induction by 14bz5m6

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									Ch – 33 Electromagnetic Induction
            Reading Quiz – Ch. 33
1. Currents circulate in a piece of metal that is
   pulled through a magnetic field. What is the
   correct name for these currents?
a. Eddy currents        c.Flux currents.
b. Induction currents d.Faraday currents.

2.    The magnetic flux is a measure of the
     magnetic field:
a.   while electromagnetic induction occurs
b.   parallel to a closed loop
c.   passing through a closed loop
d.   when it is changing
         Learning Objectives – Ch 33
• Understand how a changing magnetic field
  will induce a potential difference – this is the
  phenomenon of electromagnetic induction.
• To observe the experimental evidence for
  electromagnetic induction.
• To understand and use Lenz’s law for
  induced currents.
• To learn of Faraday’s law as a new law of
  nature.
• To understand basic applications of
  electromagnetic induction to technology.
• To gain a qualitative understanding of
  electromagnetic waves.
• To analyze circuits with inductors.
            Induced Current
• A current creates a
  magnetic field
• Is the opposite true?
• Can a magnetic field
  create, or induce a
  current?
         Faraday’s Discovery
• Faraday was hoping
  that the magnetic field
  generated by the
  current on the left
  circuit would induce a
  current in the wire on
  the right.
• But no luck, as the
  meter shows.
           Faraday’s Discovery
• So he shut down for
  lunch.
• But when he opened the
  switch to cut off the
  current in the left circuit,
  the meter suddenly
  moved, showing a
  momentary current in the
  wire on the right.
• It quickly went back to
  zero.
          Faraday’s Discovery
• Baffled, he closed
  the switch again.
• And noticed that the
  meter jumped again
  momentarily, and this
  time the meter needle
  went the other way.
• I bet he actually did it
  a bunch of times
  before he noticed that
  it went the other way..
        Faraday’s Discovery
• Faraday found that he could induce a
  current in a closed wire, but only if the
  magnetic field through the coil is changing.
• This is an informal statement of Faraday’s
  Law.
               Motional emf
An induced current in a
   circuit can be          Changing            Stationary circuit
   created 2 ways:         magnetic field
                           (since current is
                                               with meter
                           changing)
1. Change the strength
   of the magnetic field
   through a stationary
   circuit. That’s what
   Faraday did by
   opening and closing
   the switch.
                Motional emf
                                          Circuit
                                          changes
                             Constant B
                                          size due to
The other way to induce      field into
                             the page
   a current:
2. Change the size or
   orientation of the
   circuit in a stationary
   magnetic field.
We shall look at this
   second method first.
   This is motional emf.
              Motional emf
• The external B field
  causes a magnetic
  force on positive and
  negative charges
  moving to the right.
• The electron holes
  move up and the
  electrons move down.
                Motional emf
• Since the rest of the
  loop isn’t moving,
  there is no magnetic
  force on it, so the
  electron holes will
  flow along the wire to
  get back to the more
  negative side.
• Wow, electric current
  with no power source
  except a moving wire!
  Why aren’t we all fat
  and happy?
               Motional emf
• We’ll find out soon.
• Another result of the
  charge separation is
  an E field.
• This E field also
  causes a force, which
  is in the opposite
  direction as the force
  due to the B field.
                 Motional emf
• Electron holes continue to
  move up, but only until
  FE down equals FB up, at
  which charge separation
  ceases.
• As long as the wire keeps
  moving, there will be a
  charge separation.
• The magnetic force is
  doing work to maintain
  that charge separation.
        Workbook exercises
• 1 b,d,e
 Workbook exercises - answers
• b: ccw
• d: ccw
• e: 0
             Motional emf
Recall emf (motional or otherwise) is the
 work done per unit charge, i.e. a potential
 difference.
      
∆V = - Eds

When FE equals FB (assume α = 900):
          qE = qvB or E = vB
               Motional emf
For a wire of length L,
  moving in a direction
  perpendicular to an
  external B field:
         l
∆V = -
           vBds
         0
emf = v LB
                 Motional emf
• The emf due to the
  charge separation exists
  whether or not the loop is
  closed (battery analogy).
• If there is a closed loop:
          I = vLB/R
• The induced current is
  due to magnetic forces on
  moving charges.
  Why aren’t we fat and happy?
• Once there is a
  current flowing
  through the wire,
  charges move in 2
  directions.
  – They all move right
    with the moving wire.
  – Current moves up.
• Yet another magnetic
  force generated, this
  time to the left.
  Why aren’t we fat and happy?
• This magnetic force
  opposes the original
  velocity of the moving
  wire.
• The moving wire will slow
  down and stop.
• Need a constant Fpull to
  the right to make the
  contraption work.
• That’s why we’re not fat
  and happy.
 Why aren’t we fat and happy?
• To keep the wire at
  constant speed, and
  continue the emf and
  current Fpull = Fmag
    F = ILB (Ch. 32)
    Fpull = Fmag = ILB
         I = vLB/R
      F = vL2B2/R
 Why aren’t we fat and happy?
      Fpull = Fmag
      Wpull = Wmag
The rate at which work
 is done on the circuit
 equals the power
 dissipated by the
 circuit:
P = I2R = v2L2B2/R
        Numerical Example
• #3 = end of chapter
            Magnetic Flux

     m  A  B


Φm = AB cos θ

Units: 1 weber = 1Wb = 1Tm2
   Magnetic Flux in a non-uniform
               field
• Divide the loop into
  many small pieces

  d m  B  dA
• The flux is the sum of
  all these:

    m   B  dA
                 Lenz’s Law
• There is an induced
  current in a closed
  conducting loop only
  if the magnetic flux is
  changing (either B, A
  or θ). The direction of
  the induced current is
  such that the induced
  magnetic field
  opposes the change.
                   Using Lenz Law
1. Determine the direction of the external magnetic
   field.
2. Determine how the flux is changing. Is it increasing,
   decreasing, or staying the same?
3. Determine the direction of an induced magnetic field
   that will oppose the change in the flux.
   – Increasing: induced magnetic field points opposite the external
     magnetic field.
   – Decreasing: induced magnetic field points in the same direction
     as the external magnetic field.
   – Constant: no induced magnetic field.
4. Determine the direction of the induced current. Use
   the right-hand rule.
                    Faraday’s Law
Recall that a current in
  a circuit can be
  created 2 ways:
   – Change the size or
     orientation of the
     circuit in a stationary
     magnetic field.
   – Change the strength
     of the magnetic field
     through a stationary
     circuit.
Both of these create a
   changing magnetic
   flux.
                Faraday’s Law
The current exists
   because the
   changing magnetic
   flux has induced an
   emf. In a closed
   circuit with a
   resistance, R:
         I = ε/R
The current is a
   consequence of the
   emf.
The emf is a
   consequence of Φm
              Faraday’s Law
An emf is induced in a conducting coil of N turns if
  the magnetic flux through the coil changes. The
  magnitude of the emf is equal to the rate of
  change of the magnetic flux:

                 d m
             N
                  dt

The direction is given by Lenz Law.
              Faraday’s Law
Recall the expression
 for emf of a wire of
 length L, moving in an
 external B field:
         l
∆V = -
           vBds
         0
emf = v LB
              Faraday’s Law
Does Faraday’s Law
  give us the same?
ε = dΦ/dt = d(xLB)/dt
L, B are constant so
      ε = (LB)dx/dt
         ε = v LB
           Faraday’s Law
For the case shown,
 the induced B field
 will be in which
 direction?
Workbook exercise #14 (p.33-8)
Draw a graph of the current, given the graph
 of the magnetic field
                 Answer
I0-1s: negative constant
I1-2s: no current
I2-3s: positive constant, smaller magnitude
   than I0-1s
          Numerical Problem
The resistance of the
   loop is 0.10 Ω.
A. Is the magnetic field
   strength increasing
   or decreasing?
B. What is the rate of
   change of the the
   magnetic field?
              Numerical Problem
A. Induced current is
   shown moving ccw.
   RH rule indicates a
   magnetic field out of
   the page, opposing
   external field.
   Therefore, external
   magnetic field must
   have been
   increasing.
B. Rate of change is
   2.34 T/s
                 Inductors
• An inductor is a device that produces a uniform
  magnetic field when a current passes through it.
  A solenoid is an inductor.
• The magnetic flux of an inductor is proportional
  to the current.
• For each coil (turn) of the solenoid:
Φper coil = A•B
Φsol = N(A•B) = NAB = NA(u0NI/ℓ) = (Au0N2/ℓ)Isol
• This is actually a self-inductance
                  Inductors
• The proportionality constant is defined as L, the
  inductance:
                Lsol = Φsol /I = Au0N2/ℓ
• Note that the inductance, L depends only on
  the geometry of the inductor, not on the current.
• The unit of inductance is the henry
1 H = 1 Wb/Ampere
The circuit symbol for an inductor:
    Potential difference across an
               inductor
• For the ideal inductor,
  R = 0, therefore
  potential difference
  across the inductor
  also equals zero, as
  long as the current is
  constant.
• What happens if we
  increase the current?
    Potential difference across an inductor
• Increasing the current
  increases the flux.
• An induced magnetic field
  will oppose the increase
  by pointing to the right.        Induced          Induced
                                   current
• The induced current is                            field


  opposite the solenoid
  current.
• The induced current
  carries positive charge to
  the left and establishes a
  potential difference
  across the inductor.                 Potential
                                       difference
   Potential difference across an inductor
The potential difference
  across the inductor
  can be found using
  Faraday’s Law:
                                  Induced
            d m
                                                   Induced
                                  current

     N
                                                   field


             dt
Where Φm = Φper coil
Φsol = N Φper coil
We defined Φ = LI
dΦsol/dt = L |dI/dt|
                                      Potential
                                      difference
 Potential difference across an inductor
• If the inductor current is
  decreased, the induced
  magnetic field, the
  induced current and the
  potential difference all
  change direction.
• Note that whether you
  increase or decrease the
  current, the inductor
  always “resists” the
  change with an induced
  current.
  The sign of potential difference across an
                  inductor
        ∆VL = -L dI/dt
• ∆VL decreases in the
  direction of current flow if
  current is increasing.
• ∆VL increases in the
  direction of current flow if
  current is decreasing.
• ∆VL is measured in the
  direction of current in the
  circuit
      Conceptual Question - Inductors
Which of the following statements could be true?
a. I is from a to b and constant.
b. I is from a to b and increasing.
c. I is from a to b and decreasing.
d. I is from b to a and steady.
e. I is from b to a and increasing.
f. I is from b to a and decreasing.
   Faraday’s Law Problem (#34, p.1078)

The top figure shows a 5-
  turn 1.0-cm diameter coil
  with R=0.10 Ω inside a                    COIL


  2.0-cm solenoid. The
  solenoid is 8.0 cm long,
                              Solenoid cross section at t = 0.02s
  has 120 turns and carries
  the current shown in the
  graph. A positive current
  is cw when seen from the
  left. Determine the
  direction and magnitude
  of current in the coil.
               Setting up the problem
The mathematical
  representation is Faraday’s
  Law:                                              COIL




       d m
   N                                Solenoid cross section at t = 0.02s

        dt

I = ε/R
Icoil = 1/R (N dΦ/dt)
For each of the following
    variables, decide whether the
    quantity refers to the solenoid
    or the coil: R, N, Φ
              Setting up the problem
Icoil = 1/R (N dΦ/dt)
For each of the following
                                                  COIL
   variables, decide whether
   the quantity refers to the
   solenoid or the coil:            Solenoid cross section at t = 0.02s



R, N, Φ
Answers: R – coil, N – coil
Φ – coil
Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
               Setting up the problem

Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
For the above expression:
                                                  COIL

Write an expression for
   Φcoil in terms of the
   magnetic field and the           Solenoid cross section at t = 0.02s



   area. Once again, the
   pertinent question is
   solenoid or coil:
Magnetic field (B)
Area (A)
And what about cos θ?
             Setting up the problem


Area (A) – the current is
  induced in the coil, due to                 COIL


  the flux through the coil;
  therefore it’s the area of
                                Solenoid cross section at t = 0.02s
  the coil.
Magnetic field (B) – It’s the
  external magnetic field of
  the solenoid that induces
  a current in the coil
Φcoil = Acoil Bsolenoid
And what is the magnetic
  field of a solenoid?
               Setting up the problem

Φcoil = Acoil Bsolenoid

                                              COIL

Write an expression for the
 magnetic field of a
 solenoid. For any              Solenoid cross section at t = 0.02s



 variable that has a value
 for both coil and solenoid
 (e.g I,R,N,r) specify to
 which you are referring.
               Setting up the problem

Φcoil = Acoil Bsolenoid

                                              COIL

Bsol = (u0NsolIsol)/ℓ
(we weren’t given a length
                                Solenoid cross section at t = 0.02s
  for the coil).
Now we need an
  expression for dΦcoil /dt,
  so write one.
               Setting up the problem

Φcoil = Acoil Bsolenoid

                                                  COIL

Bsol = (u0NsolIsol)/ℓ

                                    Solenoid cross section at t = 0.02s

dΦcoil = (Acoil u0Nsol/ℓ)dI/dt
dt
We may actually be ready
   to solve this puppy.
Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
                           Answer


                                                  COIL




Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
                                    Solenoid cross section at t = 0.02s

Icoil = 3.7 x 10-4A
or .37 mA

Direction: clockwise when
  seen from the left
  (induces a field to the
  right).

								
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