Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.013 Electromagnetics and Applications
Lecture 1, Sept. 8, 2005
I. Maxwell’s Equations in Integral Form in Free Space
1. Faraday’s Law
d
C
E ds = - H da
dt S
0
Circulation Magnetic Flux
of E
0 = 4 ×10 -7 henries/meter
[magnetic permeability of
free space]
EQS form:
C
E ds = 0 (Kirchoff’s Voltage Law, conservative electric
field)
di
MQS circuit form: v = L (Inductor)
dt
2. Ampère’s Law (with displacement current)
d
H ds
C
= J da
S
dt
E da
S
0
Circulation Conduction Displacement
of H Current Current
MQS form:
H ds
C
= J da
S
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 1 of 7
dv
EQS circuit form: i = C (capacitor)
dt
3. Gauss’ Law for Electric Field
E da0
=
dV
S V
10-9
0
8.854 ×10-12 farads/meter
36
1 8
c= 3 10 meters/second (Speed of electromagnetic waves in
0 0
free space)
4. Gauss’ Law for Magnetic Field
S
H da0
= 0
In free space:
B = 0 H
magnetic magnetic
flux field
density intensity
(Teslas) (amperes/meter)
5. Conservation of Charge
Take Ampère’s Law with displacement current and let contour C 0
d
lim
C 0
H ds = 0 =
J da +
dt
E da 0
C S S
V
dV
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 2 of 7
d
S
J da +
dt
V
dV = 0
Total current Total charge
leaving volume inside volume
through surface
6. Lorentz Force Law
f = q E + v × 0 H
II. Electric Field from Point Charge
E da = E 4
r =q 2
0 0 r
S
q
Er =
0r
4
2
q2
T sin θ= fc =
4
0r
2
T cos θ= Mg
q2 r
tan θ= =
40
r Mg
2
2l
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 3 of 7
1
r 3Mg 2
2
q= 0
l
III. Faraday Cage
d d dq
J da = i = -
dt
dV = -
dt
-q =
dt
S
idt = q
IV. Edgerton’s Boomer
1. Magnetic Field, Current, and Inductance
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 4 of 7
N i
H
Cb
ds H 2 a = N
1 1 1
i H1 1 1
2a
N21 a2 0 N2 a 0
N1 a2 0 H1 =
2 a
i1 1
2
i1
N a 0
2
L= 1
i1 2
1
=
LC
1 1
L ip C vp ip vp C
2 2
2 2 L
C = 25 f, v p 4 k V, N1 = 50, a 7 c m
L1 0.1 mH
ip 2000 A, 20 x 103 / s f = 3k Hz
2
5
Hp 2.3 x 10 A / m Bp = 0 Hp 0.3 Teslas 3000 Gauss
2. Electrical Breakdown in Single Turn Coil with Small Gap
Bp
E0
0 Inside Metal Coil
E
E0 Small Gap
d
E
C
ds E (B
dt
0 p
R 2)
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 5 of 7
Bp Bm cos t
2
B R
E0 m sin t
Take: Bm 0.3 Tesla, 20, 000 radians/second, R 0.07 m, 0.01 mm
B R2 0.3(20, 000)(0.07)2
Em m 9 106 Volts/meter
105
6
Breakdown strength of air 3 10 Volts/meter.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
3. Force on Metal Disk
d dBp
E = B da a
2 2
ds 2 aE
a Bm sin t
Ca
dt Sa dt
a dBp a
J = E = Bm sin t
2 dt 2
F = J x 0 H , f=
F dV =
J x H dV
0
V V
Force per unit total force
volume
K JHr Hr J
F J 0 H J
i 0 Hr ir 0 JHr iz 0 J2 iz
2
a 2
Fz 0 J2
0 Bm sin t
2
2 2 4
a 2 2
fz Fz a2 0 Bm sin2 t
4
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 6 of 7
alu minum 3.7 107 Siemens/meter, a=0.07 m, =2 mm, 20, 000
radians/second, Bm 0.3 Tesla, M=0.08 kg
0
a B sin
2
2 2
fz m
t
4
2
10
7
2 10 3 3.7 10 .07 20, 000 0.3
7 2
sin
2
t
4.7 106 sin2 t
Mg (0.08)9.8 0.8 Newtons
fmax 4.7 10
6
6
5.9 10
Mg 0.8
Neglecting losses:
1 2 1 2
CV Mv (t 0) Mgh
2 2
C
v(t 0 ) V
M
C 25f , M .08 kg, Vp 4000 volts
v(t 0 ) 70.7 meters/second (Initial velocity)
2
v
h t 0 255 meters (Maximum height)
2g
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013 Electromagnetics and Applications Lecture 1
Prof. Markus Zahn Page 7 of 7