experiment 6: wireless energy transfer; displacement experiment 6: wireless energy transfer...

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  • 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

    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

    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 = 1.2×10

    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

    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 dV V ∫∫∫ (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 +Q Q

    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 ⋅ ds∫ = µ0 Ienc

    Ienc =

     J ⋅ n̂ da

    S ∫∫

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

    18

  • 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

    19

  • 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

    20

  • 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”

    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 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

    23

  • 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

    S ∫∫ (Faraday's Law)

    ! B ⋅d!s

    C #∫ = µ0

    ! J ⋅ n̂dA

    S ∫∫ + µ0ε0

    d dt

    ! E ⋅ n̂dA

    S ∫∫ (Maxwell - Ampere's Law)

    24

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