Study Materials for MIT Course [8.02T] - Electricity and Magnetism [FANTASTIC MTLS]
Experiment 6: Wireless Energy Transfer; Displacement...
Transcript of Experiment 6: Wireless Energy Transfer; Displacement...
Experiment 6: Wireless Energy Transfer;
Displacement Current
1
W13D2
Announcements
PS 11 due Week 14 Friday at 9 pm in boxes outside 26-152 Week 13 Prepset due online Friday 8:30 am Sunday Tutoring 1-5 pm in 26-152
2
Outline
Transformers Experiment 6: Wireless Energy Transfer Displacement Current
3
Transformer
4
ε p = N p
dΦdt
; ε s = Ns
dΦdt
Ns > Np: step-up transformer Ns < Np: step-down transformer
Equal flux through each turn:
ε s
ε p
=Ns
N p
Transformer: Current
5
Pp = ε p I p; Ps = ε s Is
Ns > Np: step-down current Ns < Np: step-up current
Ideal transformer has no power loss
Pp = Ps ⇒ε p I p = ε s Is
Is
I p
=ε p
ε s
=N p
Ns
Demonstrations:
26-152 One Turn Secondary: Nail H10
26-152 Many Turn Secondary: Jacob’s
Ladder H11
6
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 10&show=0
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 11&show=0
CQ: Residential Transformer
7
1. House=Left, Line=Right 2. Line=Left, House=Right 3. I don’t know
If the transformer in the can looks like the picture, how is it connected?
Marconi Coil H12
8
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 12&show=0
Transmission of Electric Power
9
Power loss can be greatly reduced if transmitted at high voltage
Example: Transmission lines
10
An average of 120 kW of electric power is sent from a power plant. The transmission lines have a total resistance of . Calculate the power loss if the power is sent at (a) 240 V, and (b) 24,000 V.
(a) I = P
V= 1.2×105 W
2.4×102 V= 500A
PL = I 2R = (5.0A)2(0.40Ω) = 10W
(b)
83% loss!!
0.0083% loss
PL = I 2R = (500A)2(0.40Ω) = 100kW
I = P
V= 1.2×105 W
2.4×104 V= 5.0A
0.40Ω
Voltage Drop Across Transmission Line
11
The voltage drop across the transmission line is Therefore the power loss across the resistor is The voltage drop across the transmission line is not 24,000 V.
( )( )5 A 0.4 2 VV IR= = Ω =
P = I 2R = V 2
R= (2.0 V)2
0.4Ω= 10 W
Experiment 6: Wireless Energy Transfer
12
James Clerk Maxwell
13
14
Maxwell’s Equations: Is There Something Missing?
!E ⋅ n̂dA
S"∫∫ = 1
ε0
ρ dVV∫∫∫ (Gauss's Law)
!B ⋅ n̂dA
S"∫∫ = 0 (Magnetic Gauss's Law)
!E ⋅d !s
C#∫ = − d
dt!B ⋅ n̂dA
S∫∫ (Faraday's Law)
!B ⋅d !s
C#∫ = µ0
!J ⋅ n̂dA
S∫∫ (Ampere's Law quasi- static)
14
Maxwell’s Equations One Last Modification: Displacement Current
15
1. Find an expression for the electric field between the plates in terms of the charge Q. Neglect edge effects.
2. Calculate the flux of the electric field between the plates:
3. Calculate the product of the with the time derivative of the flux of the electric field between the plates
Group Problem: Charging a Capacitor
R+
S
C
I+QQ
16
!E ⋅ n̂da
S∫∫
ε0
dΦE
dt
ε0
Parallel Plate Capacitor
Gauss’s Law: Flux of electric field: Time changing electric field:
ε0
dΦE
dt= dQ
dt= I
17
E = Q / Aε0
ΦE = EA = Q
ε0
Ampere’s Law: Capacitor
Consider a charging capacitor: Use Ampere’s Law to calculate the magnetic field just above the top plate
What’s Going On?
B ⋅ ds∫ = µ0 Ienc
Ienc =
J ⋅ n̂ da
S∫∫
1) Surface S1: Ienc= I 2) Surface S2: Ienc = 0
18
Displacement Current
dQdt
= ε0
dΦE
dt≡ Idis
We don’t have current between the capacitor plates but we do have a changing E field. Can we “make” a current out of that?
E = Q
ε0 A⇒ Q = ε0 EA = ε0ΦE
This is called the displacement current. It is not a flow of charge but proportional to changing electric flux
19
Displacement Current:
Idis = ε0
ddt
!E ⋅ n̂da
S∫∫ = ε0
dΦE
dt
If surface S2 encloses all of the electric flux due to the charged plate then Idis = I
20
Maxwell-Ampere’s Law
!B ⋅d!s
C"∫ = µ0
!J ⋅ n̂da
S∫∫ + µ0ε0
ddt
!E ⋅ n̂da
S∫∫
= µ0 (Ienc + Idis )
Ienc =
J ⋅ n̂da
S∫∫
Idis = ε0
ddt
E ⋅ n̂da
S∫∫
“flow of electric charge”
“changing electric flux”
21
CQ: Capacitor 1
The figures above shows a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the charge on the upper plate Q is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time.
22
E+Q
Q
view from above
EB
Group Problem: Capacitor
I I
d
r.P R
k̂r̂
ˆ
unit vectors at point P+Q(t) Q(t)
Q(t) = It
A circular capacitor of spacing d and radius R is in a circuit carrying a steady current I shown at time t. At time t = 0, the plates are uncharged. a) Using Gauss’s Law, find a vector expression for the electric
field E(t) at P vs. time t (mag. & dir.) b) Find a vector expression for the magnetic field B(t) at P
23
Maxwell’s Equations
!E ⋅ n̂dA
S"∫∫ = 1
ε0
ρ dVV∫∫∫ (Gauss's Law)
!B ⋅ n̂dA
S"∫∫ = 0 (Magnetic Gauss's Law)
!E ⋅d !s
C#∫ = − d
dt!B ⋅ n̂dA
S∫∫ (Faraday's Law)
!B ⋅d!s
C#∫ = µ0
!J ⋅ n̂dA
S∫∫ + µ0ε0
ddt
!E ⋅ n̂dA
S∫∫ (Maxwell - Ampere's Law)
24