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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.013 Electromagnetics and Applications
Quiz 1, October 20, 2005
6.013 Formula Sheet attached.
Problem 1 (35 Points)
J=J z ( r ) i z = J 0
r
iz
R1
amperes/meter 2
z
r φ
0 < r < R1
R1
R2
A coaxial cable of very long length carries a z directed current density that varies with radial
position on the inner cylinder as:
r
i z amperes/meter 2 0 < r < R1
R1
A perfectly conducting outer cylinder of radius R 2 carries all the return current so that
J=J z ( r ) i z = J 0
H = 0 for r > R 2 .
a) Find the H field for 0 < r < R1 .
b) What are the magnitude and direction of the surface current density (amperes/meter)
on the r = R 2 surface?
Problem 2 (35 Points)
V0
α
r
i
φ
ε ,σ
Depth d
z
r
0
R1
R2
Two flat electrodes at angle α extend from radius R1 to R2 and have a depth d in the z direction
(out of the paper). The electrodes enclose a lossy dielectric medium with permittivity ε and
conductivity σ . There is no free volume charge within the lossy dielectric. The electric
potentials on the electrodes are Φ (φ = 0 ) = 0 and Φ (φ = α ) = V0 .
a) The scalar electric potential Φ is of the form Φ (φ ) = Aφ + B . What values of A and
B satisfy the boundary conditions?
b) Find the electric field E(r, φ ) within the lossy dielectric.
c) What is the free surface charge density, σ sf , on the electrode at φ = α ?
d) What is the capacitance of this device? You may neglect fringing fields.
Problem 3 (30 Points)
x
ε,µ
•
θi
z
y
Ei
•
Hi
An electromagnetic wave is traveling at an angle θi with respect to the z axis within a medium
with dielectric permittivity ε and magnetic permeability µ . The magnetic field is given as:
⎡ j ⎛⎜ 2π ×108 t−π ( x+
H i = H 0 Re ⎢e ⎝
⎢⎣
3z
) ⎠⎞⎟ ⎤
⎥ iy
⎥⎦
amperes/meter .
a) Find the frequency in Hertz.
b) Find the wavelength in meters.
c) Find the numerical value of the speed of light in the medium in meters/second.
d) Find the angle θi .
6.013 Quiz 1 Formula Sheet
October 20, 2005
Cartesian Coordinates (x,y,z):
∇Ψ = x̂ ∂Ψ + ŷ ∂Ψ + ẑ ∂Ψ
∂x
∂y
∂z
∂A x ∂A y ∂A z
∇iA =
+
+
∂x
∂y
∂z
∂A y ⎞
⎛ ∂A
∂A ⎞ ⎛ ∂A y ∂A x ⎞
⎛ ∂A
∇ × A = xˆ ⎜ z −
+ ŷ ⎜ x − z ⎟ + ẑ ⎜
−
⎟
∂y
∂z
∂z
∂x
∂x
∂y ⎟⎠
⎝
⎠
⎝
⎠
⎝
2
2
2
∇ 2 Ψ = ∂ Ψ + ∂ Ψ +
∂ Ψ
∂x 2 ∂y 2 ∂z 2
Cylindrical coordinates (r,φ,z):
∇Ψ = r̂ ∂Ψ + φˆ 1 ∂Ψ + ẑ ∂Ψ
∂r
∂z
r ∂φ
∂ ( rA r ) 1 ∂A φ ∂A z
∇iA = 1
+
+
∂z
r ∂r
r ∂φ
r̂
rφ̂
ẑ
⎛ ∂ ( rA φ ) ∂A ⎞ 1
⎛ 1 ∂A z ∂A φ ⎞
∂A
∂A
⎛
⎞
1
∇ × A = r̂ ⎜
−
+ φˆ ⎜ r − z ⎟ + ẑ ⎜
− r ⎟ = det ∂ ∂r ∂ ∂φ ∂ ∂z
∂r ⎠
∂φ ⎠ r
∂z ⎟⎠
r ⎝ ∂r
⎝ ∂z
⎝ r ∂φ
A r rA φ A z
( )
2
2
∇ 2 Ψ = 1 ∂ r ∂Ψ + 1 ∂ Ψ + ∂ Ψ
r ∂r ∂r
r 2 ∂φ2 ∂z 2
Spherical coordinates (r,θ,φ):
∇Ψ = r̂ ∂Ψ + θˆ 1 ∂Ψ + φˆ 1 ∂Ψ
∂r
r ∂θ
r sin θ ∂φ
(
)
∂A φ
∂ r 2
A r
∂ ( sin θA θ )
1
∇iA =
+ 1
+ 1
∂r
∂θ
r sin θ ∂φ
r sin θ
r2
⎛ ∂ ( sin θA φ ) ∂A θ ⎞
⎛ 1 ∂A 1 ∂ ( rA φ ) ⎞
1 ⎛ ∂ ( rA θ ) − ∂A r ⎞
r −
∇ × A = r̂ 1 ⎜
−
⎟
⎟ + θˆ ⎜
⎟ + φˆ ⎜
∂θ
∂φ ⎠
r sin θ ⎝
r ⎝ ∂r
∂θ ⎠
⎝ r sin θ ∂φ r ∂r ⎠
r̂
rθ̂
r sin θ φ̂
1
=
∂ ∂φ
det ∂ ∂r ∂ ∂θ
r 2 sin θ
A r rA θ r sin θA φ
(
)
(
)
∂ sin θ ∂Ψ +
1
∂ 2Ψ
∇ 2 Ψ = 1 ∂ r 2 ∂Ψ + 1
∂r
∂θ
r 2 ∂r
r 2 sin θ ∂θ
r 2 sin 2 θ ∂φ2
Gauss’ Divergence Theorem:
∫V ∇iG dv = ∫ A Gin̂ da
Stokes’ Theorem:
∫ ( ∇ × G )in̂ da =
A
∫ C G id
Vector Algebra:
∇ = x̂∂ ∂x + ŷ∂ ∂y + ẑ∂ ∂z
A • B = A x Bx + A y By + Az Bz
∇ • ( ∇× A ) = 0
∇× ( ∇× A ) = ∇ ( ∇ • A ) − ∇ 2 A
Basic Equations for Electromagnetics and Applications
f = q ( E + v × µo H ) [ N ]
E1// − E 2 // = 0
H1// − H 2 // = J s × n̂
∇ × E = −∂ B ∂t
B1⊥ − B2 ⊥ = 0
d
∫ c E • ds = − dt ∫A
B • da
∇ × H = J + ∂ D ∂t
D1⊥ − D 2 ⊥ = ρs
Fundamentals
d
∇•D = ρ →
∫ A D • da = ∫V ρdv
∇•B = 0 →
∫ A B • da = 0
Electromagnetic Quasistatics
E = −∇Φ (r), Φ(r) = ∫V ' ( ρ ( r ) 4πε r '− r ) dv '
∇ 2Φ =
∇ • J = −∂ρ ∂t
-1
E = electric field (Vm ) H = magnetic field (Am-1) D = electric displacement (Cm-2) B = magnetic flux density (T) Tesla (T) = Weber m-2 = 10,000 gauss ρ = charge density (Cm-3)
-2
J = current density (Am )
C = Q/V = Aε/d [F]
L = Λ/I
i(t) = C dv(t)/dt v(t) = L di(t)/dt = dΛ/dt we = Cv2(t)/2; wm = Li2(t)/2 Lsolenoid = N2µA/W
τ = RC, τ = L/R Λ = ∫ B •
da (per turn)
A
J s = surface current density (Am-1) ρs = surface charge density (Cm-2) εo ≈ 8.854 × 10-12 Fm-1
µo = 4π × 10-7 Hm-1
c = (εoµo)-0.5 ≅ 3 × 108 ms-1
Electromagnetic Waves
( ∇ 2 − µε∂ 2
C
∂t 2 ) E = 0 [Wave Eqn.]
( ∇ 2 + k 2 ) E = 0, E = E o e−
jk i r
ηo ≅ 377 ohms = (µo/εo)0.5
( ∇ 2 − µε∂ 2
- ρf
ε
F = I × µ o H [ Nm -1 ]
σ = conductivity (Siemens m-1)
e = -1.60 × 10
∂t 2 ) E = 0 [Wave Eqn.]
k = ω(µε)0.5 = ω/c = 2π/λ
Ey(z,t) = E+(z-ct) + E-(z+ct) = Re{E y (z)e
jω t
}
Hx(z,t) = ηo-1[E+(z-ct)-E-(z+ct)] [or(ωt-kz) or (t-z/c)]
∫ A ( E × H ) • da + ( d dt ) ∫V ( ε E 2 + µ H 2 ) dv
= − ∫V E • J dv (Poynting Theorem)
2
2
kx2 + ky2 + kz2 = ko2 = ω2µε
vp = ω/k, vg = (∂k/∂ω)-1
θr = θi
sin θt sin θi = k i k t = n i n t
Media and Boundaries
θc = sin −1 ( n t n i )
D = εo E + P
θB = tan −1 ( ε t
εi )
0.5
for TM
∇ • D = ρf , τ = ε σ
θ > θc ⇒ E t = E i Te +αx − jk z z
∇ • ε o E = ρf + ρ p
k = k '− jk ''
∇ • P = −ρp , J = σE
Γ = T −1 )
ε = ε o 1− ωp ω2 , ωp = ( Ne 2 mεo )
2
ε eff = ε (1 − jσ ωε )
skin depth δ = (2/ωµσ)0.5 [m]
T TE = 2 (1 + [ ηi cos θt ηt cos θi ])
B = µH = µ o ( H + M )
(
1
2
0 = if σ = ∞
∫ c H • ds = ∫A J • da + dt ∫A
D • da
-19
n̂
0.5
(Plasma)
T TM = 2 (1 + [ ηt cos θt ηi cos θi ])