# experiment 6: wireless energy transfer; displacement experiment 6: wireless energy transfer...     Post on 05-Jul-2020

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• Experiment 6: Wireless Energy Transfer;

Displacement Current

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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

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• Outline

Transformers Experiment 6: Wireless Energy Transfer Displacement Current

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• Transformer

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ε p = N p

dΦ dt

; ε s = Ns dΦ dt

Ns > Np: step-up transformer Ns < Np: step-down transformer

Equal flux through each turn:

ε s ε p

= Ns N p

• Transformer: Current

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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

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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

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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

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http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 12&show=0

• Transmission of Electric Power

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Power loss can be greatly reduced if transmitted at high voltage

• Example: Transmission lines

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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 = 1.2×10

5 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 = 1.2×10

5 W 2.4×104 V

= 5.0A

0.40Ω

• Voltage Drop Across Transmission Line

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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

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R = (2.0 V)

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0.4Ω = 10 W

• Experiment 6: Wireless Energy Transfer

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• James Clerk Maxwell

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• 14

Maxwell’s Equations: Is There Something Missing?

! E ⋅ n̂dA

S "∫∫ =

1 ε0

ρ dV V ∫∫∫ (Gauss's Law)

! B ⋅ n̂dA

S "∫∫ = 0 (Magnetic Gauss's Law) ! E ⋅d !s

C #∫ = −

d dt

! B ⋅ n̂dA

! B ⋅d !s

C #∫ = µ0

! J ⋅ n̂dA

S ∫∫ (Ampere's Law quasi- static)

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• Maxwell’s Equations One Last Modification: Displacement Current

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• 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 +Q Q

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! 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

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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 ⋅ ds∫ = µ0 Ienc

Ienc =

 J ⋅ n̂ da

S ∫∫

1) Surface S1: Ienc= I 2) Surface S2: Ienc = 0

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• Displacement Current

dQ dt

= 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

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• Displacement Current:

Idis = ε0

d dt

! E ⋅ n̂da

S ∫∫ = ε0

dΦE dt

If surface S2 encloses all of the electric flux due to the charged plate then Idis = I

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• Maxwell-Ampere’s Law

! B ⋅d!s

C "∫ = µ0

! J ⋅ n̂da

S ∫∫ + µ0ε0

d dt

! E ⋅ n̂da

S ∫∫

= µ0 (Ienc + Idis )

Ienc =

 J ⋅ n̂da

S ∫∫

Idis = ε0

d dt

 E ⋅ n̂da

S ∫∫

“flow of electric charge”

“changing electric flux”

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• 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.

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E +Q

Q

view from above

EB

• Group Problem: Capacitor

I I

d

r .P R k̂

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

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• Maxwell’s Equations

! E ⋅ n̂dA

S "∫∫ =

1 ε0

ρ dV V ∫∫∫ (Gauss's Law)

! B ⋅ n̂dA

S "∫∫ = 0 (Magnetic Gauss's Law) ! E ⋅d !s

C #∫ = −

d dt

! B ⋅ n̂dA

! B ⋅d!s

C #∫ = µ0

! J ⋅ n̂dA

S ∫∫ + µ0ε0

d dt

! E ⋅ n̂dA

S ∫∫ (Maxwell - Ampere's Law)

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