • dl

(b)

π P3

3

P1

r =0

2

2

∫ F • d l = ∫ rdr + ∫ 2rd φ − ∫ 4dz = φ=

π

z =2

9 9 + 0+ 0 = 2 2

2

π P4

3

P3

r =3

0

2

∫ F • dl = ∫ ( r = 3) dr + ∫ 2( r = 3) dφ − ∫ 4 dz = 0 + 0 + 2 = 2

P2

3

P4

r =3

φ=

π

z =2

2

0

0

∫ F • dl = ∫ ( r = 3) dr + ∫ 2( r = 3) dφ − ∫ 4 dz = 0 − 3π + 0 = − 3π φ=

π

z =0

2

P2

P3

P4

P2

P1

P1

P3

P4

∫ F • dl = ∫ F • dl + ∫ F • dl + ∫ F • dl =

9 13 + 2 − 3π = − 3π 2 2

z P1(0, 0, 2)

P3(0, 3, 2) 2

P4(0, 3, 0) 3

y

P2(3, 0, 0) 3 x

2.8.6 P2

∫ F • dl = ∫ 2rdr + ∫ 3rdθ + ∫ 2r sin θ dφ . The two paths are sketched below.

P1 P2

P0

P2

P1

P1

P0

(a) ∫ F • dl = ∫ F • dl + ∫ F • dl

2-12

P0

0

P1

r =3

P2

3

P0

r =0

P2

P0

P2

P1

P1

P0

0

0

∫ F • dl = ∫ 2 rdr + ∫ 3rdθ + ∫ 2 r sin θ dφ = − 9 + 0 + 0 = − 9 θ=0 π

∫ F • dl = ∫ 2rdr +

φ= 0 0

2

∫ 3rdθ + ∫ 2r sin θ dφ = − 9 + 0 + 0 = 9 θ =π 2

φ =0

∫ F • dl = ∫ F • dl + ∫ F • dl = −9 + 9 = 0

(b)

P2

P3

P2

P1

P1

P3

∫ F • d l = ∫ F • d l + ∫ F • dl π

P3

3

P1

r =3

P2

3

P3

r= 3

P2

P3

P2

P1

P1

P3

π

2

∫ F • d l = ∫ 2(r = 3)dr + ∫ 3(r = 3)d θ +

φ =π 2

θ =0 π

∫ F • d l = ∫ 2 (r = 3 )dr +

2

∫ 3 (r = 3)dθ ...