Transcript
Chapter 3 Problem Solutions
311 .. π 1
Q = ∫∫∫ ρ v dv = ∫
r =0
2
2
∫
φ=
∫ 2 zrdzdφdr =
π z =0
π 2
C
4
31 . .2 2 −2 x + 5
∫ 3xy( z = 2) dxdy = 13C .
The surface is shown below. Q = ∫∫ ρ s ds = ∫
x =1 y =1
z (1, 1, 2)
(1, 3, 2) y = –2x + 5
(2, 1, 2)
y x
31 . .3 The problem is sketched below. A vector from the second charge to the first is
R 21 = (1 − ( −1)) a x + (1 − 0)a y + (1 − ( −2)) a z = 2a x + a y + 3a z whose length is R21 = 14 . A unit vector pointing from the second charge to the first is a 21 = Coulomb’s law yields F21 = 9 × 10
9
(100 × 10−6 )(50 × 10−6 ) a 21 = 1718 . a x + 0.859a y + 2.577a z 2 R21
z Q2 = 50 C (–1, 0, 2) Q1 = 100 C (1, 1, 1) y
x
3-1
N.
R 21 . R21
31 . .4
100 × 10 −6 ) 9( (a) F=0, (b) F = 4 × 9 × 10 100 × 10 −6 ) 9( (c) F1 = 9 × 10
(1) 2
(
2
( 2)
2
2
× cos 45o a z ,
100 × 10 −6 ) 9( a x = 90a x , F2 = 9 × 10
)
(
)
(1)2
2
a y = 90a y ,
90 90 2 and cos θ a x + sin θ a y , F4 = sin θ a x + cos θ a y . But cos θ = 5 5 5 1 . Hence F = F1 + F2 + F3 + F4 = 11415 . a x + 11415 . ax . sin θ = 5 F3 =
F2
y
F4
1
F3
1
F1
5
3 –1
1
5 –1
2 x
4
31 . .5 The problem is sketched below. For the forces exerted on Q3 by the other two charges to be equal and oppositely directed, we must have 9 × 10
9
(18 × 10−6 )(8 × 10−6 ) = 9 × 109 (72 × 10−6 )(8 × 10−6 ) . Solving for d gives d2
(0.03 − d ) 2
d=1cm.
3 cm d Q1 = 18 C
Q3 = –8 C
3-2
Q2 = 72 C
31 . .6 . × 10 −19 C. Placing the positive charge on the left The charge of an electron is e = −16
and the negative charge on the right, the force exerted on the electron by the positive charge is directed to the left and is F1 = 9 × 10
9
(35 × 10−6 )(16. × 10−19 ) = 5.04 × 10−12 2 (10 × 10−2 )
N . The force exerted on the
electron by the negative charge is of the same magnitude and in the same direction so that the net force is 1 × 10 −11 N directed toward the positive charge. 31 . .7 The problem is sketched below. In order that the forces balance, the Coulomb force acting in the horizontal direction is F =
Q2 4π ε o (2l sin θ )
2
. The component of the
restraining force along the string that is horizontally directed is T sin θ and the force of sin θ gravity acting downward on the charges is T cos θ = mg . Hence F = mg . cos θ 1 16πε o l 2 mg sin 3 θ . Equating the two and solving gives Q 2 = cos θ
(
T F
)
l
Q
Q
2l sin mg
3.2.1 The problem is sketched below. The electric field due to the positive charge is Q E1 = 9 × 109 12 a y = 2,812.5a y . The electric field due to the negative charge is (4) Q E 2 = −9 × 109 22 a z = −22,500a z . Hence the total electric field is (2) V E = E1 + E 2 = 2,812.5a y − 22,500a z . m 3-3
z Q1 = 5 C
E1
(0, 4, 2) z=2 E2
E
y=4 y
Q2 = –10 C
3.2.2 The problem is sketched below. The angle θ is θ = 60 o . The distances from the triangle vertices to the center is, according to the law of cosines, 52 = d 2 + d 2 − 2d 2 cos( 2θ ) giving d=2.887 m. The vector contributions are Q Q kV o E= + 2 cos 60 = 108 . m 4π ε o d 2 4π ε o d 2
( )
Q
d
5m d
—Q
5m
5m
d
—Q
3.2.3 The problem is sketched below. (a) First we determine the electric field along the z axis. Superimposing the fields due to the two charges gives 1 Q 1 Q Ql E= a − a = z z 2 2 4π ε o 4π ε o 2π ε o l l z − z + z − 2 2
3-4
z
Ql z a = az z 2 2 2π ε o 22 l l l z + z 2 − 2 2 4
(b) Now we determine the electric field along the y axis. Superposing the fields as shown l 1 Ql Q 2 gives E = −2 az = − az . 1 3 2 4π ε o 2 l 2 2 2 2 2 l y + 2 l y + y + 4 π ε 4 o 4 4 14 4244 3 sin θ
z z
E+
E–
l 2
y
l 2
E–
y E+
E
3.2.4 The problem is sketched below. Divide the charge into chunks of charge, dQ = ρ l ad φ. { dl
At a distance d from the center and on a line perpendicular to the ring, the horizontal components cancel out leaving only the vertical components so that 2π 1 ρ l adφ E= ∫ cos(α ) a z 2 π ε 4 R o φ =0 d . Substituting these gives R 2π ρ l ad 1 E= ∫ dφ a z 3 φ = 0 4π ε o 2 2 2 d +a
where R = d 2 + a 2 and cos(α ) =
(
=
1 2ε o
ρ l ad
)
3
(d 2 + a 2 ) 2
az
z>0
At a large distance from the center, d >> a , this result reduces to 1 ρla E= a z d >> a 2ε o d 2 2π a ρ l = a z d >> a 4π ε o d 2
3-5
z ␣
E
z=d R
␣
y
a l ad x
3.2.5 The problem is sketched below. The chunks of charge are dQ = ρ s dr rdφ and 123 ds
again, by symmetry, the horizontal components cancel leaving only the vertical (zdirected) components. Summing these contributions gives a 2π ρ rdφ dr d 1 s E= ∫ ∫ az 2 2 2 2 π ε 4 d r + o 12 r = 0φ = 0 d 2+4 r3 4 4 3 14 1
cos(α )
R2
=
ρs d 1 − a z 2 ε o d 2 + a 2
=
π a2ρ s az 4π ε o d 2
z>0
d >> a
z ␣
E
z=d ␣
R
a y
r s rdrd x
3-6
3.2.6 The problem is sketched in the xy plane below. Using the results of Example 3.3 and superpositioning the fields gives (a) ρl ρl 1 1 E= ay − l 2π ε o 2π ε o y− y+ 2
l 2
ay =
ρl 2π ε o
l
l2 y 2 − 4
Similarly the fields along the x axis become E = −2
ay.
l
ρl
2
2
l l2 x2 + 2π ε o x 2 + 4 14243 4
ay.
cos θ
l 2
—
y
–
R
Er+
l 2
Er
y Er–
R Er+
x x
3.2.7 The problem is sketched below. Place the strip in the xz plane centered on the origin. C Divide the strip into infinite line charges with distribution ρ l = ρ s dz . Use the result m W
ρ s dz cos α z = 0` 2πε o R
in Example 3.3, equation (3.10). Accounting for symmetry, E = 2 ∫ where R =
z 2 + d 2 and cos α =
ay
d . Hence, using the integral R W
ρ sd 2 1 1 −1 z dz = tan , E = dz ∫ 2 ∫ 2 2 d d πε o z = 0` d + z 2 d +z 1
2
(
3-7
)
ay =
ρs W tan −1 2d πε o d
ay.
z
z
W 2 R
E– ␣
W
␣
d
E+
R —z
y
W — 2
3.3.1
The electric field intensity vector is E =
V V = 105 . The polarization vector is d m
P = D − ε o E . Substituting D = ε r ε o E gives µC 1 P = ε o (ε r − 1) E = × 10 −9 (5.4 − 1) × 105 = 389 . . 36π m2 3.4.1
The problem is sketched below. (a) the total charge enclosed is a
Qenc = ∫
2π π
∫
a
2 sin θ drdφ dθ = ∫ ∫ ρ v r144 2443
r = 0 φ = 0θ = 0
dv
2π π
∫
3 4 ∫ kr sin θ drdφ dθ =π ka . (b) Using
r = 0 φ = 0θ = 0
Gauss’ law, ∫ ε o E • ds = ε o E 4π r 2 = Qenc = π ka 4 giving E = enclosed is Qenc = π kr 4 . Hence the electric field is E =
ka 4
4ε o r 2
kr 2 ar . 4ε o
z
v = kr r
dv y
a
x
3-8
a r . (c) The charge
3.4.2
No. No closed surface can be found for which the electric field is perpendicular to all sides and hence no simplification of ∫ D • ds can be obtained. 3.4.3
The problem is sketched below. Since D is directed in the z direction, there is no flux through the sides. Hence Gauss’ law gives 2π
top
bottom
2π
a
φ− ∫ ( z = 4)rrdrd 123
Qenc = ∫ D • ds + ∫ D • ds = ∫ 123 123
φ = 0r = 0
ds
∫
a
φ= ∫ ( z = 0)rrdrd 123
φ = 0r = 0
ds
8πa 3 64π = C. 3 3
z Dz a
4
y
Dz
x
3.4.4 b 2π π
(
)
k 2 r sin θ drdφ dθ = 2π k b 2 − a 2 . By symmetry, 1442443 r = a φ = 0θ = 0 r
For r ≥ b Qenc = ∫ ρ v dv = ∫
∫
∫
dv
(
)
the field is radially directed. Hence ∫ D • ds = Qenc so that ε o Er 4π r 2 = 2π k b 2 − a 2 . Thus Er =
(
k b2 − a 2
2ε o r
2
) . For r ≤ a , Er = 0 since no charge is enclosed. For a ≤ r ≤ b ,
k 1 − Qenc = 2π k r 2 − a 2 . Hence ε o Er 4π r 2 = 2π k r 2 − a 2 . Thus Er = 2ε o
(
)
(
)
3.5.1
The problem is sketched below. The work required to move q around the paths is W = − q ∫ E • dl = − q ∫ xdx − q ∫ ydy − q ∫ zdz . (a)
3-9
a2 . r2
1
2
0
0
z =0
y =0
z =1
y =2
1
0
1
0
z =0
z =1
x =0
x =1
W = − q ∫ zdz − q ∫ ydy − q ∫ zdz − q ∫ ydy = 0 . (b) W = − q ∫ zdz − q ∫ ydy − q ∫ xdx − q ∫ xdx = 0 . z
(0, 0, 1 m)
(0, 2 m, 1 m)
z=1
(0, 2 m, 0) y
y=2 x=1 x
3.5.2
The problem is sketched below. Applying superposition and equation (3.37) yields Q 1 1 V = − = 15kV . 4π ε o 2 3 z
+
(0, 3, 2)
V
– (0, 0, 0)
y=3 Q = 10 C
y
3.5.3
The problem is sketched below. Applying superposition and equation (3.37) yields Q1 Q2 1 1 1 1 V = − + − = −13,287.42 − 14782.56 = −28,070V 4π ε o 5 2 + 5 − 3 2 3 4π ε o 5 2 + 5 − 2 2 2 ( ) ( ) ( ) ( )
3-10
z
2
(5)
Q1 = 10 C
+ (5
z=3
–
2 = – 3)
29
(0, 5, 5)
+
(5 – 2)2 + (5)2 =
V
34
y=2 y
Q2 = 5 C
3.5.4
The problem is sketched below. Applying superposition and equation (3.37) yields Q1 Q2 1 1 1 1 V = − + − = −28,287.42 − 7,282.56 = −35,570V 4π ε o 52 + 22 2 4π ε o 52 + 32 3 z (0, 0, 5)
+ 52 + 22
52 + 32 V
y = –2
y=3
–
Q1 = 10 C
Q2 = 5 C
y
3.5.5
The problem is sketched below. Applying superposition and equation (3.37) yields Q1 Q2 1 1 1 1 V = − + − = −7,282.56 + 28,287.42 = 21,005V . 4π ε o 52 + 32 3 4π ε o 52 + 22 2
3-11
z
Q2 = –10 C
z=2 52 + 22
(0, 5, 0)
y
52 + 32
x=3 Q1 = 5 C x
3.5.6
The problem is sketched below. Applying superposition and equation (3.38) yields ρ1 ρ2 3 2 + = −89,145 + 119,622 = 30,477V . ln ln V = 2π ε o 52 + 22 2π ε o 52 + 32 z z=5 + 52 + 32
y = –3 l = –10 C/m
52 + 22
V
y=2
–
1 = 5 C/m
y
3.5.7
The problem is sketched below. By the law of cosines, the distance from each charge to
( )
the center of the triangle is related to the side length as l 2 = d 2 + d 2 − 2d 2 cos 120o so l . Applying superposition and equation (3.42) yields 1732 . Q Q = 4.677 × 1010 = 31177V , . V =3 l l 4π ε o 1732 .
that d =
3-12
Q
d l
l 120°
120° d
d 120°
Q
Q l
3.5.8
The problem is sketched below. Using (3.44) gives 2π ρ l adφ ρ la . The potential is only a function of z and hence V = ∫ = φ = 0 4π ε o z 2 + a 2 2ε o z 2 + a 2 the gradient is aρ l ∂ 2 ∂V E = − gradientV = − az = − z + a2 2ε o ∂ z ∂z
(
)
− 12
az =
aρ l 2ε o
agrees with the results of Problem 3.2.4.
z
R
y
a l adC
x
3-13
z
(z
2
+a
2
)
3
2
a z which
3.5.9
The problem is sketched below. Using (3.44) gives a 2π ρ s rdrdφ ρ V = ∫ ∫ = s z 2 + a 2 − z . The potential is only a function of z r = 0 φ = 0 4π ε z 2 + r 2 2ε o o
and hence the gradient is E = − gradientV = −
∂V ρ ∂ 2 ρ z az = − s z + a 2 − z a z = s 1 − a z which 2ε o ∂ z 2ε o ∂z z 2 + a 2
agrees with the results of Problem 3.2.5. z
R a s
y
r s rdrd x
3.5.10
(a) E = − gradientV = − =
∂V ∂V ∂V ax − ay − a ∂x ∂y ∂z z
x
(x
2
2
+y +z
2
)
3
(b)
E = − gradientV = −
2
ax +
y
(x
2
2
+y +z
2
)
3
2
ay +
∂V ∂V 1∂V ar − aφ − a r ∂φ ∂r ∂z z
= − e − z cos φ a r + e − z sin φ a φ + re − z cos φ a z
3-14
z
(x
2
2
+y +z
2
)
3
2
az
(c) E = − gradientV = − =2
sin θ cos φ r3
∂V 1∂V 1 ∂V ar − aθ − aφ r ∂θ r sin θ ∂ φ ∂r cos θ cos φ sin φ
ar −
r3
aθ +
r3
aφ
3.6.1 µC µC kV V , D = ε oε r E = 0.478 2 , P = D − ε o E = 0.389 2 , E = = 10 m d m m A C = ε r ε o = 477.5 pF . d 3.6.2 2
π (01 .) A C = εo = εo = 277.8pF . − d 10 3 3.6.3 There are two capacitors in series: C1 = ε r1ε o
A A and C2 = ε r 2ε o . Capacitors in d1 d2
ε r1ε r 2 series add like resistors in parallel so that C =
C1C2 d1d 2 = εoA . The total free ε r1 ε r 2 C1 + C2 + d1 d2
charge on the upper (and lower) plate is Q f = CV . Using Gauss’ law and surrounding Qf CV = . In the the upper plate with a closed surface yields DA = Q f . Therefore D = A A D CV = and in the lower dielectric, upper dielectric, E1 = ε r1ε o ε r1ε o A D CV V E2 = = . Evaluating these gives C = 70.74pF , E1 = 4000 , m ε r 2ε o ε r 2ε o A V E2 = 2000 . m 3.6.4 A A There are two capacitors in parallel: C1 = ε r1ε o 1 and C2 = ε r 2ε o 2 . Capacitors in d d (ε A + ε r 2 A2 ) . Using parallel add like resistors in series so that C = C1 + C2 = ε o r1 1 d Gauss’ law and surrounding each portion of the upper plate with a closed surface yields D1 C1V = and D1 A = Q f 1 and D2 A = Q f 2 . Therefore E1 = ε r1ε o ε r1ε o A1 3-15
D2
C2V . Evaluating these gives C1 = 88.42pF , C2 = 707.4pF , ε r 2ε o ε r 2ε o A2 V V C = 795.8pF , E1 = 5000 , E2 = 5000 . Observe that V = E1d = E2 d = 10V and m m E2 =
=
Q f = Q f 1 + Q f 2 = CV = 7.96nC . 3.6.5 We observe that there are essentially two spherical capacitors in series. First we obtain the capacitance of a spherical capacitor with a homogeneous dielectric filling the interior. Using Gauss’ law and surrounding the inner sphere with a sphere of radius r gives Qf Qf D= . Hence the electric field is radially directed and is E = . The voltage 2 4π r 4π ε r 2 a Qf Q f 1 1 between the spheres is V = − ∫ dr = − . Hence the capacitance is 2 4π ε a b b 4π ε r C=
Qf V
=
4π ab . This result was derived in Exercise Problem 3.8. Hence the (b − a )
capacitances are C1 =
4π ar1 4π br1 and C2 = . Since these are in series and capacitors (r1 − a ) (b − r1 )
in series add like resistors in parallel we obtain CC 1 . Substituting the values gives C = 1 2 = 4π ε o C1 + C2 1 1 1 1 1 1 − + − ε r 2 r1 b ε r1 a r1 0.303pF. 3.6.6 The per-unit-length capacitance for a coaxial cable filled with a homogeneous dielectric 2π ε F . For this problem we observe that there was obtained in Example 3.15 as c = b m ln a 2π ε oε r1 2π ε oε r 2 and c2 = . Since capacitors in ae two such capacitors in series: c1 = r1 b ln ln a r1
3-16
series add like resistors in parallel, c =
c1c2 = c1 + c2
2π ε oε r1ε r 2 . Evaluating this b r1 ε r1 ln + ε r 2 ln a r1
for the given dimensions yields 82.06 pF m . 3.7.1 Converting the radius from mils to meters gives 1inch 2.54cm 1meter 16mils × × × = 4.064 × 10 −4 m . Hence the resistance is 1000mils 1inch 100cm 1000m . Ω. R= = 3323 7 −4 2 σ = 5.8 × 10 × A = π 4.064 × 10
(
)
3.7.2 V V . Hence the current density is J = σ E = σ . The total d d V V d current is I = JA = Aσ . Hence the resistance is R = = Ω . Evaluating this d I σA The electric field is E =
gives 50mΩ. 3.7.3 The electric field intensity at a radius r was determined in Problem 3.6.5 as E =
Qf
. 4π ε r 2
The voltage between the spheres was also determined as a Qf Q f 1 1 V = −∫ dr = − . The current flowing between the two spheres is 2 a b π ε 4 4 π ε r b I = JA = σ EA = σ R=
Qf
Qf 2 π r = σ . Hence the resistance is 4 ε 4π ε r 2
1 1 1 V = − . Evaluating this for the given dimensions yields 106.1Ω. I 4π σ a b
3.7.4 The voltage between the inner wire and the shield was determined in Example 3.15 in ρ l b C as V = terms of the charge distribution on the inner wire of ρ l ln . m 2π ε a Similarly, the electric field in this region was determined in Example 3.7 to be
3-17
E=
ρl 2π ε r
. Hence the current flowing from one cylinder to the other (per-unit-of line
length) is I = JA = σ r=
1 V b = ln I 2π σ a
ρl 2π ε r
2π r = σ
ρl ε
. Thus the resistance per unit length is
Ω / m . Evaluating this for the given dimensions yields 01 .Ω/m
3.7.5 From the results of Example 3.16, the magnetic flux density vector at a perpendicular L µ I 2 distance from the midpoint of the current element is B = o a φ . (a) At 2 2π r L r2 + 4 (0,3m,0) the field is 0.2858 a φ µ T . (b) Off the ends of the curent element dl × a R is zero and hence the field is zero off the ends.
3.7.6 Utilizing the result obtained in Example 3.16 for an infinite current, the magnetic flux µ I Wb . Hence the total magnetic flux penetrating the loop density at a distance r is B = o 2π r m 2 is ψ = ∫ B • ds . By the right-hand rule the B field is directed into the page and hence the dot product can be removed. However, the B field depends on distance away from the current and cannot be removed from the integral. Hence l r2 µ I o drdz = µ o Il ln r2 Wb . For the given parameters we obtain ψ= ∫ ∫ 2π r1 z = 0 r = r1 2π r
ψ = 0139 . µWb . 3.7.7 The problem is sketched below. Using the results of Example 3.16 we may superimpose the contributions from the four identical sides. Observe that the contributions from opposite sides cause the net B from those two to be directed in the +z direction. similarly l µoI 2 the other two sides cause a similar result. Hence B = 4 cos α a z where 2π R 2 l2 R + 4
3-18
l 2 l z + 4 and cos α = 2 . Combining these gives R 2
R= B=
l2
2µ o I
π
4
a z . For l = 2m and I = 10A , at the center of the
z2 + l z2 + l2 4 2 2
loop, z = 0 , we obtain B = 5.66µ Ta z .
z ␣
␣
R
R l l (– , , 0) 2 2
I ␣
␣
I l 2
l 2 I
I y
l l ( , , 0) 2 2
x
3.7.8 The problem is sketched below. Using the results of Example 3.16 we may superimpose the contributions from the three identical sides. The radial distance from the center of l each side to the center of the triangle is r = 0.577 . Using (3.60) of Example 3.16 we 2 l 9µ o I µ I 2 = into the page. For l = 5cm and I = 4A we obtain obtain B = 3 o 2π l 2π r 2 l2 r + 4 B = 144µ T .
3-19
30° l
I
l
I r
r r I l
3.7.9 The problem is sketched below. According to the Biot Savart law, the magnetic field off the ends of a current is zero since dl × a R is zero. Hence there are no contributions to the magnetic field at point P due to sides DA and BC. For side CD dl × a R = rdθ and is out of the page. Hence the Biot Savart law gives for this contribution θ µ I µoI µ I r dθ = o θ . Similarly the contribution from the segment AB is o θ ∫ 2 2 4π r1 4π r2 θ = 0 4π (r2 ) which is into the page. Hence the total is B =
µoI θ 1 1 − which is directed into the 4π r1 r2
page since r2 > r1 . C I B I I D
I
P
A r1 r2
3-20
3.9.1 The problem is sketched below. Use the result for the magnetic field from an infinitely I . (a) At the center, the H fields long current filament obtained in Example 3.19: H = 2π r I 2I = . (b) At a distance D from the center and in a plane add giving H = 2 d π d 2π 2 containing the currents we superimpose the fields to give I I I d H= − = . 2π 2 d 2 2π D − d 2 2π D + d 2 D − 4
(
)
(
)
I
I D
d 2
d 2
3.9.2 The problem is sketched below. Place the strip on the z axis centered about the origin. Treat the linear current density as filaments of current Kdz A . Superimpose the magnetic fields using equation (3.81) of Example 3.19. The magnetic field intensity vector is w
2 Kdz
H=2 ∫
cos α a z where R = d 2 + z 2 and cos α =
z = 0 2π R w w 1 Kd 2 K −1 z 2 K −1 w H= dz = tan = tan ∫ π z =0 d 2 + z 2 π 2d d z = 0 π
(
w → ∞, H →
)
d . Hence R . For an infinite strip as
K which agrees with the result for an infinite strip in Example 3.21. 2
3-21
z z= Kdz
w 2
␣
␣
R ␣ d
y R
z=_
w 2
3.9.3 2π
For (b) a < r < b , ∫ Hφ rdφ = NI . Hence Hφ = φ =0
NI . For (a) r < a and (c) r > b the 2π r
net current penetrating this loop of radius r is zero and hence H=0.
3.9.4 The problem is sketched below. Construct a rectangular contour as shown. By symmetry, the magnetic field is directed along the solenoid and is directed right to left. Applying Ampere’s law to this yields ∫ H • dl = HL = I enclosed = InL . Hence H = nI and B = µ r µ o H = µ r µ o nI . L C H
H
I
I
3.11.1 The magnetic flux density along the axis of the solenoid was obtained in Problem 3.9.4 as B = µ r µ o nI . Since this is uniform over the cross section of the core and is axially 2 directed, the flux is ψ = ∫ B • ds = Bπ a . The flux linkages per unit length are core cross section
3-22
Λ = nψ . Hence the per-unit-length self inductance is l =
Λ = µ r µ o n 2π a 2 I
H . For m
2
turns H 2 . . the given dimensions, l = 1000 × 4π × 10 −7 × 2000 π ( 0.01) = 158 m m
3.11.2 Treating this as a long parallel wire line of length l and separation w and neglecting the contribution from the end segments gives, using the result of Example 3.25 for an infinite µ l w line, L ≅ o ln . π a
3.11.3 The problem is sketched below. Assuming the plate width is much greater than the plate separation we can use the result in Example 3.18 for the magnetic flux density from an A . The magnetic flux density is infinite plate carrying a linear current density of K m µ K parallel to the plate and opposite to the direction of the current: B = o . 2 Superimposing the fields due to both plates gives B = µ o K which is constant across the cross section between the plates. The flux penetrating the surface between the plates is
ψ = ∫ B • ds = Bs∆l = µ o Ks∆l . The total current on each plate is I = Kw . Hence the ψ s s flux is ψ = µ o I∆l . Thus the inductance per unit length is l = I = µ o . ∆l
w
w B
s
3-23
w
3.11.4 First we solve the basic problem shown below. The magnetic field intensity due to the µ I infinitely long current is B = o . The flux through the loop is obtained by integration 2π r b
µoI µ Il b dr = o ln . Applying this to the original problem and a 2π r = a 2π r
as ψ = l ∫
superimposing the fluxes due to each current gives s w s w µ o Il D − 2 + 2 µ o I D + 2 + 2 ψ2 = ln ln − D − s − w 2π D + s − w 2π 2 2 2 2 s w s w ψ 2 µ ol D − 2 + 2 D + 2 − 2 is M = . = ln I 2π D − s − w D + s + w 2 2 2 2
( (
( (
) )
)( )(
( (
) )
a
) . Hence the mutual inductance )
l
I b
3.12.1 The particle is traveling with constant velocity so that F = ma = 0 . In order for the vertical forces to balance so that the particle passes through the hole we must have the electric force, Fe = qE , equal the magnetic force, Fm = qvB . Solving gives the critical E velocity of the particle as v = . Evaluating this for the given conditions yields B 2000 m v= = 2 × 10 6 . −3 s 1 × 10
3.12.2 The magnetic flux density vector is radially directed about each wire and is given by µ I B = o . The magnetic flux density at wire 2 due to the current of wire 1 is therefore 2π r
3-24
µ o I1 and is perpendicular (into the page) to current I 2 . The force exerted on a ∆l 2π s µ I section of wire 2 is, according to the Lorentz force equation, F21 = I 2 ∆lB21 = I 2 ∆l o 1 . 2π s F µ II N . Hence the force per unit length exerted on wire 2 is f 21 = 21 = o 1 2 ∆l 2π s m B21 =
3.12.3 From the previous problem, the force exerted on the left side of the loop is µ II µ II F1 = o 1 2 w , and the force exerted on the right side of the loop is F3 = o 1 2 w . 2π a 2π b Along the upper and lower segments of the loop, the B field varies along the wire. Hence b µ I I o 1 2 dr = µ o I1 I 2 ln b = F . the force along the upper segment is F2 = ∫ 4 a 2π r = a 2π r
3.12.4 The sliding bar cuts the magnetic field resulting in a voltage source , Bvw , inserted in it Bvw . as shown below. Hence the current is I = − R
R
+ –
Bvw
I
3.12.5 The vertical side of the loop cuts the magnetic field so that a voltage is induced in it. The horizontal sides have no inserted source since v × B is perpendicular to the wire. Shown below is a view in the xy plane. The tangential velocity is v = lω and
v × B = lω B sin(ω t ) assuming that the loop starts at the x axis at t=0. The voltage source is inserted as shown below so that the current through the resistor is wlω B sin(ω t ) I=− . R
3-25
B
y
t
l v
x B
I
– +
R
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l B sin t
