Electrodynamics Dr. Ray Kwok sjsu
Electrodynamics
Dr. Ray Kwok
sjsu
Electrodynamics – Dr. Ray Kwok
Static Fields (stationary charges, steady current)
=×∇
=⋅∇
=×∇
ρ=⋅∇
f
f
JH
0B
0E
D
rr
r
r
r
µ=⋅
=⋅
=⋅
ε=⋅
∫∫∫
∫
to
o
t
IdB
0adB
0dE
QadE
lrr
rr
lrr
rr
)MH(HHB
PEEED
oro
ororrrrr
rrrrr
+µ=µµ=µ=
+ε=εε=ε=
Gauss’s Law
Conservative
No magnetic charge
Ampere’s Law
f
r
fbft
b
o
t
P
E
ρ≤ερ
=ρ+ρ=ρ
ρ−=⋅∇
ερ
=⋅∇
r
rlinear, homogeneous, isotropic
Electrodynamics – Dr. Ray Kwok
Orsted’s Discovery
toIdB µ=⋅∫ lrr
V
S
R
compass
I
André-Marie Ampère(1775–1836) France
SI unit for current
Hans Christian Ørsted(1777–1851) Danmark
cgs unit for B
4/21/1820
Danmark
Ampere’s Law, 9/18/1820,
after he learned about Orsted’s
discovery on 9/11/1820 !!
Electrodynamics – Dr. Ray Kwok
Faraday’s Experiment
Michael Faraday(1791–1867) England
SI unit for capacitance
Joseph Henry(1797–1878) USA
SI unit for inductance
With stationary magnet,
no current induced (1831)
Electrodynamics – Dr. Ray Kwok
Magnetic induction
Electrodynamics – Dr. Ray Kwok
Faraday’s Law
∫
∫⋅=Φ
Φ−=⋅≡
adB
dt
ddEVemf
rr
lrr
R
uB
R
VI
VuBdt
dxB
dt
d
xBdxB
emf
emf
x
0
l
ll
ll
==
−===Φ
==Φ ∫
magnetic flux
oppose the “change” of
direction given by Lenz’s Law
u
B
IR
x
l
Electrodynamics – Dr. Ray Kwok
u
B
IR
x
Heinrich Friedrich Emil Lenz
(1804-1865) Italy
Lenz’s Law (1834)“Back emf” to oppose the “change” of magnetic flux
Electrodynamics – Dr. Ray Kwok
Example – moving magnet
Electrodynamics – Dr. Ray Kwok
Example - jumping ring
V
S
R
Al ring
I
FB B
I
FB
Electrodynamics – Dr. Ray Kwok
e.g. magnet in a copper tube
N
S
Bind
Electrodynamics – Dr. Ray Kwok
Eddy current - disc brake
BLIFBrrr
×=
Lenz’s Law
metal plate
Electrodynamics – Dr. Ray Kwok
To reduce eddy current
transformer
Electrodynamics – Dr. Ray Kwok
Example – metal detector
Electrodynamics – Dr. Ray Kwok
Example – guitar pickup
Electrodynamics – Dr. Ray Kwok
Question
Right, 0, Left, Left, Left
V
S r
R
Find the direction of the current in the resistor R shown in Figure at
each the following steps: (a) at the instant the switch is closed, (b)
after the switch has been closed for several minutes, (c) when the
variable resistance r increases, (d) when the circuit containing R
moving to the right, away from the other circuit, and (e) at the instant
the switch is opened.
Electrodynamics – Dr. Ray Kwok
Moving Conductor
u
B
+FB
−−−−FB
l
l
rrr
rrr
rr
uBV
uBV
BuE
0BuqEq
0FF BE
−=
−=
×−=
=×+
=+
saturate when
emf
u
B+
-
IR
induced E
similar to Hall Effect
induced
separate the charges
direction agrees w/ Lenz’s Law
emf is “+” u x B
induced current
I = |V/R| = B u/Rl
Electrodynamics – Dr. Ray Kwok
Generalized Faraday’s Law
∫∫
∫∫∫∫
∫∫
∫∫
⋅∂∂
−⋅×=Φ
−=
⋅×−⋅∂∂
=⋅
××∇−
∂∂
=⋅
⋅×=⋅≡
⋅∂∂
−=Φ
−=⋅≡
adt
BdBu
dt
dV
dBuadt
Bad)Bu(
t
BadB
dt
d
dBudq
FV
adt
B
dt
ddEV
emf
Bemf
emf
rr
lrrr
lrrrr
rrrr
rrr
lrrr
lr
r
rr
lrr
changing B(t), stationary loop
fixed B, moving loop
induced emf in a moving loop w.r.t. “stationary” B(t)
u
loop
magnet
Einstein’s Relativity:
move the loop, Lorentz force (magnetic) [motional emf]
move the magnet, induced emf – electric [transformer emf]
hw
Electrodynamics – Dr. Ray Kwok
Example – rotating loop
tsinBAtsinBh2
w2
h)tsinuB(2dBuVemf
ωω=ω
ω=
ω=⋅×= ∫ lrrr
tsinBA)tcosBA(dt
d
adBdt
dVemf
ωω=ω−=
⋅−= ∫rr
θ = ωt uniform B
OR
Direction of current (at this instance)??
Electrodynamics – Dr. Ray Kwok
Group Exercise
Find the induced emf in a
rectangular loop rotating at
an angular velocity ω in a
magnetic field Bosinωt.
& the direction of induced
current?
(Can you do this problem in
2 ways?)
−ωBoAcos2ωt
Electrodynamics – Dr. Ray Kwok
tsinBAN
dt
dNVemf
ωω=
Φ−=
AC generator
Electrodynamics – Dr. Ray Kwok
DC generator
Electrodynamics – Dr. Ray Kwok
Electromotor
To turn faster, should we
1. use thicker wire?
2. use more turns?
3. make bigger loop?
4. use stronger magnet
Electrodynamics – Dr. Ray Kwok
Answer… (not counting friction)
e.g. flash lights
l
l
rr
r
rrrrr
rlrr
'AVm
'AR
I
nNIAm
BmFr
BIF
R
BAN|I|
tsinBANdt
dNV
B
B
emf
ρ=ρ=
ρ=
α=τ
=
×=×=τ
×=
ω=
ωω=Φ
−= 1. use thicker wire?
2. use more turns?
same.
e.g. half R, double mass, same α
same.
double N, double R, same I,
double mm, but double inertia, same α.
3. use bigger loop? (same N)
same.
bigger A, more Φsomewhat higher R, but still more I ~A/R~ r
mm~A2/R~r3, inertia~r3, same α
torque
magnetic moment
I = moment of inertia
4. use stronger magnet?
YES
more I, m, τ, α
resistivity
mass density
Electrodynamics – Dr. Ray Kwok
Another important application – telephony / loudspeaker
(?? before him)
Alexander Graham Bell(1847 –1922) UK
(practical improvement)
Thomas Alva Edison
(1847 –1931) USA
Electrodynamics – Dr. Ray Kwok
Ideal Transformer (AC)
Φ is confined in the core (µ = ∞)
L
2
2
1
2
2
1
2
2
1
1in
2
1
1
2
2
1
2211
21
2
1
2
1
22
11
RN
N
N
N
I
V
I
VR
N
N
I
I
V
V
IVIV
PP
N
N
V
V
dt
dNV
dt
dNV
=
==
==
=
=
=
Φ−=
Φ−=
RL
Electrodynamics – Dr. Ray Kwok
Self-Inductance
dt
dIL
dt
dNV
I
NL
−=Φ
−=
Φ≡
back emf causes I lags V
V = IZ = I(R + jωL) = I |Z|ejθ
RL
V
Electrodynamics – Dr. Ray Kwok
Ideal Solenoid
ideal:
• large N tightly wound
• no end effect
• uniform internal B
• zero external B in the vicinity
AnI
nIAN
I
NL
nIAadB
nINI
B
2
oo
o
oo
l
rrl
µ=µ
=Φ
=
µ=⋅=Φ
µ=µ
=
∫
Ampere’s Law:
Electrodynamics – Dr. Ray Kwok
π
µ=
Φ=
π
µ=φ⋅φ
πµ
=⋅=Φ
φπ
µ=
µ=⋅
∫∫
∫
a
bln
2
h
IL
a
bln
2
Ihhdrˆˆ
r2
IadB
ˆr2
IB
IdB
o
o
b
a
o
o
o
rr
r
lrr
Coaxial Cable
Electrodynamics – Dr. Ray Kwok
LC Resonator (Lenz’s Law, later resonator…)
Electrodynamics – Dr. Ray Kwok
Mutual-Inductance
1
1212
IM
Φ≡
Φ12 is the flux through loop 2 due to the
B generated by loop 1
21
12
21o
1
1212
21
12
1o12
oo
21212112
Mr
dd
4IM
ddr
I
4
'd|'rr|
)'r(I
4'dV
|'rr|
)'r(J
4)r(A
dAadAadB
∫∫
∫∫
∫∫
∫∫∫
=⋅
πµ
=Φ
=
⋅π
µ=Φ
−πµ
=−π
µ=
⋅=⋅×∇=⋅=Φ
lr
lr
lr
lr
lr
rrrr
rrrr
lrrrrrr
more than 1 turn?
1
12212
I
NM
Φ=
Electrodynamics – Dr. Ray Kwok
Transformer - Primary / Secondary coils
same cross-section?
12
12
2
1
1
2
2
2
1
1
M
I
N
I
N
tN
V
N
V
Φ==
∂Φ∂
−==
same equations different cross-section? h.w.
Electrodynamics – Dr. Ray Kwok
Magnetic Energy
21
2
22
2
11m
222121
111212
2222
1111
122222211111m
12222
21111
2211m
IMIIL2
1IL
2
1W
MIIM
MIIM
IL
IL
I2
1I
2
1I
2
1I
2
1W
I2
1I
2
1W
++=
==Φ
==Φ
=Φ
=Φ
Φ+Φ+Φ+Φ=
Φ+Φ=Φ
Φ+Φ=Φ
Φ+Φ=
For coupling circuits 1 & 2…For an inductor …
Φ=
=
µ= ∫
I2
1W
LI2
1W
dVB
2
1W
m
2
m
o
2
m
Electrodynamics – Dr. Ray Kwok
Faraday’s Law - differential
Stokes’ theorem
E is induced in moving medium
B is measured in stationary frame
curl operates in moving frame
t
)r(B)r(E
))r(Bu('t
)r(B)'r(E
)Bu(t
BE
ad)Bu(t
BadE
adBdt
ddEVemf
∂∂
−=×∇
××∇+∂
∂−=×∇
××∇+∂∂
−=×∇
⋅
××∇−
∂∂
−=⋅×∇
⋅−=⋅≡
∫∫
∫∫
rrrr
rrrrr
rr
rrr
r
rrrr
rr
rrlrr
for stationary loop (rest frame)
or localized, RF EM wave (microscopic)
relativity, low f
Electrodynamics – Dr. Ray Kwok
Electrodynamics
=×∇
=⋅∇∂∂
−=×∇
ρ=⋅∇
f
f
JH
0Bt
BE
D
rr
r
rr
r
µ=⋅
=⋅
Φ−=⋅
ε=⋅
∫∫
∫
∫
to
o
t
IdB
0adB
dt
ddE
QadE
lrr
rr
lrr
rrGauss’s Law
Faraday’s Law
No magnetic charge
Ampere’s Law
Electrodynamics – Dr. Ray Kwok
Electric Potential
0V
VE
0E
=∇×∇
−∇≡
=×∇r
rstatic
define
such that
t
AVE
Vt
AE
0t
AE
Att
BE
∂∂
−−∇=
−∇≡∂∂
+
=
∂∂
+×∇
×∇∂∂
−=∂∂
−=×∇
rr
rr
rr
rr
rdynamic
Need both V and A to find E !!!!!
0A
AB
0B
=×∇⋅∇
×∇≡
=⋅∇
r
rr
r
define
such that
Electrodynamics – Dr. Ray Kwok
Group Exercise
I1(t)
I2
1 m
3 m
5 m
N = 20
Rectangular loop of 20 turns is placed at
1 m away from a long current-carrying
wire as shown.
I1(t) = 2 cos(60t) (A)
Resistivity of the wire in the loop is 4 Ω/m.
Find I2.
+2.6 sin(60t) µA
independent of N