6.013 Content:
Fundamentals and Applications
Fields, media, and boundaries
Circuits
Motors, generators, and MEMS
Limits to computation speed
Microwave communications and radar
Wireless communications and waves
Optical devices and communications
Acoustics
Prerequisites:
6.003 or 6.02+6.007
[6.002+8.02+18.02]
Appendices B-E
Handouts:
Administration sheet, Equations
Subject outline, lecture notes
Prob. set 1, Objectives & Outcomes
L1-1
WHAT ARE E AND H?
Lorentz Force Law:
f = q ( E + v × μ o H ) [Newtons]
4π×10-7
Velocity [m/s]
Charge [Coulombs]
“EM fields were invented to explain forces”
Find⎯E by setting ⎯v = 0 and measuring ⎯f
Find⎯H by setting ⎯v = 0, ⎯E = 0, and measuring ⎯f
(requires 2⎯f measurements using 2⎯v tests)
L1-2
INTEGRAL MAXWELL’S EQUATIONS
Graphical Equations:
ˆ = ∫∫∫ V ρ dv
∫∫ S D • nda
Gauss
ρ
ˆ =0
∫∫ S B • nda
Gauss
0
D
B
0 (statics)
∂
ˆ
•
=
−
E
d
s
∫∫A B • nda
∫
t
∂
c
Faraday
E
B
0 (statics)
H
ˆ + ∂ ∫∫A D • nda
ˆ
Ampere
∫ H • ds = ∫∫A J • nda
∂t
c
J
D = εE,
B = μH
D
Constitutive relations
L1-3
EXAMPLE: SPHERICAL CHARGE
Example I. Sphere of radius R, charge density ρo
Spherical symmetry precludes θ and φ components for E
Solution:
ˆ = ∫∫∫ V ρ dv = Q (Gauss’s Law)
∫∫ S D • nda
E
Q
2 [V/m]
4πεr
ρ r
r < R ⇒ εEr(4πr2) = ρo( 4 πr 3) ⇒ Er = o
[V/m]
3
3ε
r > R ⇒ εEr(4πr2) = Q
ρο
r
R
Surface =
Er
∝r
4πr2
ρoR3
Equivalently, r > R ⇒ Er =
3εr 2
⇒ Er =
∝ 1/r2
0
R
r
L1-4
EXAMPLE: CYLINDRICAL CHARGE
Example 2. Cylinder of radius R, charge density ρo
Cylindrical symmetry precludes θ and z components for E
Solution:
L
ˆ = ∫∫∫ V ρ dv
∫∫ S D • nda
E
(Gauss)
2
ρ
R
r > R ⇒ εEr(2πrL) = ρo(πR2L) ⇒ Er = 2oεr [V/m]
r < R ⇒ εEr(2πrL) = ρo(πr2L) ⇒ Er =
ρο
r
R
∞
Er
∝r
Surface area = 2πrL
0
ρor
2ε
[V/m]
∝ 1/r
R
r
L1-5
HOLLOW SPHERICAL CHARGE
Example 3:
Hollow sphere of inner radius Ro, charge density ρo
Spherical symmetry precludes θ and φ components for E
Solution:
ˆ = ∫∫∫ V ρ dv = Q (Gauss’s Law)
∫∫ S D • nda
E
r > R ⇒ εEr(4πr2) = Q ⇒ Er =
Ro
ρο
r
Q
2 [V/m]
4πεr
r < Ro ⇒ εEr(4πr2) = 0 ⇒ Er = 0
R
Er
Surface = 4πr2
Hollow cylinders also have zero⎯E
[V/m]
∝ 1/r2
0
Ro R
r
L1-6
CYLINDRICAL CURRENT J[A/m2]
Example 4. Uniform current J z for r < R:
L
H
Jο
z
r
0
∂
ˆ + ∫∫A D • nda
ˆ
∫ H • ds = ∫∫A J • nda
∂t
c
r > R: 2πr Hθ = I = JοπR2 ⇒ Hθ = I/2πr [A/m]
r < R: 2πr Hθ = Jοπr2
R
⇒ Hθ = Jοr/2 [A/m]
Hθ
θ
∝ 1/r
0
r
R
Example 5. Hollow current Jz :
Jοο
z
Roo
r
Hollow cylinders
also have zero H
for r < Roo
Hθθ
∝ 1/r
0
Roo R
r
L1-7
STATIC POTENTIALS Φ(Volts)
Example 6. Spherical charge density ρo, radius R:
What is the electric potential Φ(r) [volts] for r > R?
Volts = [volts/meter]×[meters]
ρο
Φ(r = a)
Φ(r = b)
R E = rˆ Q
4 πr 2
Solution:
b
Φ a − Φ b = ∫ Eid r (Volts)
a
∞
Φ(r) − Φ ∞ = ∫ Eid r = Φ(r)
r
∞
∞
Q
Q
−2
−1
r rˆid r = −
r
Φ(r) = ∫
r 4π
4π |r
Φ(r) = Q Volts, (r > R)
4 πr
Absolute potentials are defined relative to infinity: Φ(∞)
0
L1-8
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6.013 Electromagnetics and Applications
Spring 2009
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