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Fall 10
EEE-161
Final
12-13-10
General instructions:
 12/13/2010 Monday, 5:15 to 7:15 pm, RVR-1006
 Open book/note
 Use bluebook (10 points deduction)
 Use engineering units
 5 problems (5 out of 6) – 20 points each
 No computer (laptop, palmtop, …)
 Show details
 Include units in your answers
 Random seating assignments
1) 0) A 2.5 m long cylindrical wire’s diameter is 5 mm and a uniform electric field of 3
V/m is set up inside the wire. Determine the total current flow if the conductivity of
the wire is 500 S/m.
L  2.5m
E  3
d  5mm
V
  500
m
J   E
r 
3
d
r  2.5  10
2
2
A  r
m
5 2
A  1.963  10
m
S
m
3 A
J  1.5  10
I  J A
I  0.029 A
2
m
2) 1) A conductor occupying the region z  0 has a surface charge distribution (in
cylindrical coordinates) s = 20 exp(-r2) nC/m2 at z = 0 m, determine
a) the electric field (in rectangular coordinates) at (0.2 , 0.3 , 0+).
b) the electric field (in rectangular coordinates) at (0.2 , 0.3 , 0-).
x  0.2
y  0.3
2
 r nC
2
 ( r)  20 e
r 
2
2
x y
r  0.361
9 C
2
 ( r)  17.562  10
m
m
 ( r)
at z = 0+
Ex  0
Ey  0
Ez 
at z = 0-
Ex  0
Ey  0
Ez  0
0
3V
Ez  1.983  10
m
Fall 10
EEE-161
Final
12-13-10
3) 2) A microstrip test fixture (d = 2 cm and r = 3.5) is used to measure the dielectric
strength of materials.
a) Determine the dielectric strength (MV/m) of the test material if it breaks down at
2 MV.
b) Determine the surface charge density just before the occurrence of breakdown.
V0
d
w

d  2cm
 r  3.5
Dn  Emax r  0
Vmax  2MV
 s  Dn
Emax 
 s  3.099  10
Vmax
d
MV
Emax  100
m
3 C
2
m
4) 2) The inner radius of a coaxial transmission line is 5 cm. The insulation has a
dielectric strength = 100 MV/m and an r = 2.5. Determine the minimum outer radius
(in cm) of the transmission line if it is expected to safely handle 4.2 MV.
 r = 2.5
a = 5 cm
b=?
a
b
a  5cm



2  r  0 Vmax
 b
a ln 
a
Emax
 b
ln 
a
Vmax
Emax
MV
Emax  100
m
Vmax  4.2MV
 b
ln 
a



2    r  0 a
Vmax
b
a Emax
a
Emax
 r  2.5
Vmax
 b
a ln 
a
 Vmax 

 a Emax
exp 
 Vmax 

 a Emax
b  a exp 
b  11.582 cm
Fall 10
EEE-161
Final
12-13-10
5) 2)If the charge density between the plates is v(z) = sin(z), find the E-field between
the following infinite parallel-plate system by solving the Poisson’s equation.
z
V0
r
d
2
2

V( z)
dz
d
V ( z)
dz
V ( z)
V( d)

E



( cos( z)  A) d z  sin( z)  Az


 A z  B
V ( 0)
V0
V0 
A


sin( z) d z  cos( z)


sin( z)
1
 ( cos( z) )  A

sin( z)
sin( d)
V0


E

cos( z)



B
0
 V  sin( d) 
 0


z

d


sin( z)
V( z)
d
0
 A d
sin( d)
d
 V( z)
dz
d
V0 
sin( d)

d
6) 3) An infinite coaxial line (a = 2 cm, b = 4 cm, and c = 6 cm) carries 200 mA in the
inner conductor and -100 mA in the outer conductor.
a) Determine theH-field at r = 1 cm.
b) Determine theH-field at r = 3 cm.
c) Determine theH-field at r = 5 cm.
d) Determine theH-field at r = 7 cm.
a
c
b
Fall 10
EEE-161
Final
a  2cm
a)
b  4cm
r  1cm
c  6cm
12-13-10
Iin  200mA
Iout  100 mA
Iinr
H 
A
H  0.796
m
2
2 a
b)
c)
d)
r  3cm
r  5cm
r  7cm
H 
H 
H 
Iin
A
H  1.061
m
2 r
Iin
2  r


Iout
2

2
 c  b

2
 r  b

2
A
H  0.493
m
2  r
Iin  Iout
A
H  0.227
m
2 r
7) 3) Charged particles traveling at u = 2aˆ x  3aˆ y   105 m/s enter a region having a
uniform B-field: B = 3.6âx  T. Determine the magnitude and direction of the E-field
that is required to permit the undeflected passage of the charged particles.
Answer: F = q(E + u  B) = 0
u  B = - âz (3.6  3  105) = - âz (1.08  106)
E = âz (1.08  106) V/m
8) 3) Determine the force exerted on the wire if the magnetic field B = ( 2 â x – 3 â y )
10-5 T. Express the answer in rectangular coordinates.
z
2
I=2A
3
x
 0 
 2 
5
2 mA 3  m   3  10 T 
 2 
0 
 
 
 120  10  9 



9 N
 80  10 

9
 120  10 
y
Fall 10
EEE-161
Final
12-13-10
9) 4) For a coil with radius a = 1/8 inch, length l = 1 inch, and N = 15 turns:
a) Calculate the inductance of coil in terms of H using the long coil approximation
given in the text.
b) Calculate the inductance of coil in terms of H using the modified formula given
below:
L
 N 2 a 2
l 1  2a l 
2
N 
a 
15
1
8
N  a  
l
2
L 
2
N  a  
2
L 
l
l 
in
1
L
0.353
H
L
0.342
H
2
2a
  
 l 
2
  4    10  7
1in
H
m
10) 4) A coaxial transmission line has a Z0 = 75 Ω. If the radius of the inner conductor is
2 cm and r = 2.5 for the dielectric material, determine the radius of the outer
conductor. Also compute the inductance (H/m) and capacitance (pF/m).
Z0  75
Z0
L 
 r  2.5
 b
ln 
r  a 
60
0
 b
ln 
2  a 
e
Z0   r
60
L  0.395
a  2cm
b
b  a e
a
H
m
C 
Z0   r
60
2  r  0
 b
ln 
a
b  14.434 cm
C  70.369
pF
m
11) 0) Determine the total inductance (H/m) for a cylindrical conductor (r = 5 mm) 1.5
feet above a conducting plane.
d  1.5ft
a  5mm
7H
Lint  0.5 10
m
Lext 
0
 d
ln 
2   a 
Ltotal  Lint  Lext
7H
Lext  9.031  10
H
Ltotal  0.953
m
m
Fall 10
EEE-161
Final
12-13-10
12) 5) The core of a magnetic circuit is of mean length 50 cm and uniform cross-sectional
area 23 cm2. The relative permeability of the core material is 1500. An air gap of 1.5
mm is cut in the core. Determine the total reluctance of the magnetic circuit.
Reluctance
ltotal  50cm
lair  1.5mm
Acore  a b
Rcore 
l
Acore  6 cm
lcore
 r  1500
 A
 0  4 10
lcore  ltotal  lair

2
m
lcore  0.499 m


Aair  a  lair  b  lair
31
Rcore  440.771  10
H
 r  0  Acore
7H
a  2cm
Aair  6.772 cm
Rair 
lair
b  3cm
2
61
Rair  1.763  10
H
 0  Aair
61
Rtotal  2.203  10
H
Rtotal  Rcore  Rair
2
13) 5) Determine the total magnetic flux through the following magnetic circuit.
Area = 22 cm2
N = 100 turns
I = 1.5 A
mean length:
l1 = l3 = 5 cm and l2 = l4 = 7 cm
I
1
r1 = 500, r2 = 1000, r3 = 700, and r4 = 1500.
Area  2cm 2cm
N I  150 A
2
4
N  100
 0  4 10
I  1.5 A
 r1  500
l1  5cm
 r2  1000
l2  7cm
 r3  700
l3  5cm
 r4  1500
l4  7cm
B1 l1
F
N I
F
l2
l3
l4  
 l1




  r1  0  r2  0  r3  0  r4  0  Area


 
Or:
Area  4 cm
H1 l1  H2 l2  H1 l1  H4 l4
 r1  0

B2 l2
 r2  0

B3 l3
 r3  0

 r4  0
  261.713  10
l2
6
Wb
l
  Area
l1
l2
 r1  0  Area
51
R1  1.989  10
H
R2 
 r2  0  Area
51
R2  1.393  10
H
l3
R3 
 r3  0  Area
51
R3  1.421  10
H
l4
R4 
 r4  0  Area
41
R4  9.284  10
H
R1 
m
N I
l3
l4 





  r1  0  r2  0  r3  0  r4  0 


Reluctance
7H
B4 l4
N I  Area
l1
Rtotal  R1  R2  R3  R4
51
Rtotal  5.731  10
H
 
N I
Rtotal
3
F
  261.713  10
N I
6
 Rtotal
Wb
6) One of the 3 practice problems from Chapter 12 (p. 198-199, 12.3, 12.4, 12.7).
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