INDUCTION

							INDUCTION
   Spring ‘09
DIS HERE WEEK
 We begin the study of magnetic induction
 There will be a quiz on Friday
 There is a new WebAssign.
 Exam #3 is Wednesday4/8
     It will include the material covered through
      April 6th.
 The   end is in sight …
     Check the website for the Final Exam
      Schedule
AMPERE’S LAW


                        B  ds  0ienclosed

                        B  ds  0 (i1  i2 )

 USE THE RIGHT HAND RULE IN THESE CALCULATIONS
LAST TIME: A SIMPLE EXAMPLE
FIELD AROUND A LONG STRAIGHT WIRE

                  B  ds   i
                             0 enclosed


                 B  2r   0i
                     0i
                 B
                    2r
THE FIGURE BELOW SHOWS A CROSS SECTION OF AN INFINITE CONDUCTING SHEET
CARRYING A CURRENT PER UNIT X-LENGTH OF L; THE CURRENT EMERGES
PERPENDICULARLY OUT OF THE PAGE. (A) USE THE BIOT–SAVART LAW AND
SYMMETRY TO SHOW THAT FOR ALL POINTS P ABOVE THE SHEET, AND ALL POINTS
P´ BELOW IT, THE MAGNETIC FIELD B IS PARALLEL TO THE SHEET AND DIRECTED AS
SHOWN. (B) USE AMPERE'S LAW TO FIND B AT ALL POINTS P AND P´.
FIRST PART

             Vertical Components
             Cancel
APPLY AMPERE    TOL CIRCUIT
                  B


                                          Infinite Extent


                  B

         current per unit length
       Current inside the loop is therefore :
       i  L
THE “MATH”
              B


                                      Infinite Extent


             B

              B  ds   i
                         0 enclosed

             BL  BL   0 L
                   0
             B
                   2
Infinite Sheet of Charge


           
       E
          2 0

Infinite sheet of current

          0
       B
           2
COIL OR SOLENOID




  Valve Application

                      Switches
THE MAGNETIC FIELD
A PHYSICAL SOLENOID/COIL MODEL
INSIDE THE SOLENOID




   For an “INFINITE” (long) solenoid the previous
   problem and SUPERPOSITION suggests that
     the field OUTSIDE this solenoid is small!
MORE ON LONG SOLENOID
Field is ZERO far from coil!




                               Field looks “UNIFORM”




Field is ZERO far from coil
THE REAL THING…..
      Finite Length

                      Weak Field



                            Stronger - Leakage
ANOTHER WAY


    Far away
                                                Ampere :

                                                 B  ds   i0 enclosed

                                                0h  Bh   0 nih
                                                B   0 ni


   n=number of turns per unit length. nh=total number of turns.
THE LENGTH OF A SOLENOID IS DOUBLED BUT
THE NUMBER OF TURNS REMAINS THE SAME.

 A.   The magnetic field is doubled.
 B.   The magnetic field is cut to a quarter.
 C.   The magnetic field does not change.
 D.   The magnetic field is quadrupled.
 E.   The magnetic field is cut in half.
THE NUMBER OF TURNS IS DOUBLED
BUT THE LENGTH REMAINS THE SAME.
 A.   The magnetic field is quadrupled.
 B.   The magnetic field is cut to a quarter.
 C.   The magnetic field is doubled.
 D.   The magnetic field does not change.
 E.   The magnetic field is cut in half.
B   0 ni
APPLICATION
 Creation of Uniform Magnetic Field Region
 Minimal field outside
       except at the ends!
TWO COILS
“REAL” HELMHOLTZ COILS

                    Used for
                    experiments.

                    Can be aligned to
                    cancel
                    out the Earth’s
                    magnetic
                    field for critical
                    measurements.
THE TOROID




              Slightly less
              dense than
             inner portion
THE TOROID
                Ampere again. We need only worry
                 about the INNER coil contained in
                       the path of integration :

              B  ds  B  2r   Ni (N  total # turns)
                                   0

                                   so
                                    0 Ni
                              B
                                   2r
MAGNETIC FLUX

 NEW CONCEPT – WELL SORTA
 NEW
MAGNETIC FLUX




                           B  dA     M




     M   B  dA Like Gauss - Open Surface
MAGNETIC FLUX IS A


            A.   Vector
            B.   Scalar
            C.   Tensor
MAGNETIC FLUX

         M   B  dA Like Gauss - Open Surface


For a CLOSED Surface we might expect this to be equal to
                some constant times the
      enclosed poles … but there ain’t no such thing!




               B  dA  0
       CLOSED SURFACE
A PUZZLEMENT ..


              closed
               path
                        B  ds  0ienclosed


    Let’s apply this to the gap of a capacitor.
CONSIDER THE POOR LITTLE CAPACITOR…



        i                            i




    CHARGING OR DISCHARGING …. HOW CAN CURRENT
              FLOW THROUGH THE GAP
                In a FIELD description??
THROUGH WHICH SURFACE DO WE
MEASURE THE CURRENT FOR
AMPERE’S LAW?




                              I=0
IN THE GAP…
DISPLACEMENT CURRENT
       The ELECTRIC FLUX through S2
                            q
                E  EA 
                            0
                d E 1 dq
                    
                 dt    0 dt
                    Let
               dq
                   I d (in gap)
               dt
            Displacement Current
                  dq      d E
             Id      0
                  dt       dt
SO FAR ..
 We found that currents create magnetic fields.
 Stationary charges do not.

 Static magnetic fields do not create currents.

 What about CHANGING magnetic fields??
FROM THE DEMO ..
FARADAY’S EXPERIMENTS




                   ?
                   ?
INSERT MAGNET INTO COIL
REMOVE COIL FROM FIELD REGION
Summary
THAT’S STRANGE …..




    These two coils are perpendicular to each other
REMEMBER THE DEFINITION OF TOTAL
ELECTRIC FLUX THROUGH A CLOSED
SURFACE:



              E             d E
                      surface

         Total Flux of the Electric
        Field LEAVING a surfaceis
              E   E  n out dA
MAGNETIC FLUX - REMINDER
 Applies to an OPEN SURFACE only.
 “Quantity” of magnetism that goes through a surface.

 A Scalar




                B   B  dA
                        surface
CONSIDER A LOOP       Magnetic field passing
                       through the loop is
                       CHANGING.
   xxxxxxxxxxxxxxx    FLUX is changing.
   xxxxxxxxxxxxxxx    There must be an emf
   xxxxxxxxxxxxxxx     developed around the
   xxxxxxxxxxxxxxx     loop.
                          A current develops (as we
   xxxxxxxxxxxxxxx         saw in demo)
   xxxxxxxxxxxxxxx    Work has to be done to
   xxxxxxxxxxxxxxx    move a charge
                      completely around the
                      loop.
FARADAY’S LAW (MICHAEL FARADAY)
                      Again, for a current to
                       flow around the circuit,
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx     there must be an emf.
   xxxxxxxxxxxxxxx    (An emf is a voltage)
   xxxxxxxxxxxxxxx    The voltage is found to
   xxxxxxxxxxxxxxx     increase as the rate of
   xxxxxxxxxxxxxxx     change of flux
   xxxxxxxxxxxxxxx
                       increases.
FARADAY’S LAW (MICHAEL FARADAY)


   xxxxxxxxxxxxxxx          Faraday's Law
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx                    d
   xxxxxxxxxxxxxxx            emf  
   xxxxxxxxxxxxxxx                    dt

     We will get to the minus sign in a short time.
FARADAY’S LAW (THE MINUS SIGN)


   xxxxxxxxxxxxxxx   Using the right hand rule, we
   xxxxxxxxxxxxxxx   would expect the direction
   xxxxxxxxxxxxxxx   of the current to be in the
   xxxxxxxxxxxxxxx   direction of the arrow shown.
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx
FARADAY’S LAW (MORE ON THE
MINUS SIGN)

                                             The minus sign means
        xxxxxxxxxxxxxxx                      that the current goes the
        xxxxxxxxxxxxxxx                      other way.
        xxxxxxxxxxxxxxx
        xxxxxxxxxxxxxxx                      This current will produce
        xxxxxxxxxxxxxxx                      a magnetic field that
        xxxxxxxxxxxxxxx                      would be coming OUT of
        xxxxxxxxxxxxxxx                      the page.

The Induced Current therefore creates a magnetic field that OPPOSES the attempt to
INCREASE the magnetic field! This is referred to as Lenz’s Law.
HOW MUCH WORK?


   xxxxxxxxxxxxxxx      Work/Unit Charge 
   xxxxxxxxxxxxxxx
                                                 d
   xxxxxxxxxxxxxxx     W / q  V   E  ds  
   xxxxxxxxxxxxxxx                               dt
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx
   xxxxxxxxxxxxxxx

     A magnetic field and an electric field are
     intimately connected.)
MAGNETIC FLUX

                    B   B  dA

 This is an integral over an OPEN Surface.
 Magnetic Flux is a Scalar


   The UNIT of FLUX is the   weber
       1 weber = 1 T-m2
WE FINALLY   STATED


 FARADAY’s LAW
                        d
emf  V   E  ds  
                        dt
  FROM THE EQUATION    Lentz

                        d
emf  V   E  ds  
                        dt



                           B   B  dA
FLUX CAN CHANGE


             B   B  dA
     If B changes
     If the AREA of the loop changes
     Changes cause emf s and currents and consequently
      there are connections between E and B fields

     These are expressed in Maxwells Equations
               MAXWELL’S FOUR EQUATIONS



Ampere’s Law



        Gauss

                         closed
                          surface
                                    B  dA  0
        Faraday



      No Monopoles
                           B  dA  0
ANOTHER VIEW OF THAT DAMNED
MINUS SIGN AGAIN …..SUPPOSE
THAT B BEGINS TO INCREASE ITS
MAGNITUDE INTO THE PAGE
 xxxxxxxxxxxxxxx      The Flux into the page
                       begins to increase.
 xxxxxxxxxxxxxxx
                      An emf is induced around a
 xxxxxxxxxxxxxxx       loop
 xxxxxxxxxxxxxxx      A current will flow
 xxxxxxxxxxxxxxx      That current will create a
 xxxxxxxxxxxxxxx       new magnetic field.
 xxxxxxxxxxxxxxx      THAT new field will
                       change the magnetic flux.
THE STRANGE WORLD OF DR. LENTZ
LENZ’S LAW
Induced Magnetic Fields always FIGHT to stop
what you are trying to do!
i.e... Murphy’s Law for Magnets
EXAMPLE   OF   NASTY LENZ




          The induced magnetic field opposes the
          field that does the inducing!
DON’T HURT YOURSELF!




   The current i induced in the loop has the direction
   such that the current’s magnetic field Bi opposes the
   change in the magnetic field B inducing the current.
Let’s do the
Lentz Warp
  again !
AGAIN: LENZ’S LAW

    An induced current has a direction
    such that the magnetic field due to
    the current opposes the change in
    the magnetic flux that induces the
    current. (The result of the
    negative sign!) …
                         OR
      The toast will always fall buttered side down!
AN EXAMPLE

             The field in the diagram
             creates a flux given by
             B=6t2+7t in milliWebers
             and t is in seconds.

             (a) What is the emf when
             t=2 seconds?

             (b) What is the direction
             of the current in the
             resistor R?
THIS IS   AN EASY ONE   …

                             B  6t  7t
                                    2


                                     d
                            emf          12t  7
                                      dt
                            at t  2 seconds
                            emf  24  7  31mV
    Direction? B is out of the screen and increasing.
    Current will produce a field INTO the paper
    (LENZ). Therefore current goes clockwise and R
    to left in the resistor.
The diagram shows two parallel loops of wire having a common
axis. The smaller loop (radius r) is above the larger loop (radius
R) by a distance x >> R. Consequently, the magnetic field due
to the current i in the larger loop is nearly constant throughout
the smaller loop. Suppose that x is increasing at the constant rate
of dx/dt = v. (a) Determine the magnetic flux through the area
bounded by the smaller loop as a function of x. In the smaller
loop, find (b) the induced emf and (c) the direction of the
induced current.

                                     v
DOING IT

               B is assumed to be
               constant through the
               center of the small
               loop and caused by
               the large one.
           q
THE CALCULATION OF BZ
                                                  0 ids
                  dBz  dB cosq  cosq
                                                 4 R 2  x 2 
                                     R
                  cosq 
                            x2 
                            R   2        1/ 2


              q        0 ids               R
                dBz 
                      4 R 2  x 2  R 2  x 2 1/ 2
                  ds  Rd
                              0iR 2
                  Bz 
                         2R 2  x 2 
                                         3/ 2
MORE WORK

    In the small loop:

                                  r 2 0iR 2
             Bz A  r 2 Bz 
                             2R x   2
                                           
                                          2 3/ 2


           For x  R (Far Away as prescribed)
                r 2 0iR 2
                    3                            dx/dt=v
                 2x
                   d 3r 2  0iR 2
           emf            4
                                    V
                   dt    2x
WHICH WAY IS CURRENT IN SMALL
LOOP EXPECTED TO FLOW??




               B

                       q
WHAT HAPPENS HERE?
                      Begin  to move handle
                       as shown.
                      Flux through the loop
                       decreases.
                      Current is induced
                       which opposed this
                       decrease – current
                       tries to re-establish
                       the B field.
MOVING THE BAR


                 Flux  BA  BLx
                 Dropping the minus sign...
                       d     dx
                 emf      BL  BLv
                       dt     dt
                    emf BLv
                 i     
                     R     R
MOVING THE BAR TAKES WORK

                                    BLv
                   F  BiL  BL 
                                     R
                   or
                      B 2 L2 v
                   F
                         R
       v
                                      Fx  Fv
                                 dW d
                   POWER 
                                  dt  dt
                      B 2 L2 v
                   P          v
                         R
                      B 2 L2 v 2
                   P
                          R
WHAT ABOUT A SOLID LOOP??

                      Energy is LOST
                      BRAKING SYSTEM




             METAL          Pull
Back to Circuits for a bit ….
DEFINITION




     Current in loop produces a magnetic field
     in the coil and consequently a magnetic flux.

     If we attempt to change the current, an emf
     will be induced in the loops which will tend to
     oppose the change in current.

     This this acts like a “resistor” for changes in current!
REMEMBER FARADAY’S LAW

                        d
emf  V   E  ds  
                        dt
          Lentz
LOOK AT THE FOLLOWING CIRCUIT:




 Switch is open
 NO current flows in the circuit.

 All is at peace!
CLOSE THE
CIRCUIT…




 After the circuit has been close for a long time, the
  current settles down.
 Since the current is constant, the flux through the coil
  is constant and there is noEmf   .

   Current is simply E/R (Ohm’s Law)
CLOSE THE
CIRCUIT…




 When  switch is first closed, current begins to flow
  rapidly.
 The flux through the inductor changes rapidly.
 An emf is created in the coil that opposes the
  increase in current.
 The net potential difference across the resistor is
  the battery emf opposed by the emf of the coil.
CLOSE THE
CIRCUIT…




         d   Ebattery  V (notation)
 emf  
         dt              d
               V  iR     0
                         dt
             Ebattery  …
MOVING RIGHT ALONG V (notation)
                             d
                   V  iR      0
                             dt
        The flux is proportional to the current
        as well as to the number of turns, N.
                   For a solonoid,
                    i  Li  N B
                     d      di
                         L
                      dt     dt
                             di
                  V  iR  L  0
                             dt
DEFINITION OF INDUCTANCE L


                       N B
                    L
                        i

       UNIT of Inductance = 1 henry = 1 T- m2/A

  B is the flux near the center of one of the coils
                 making the inductor
CONSIDER A SOLENOID

                  l
                                 B  ds   i
                                             0 enclosed

                                 Bl   0 nli
                                or
      n turns per unit length
                                B   0 ni
SO….

              N B nlBA nl 0 niA
           L         
               i     i     i
           or
            L   0 n 2 Al
           or
                  inductance
            L/l               n 2 A
                  unit length
       Depends only on geometry just like C and
       is independent of current.
INDUCTIVE CIRCUIT
                       Switch to “a”.
          i            Inductor seems like a
                        short so current rises
                        quickly.
                       Field increases in L and
                        reverse emf is generated.
                       Eventually, i maxes out and
                        back emf ceases.
                       Steady State Current
                        after this.
THE BIG INDUCTION
 As  we begin to increase the current in the coil
 The current in the first coil produces a magnetic
  field in the second coil
 Which tries to create a current which will reduce
  the field it is experiences
 And so resists the increase in current.
BACK TO   THE REAL WORLD…

           Switch to “a”
                  i        sum of voltage drops  0 :
                                       di
                            E  iR  L  0
                                       dt
                           same form as the
                           capacitor equation
                              q   dq
                           E R    0
                              C   dt
SOLUTION


              E        Rt / L
           i  (1  e          )
              R
           time constant

             L
               R
SWITCH POSITION “B”

                      E0
                        di
                      L  iR  0
                        dt
                         E t / 
                      i e
                         R
           Max Current Rate of
VR=iR      increase = max emf
~current
   E
i   (1  e  Rt / L )
   R
   L
  (time constant)
   R
IMPORTANT QUESTION
                  Switch closes.
                  No emf

                  Current flows for a while

                  It flows through R

                  Energy is conserved (i2R)




   WHERE DOES THE ENERGY COME FROM??
FOR AN ANSWER
RETURN TO THE BIG C
       E=0A/d         We  move a charge dq
                        from the (-) plate to the
                        (+) one.
                       The (-) plate becomes
                        more (-)
                       The (+) plate becomes
        +dq             more (+).
  +q                   dW=Fd=dq x E x d
                 -q
THE CALC

                                 q
    dW  (dq) Ed  (dq) d  (dq)      d
                       0        0 A
       d             d q2
   W
      0 A  qdq   0 A 2
    or
          d         1  2 Ad 1   2      1
   W        (A) 
                 2
                                  2  Ad   0 E 2 Ad
                             0 
      2 0 A        2 0     2  0      2
         energy    1
    u             0 E           The energy is in
                         2

       unit volume 2                the FIELD !!!
WHAT ABOUT POWER??

                            di
                      E  L  iR
                            dt
                      i :
                             di 2
                      iE  Li  i R
                    power
                             dt power
                      to      dissipated
                    circuit   by resistor
   Must be dWL/dt
SO


     dWL      di
          Li
      dt      dt
                   1 2
     WL  L  idi  Li
                   2
          1               Energy
     WC  CV     2
                          stored
          2                in the
                         Capacitor
WHERE IS   THE ENERGY??

           l
                     B  ds   i  0 enclosed

                    0l  Bl   0 nil
                    B   0 ni
                    or
                           0 Ni
                    B
                            l
                                    0 Ni
                      BA                 A
                                     l
REMEMBER   THE   INDUCTOR??

                    N
                L
                      i
      N  Number of turns in inductor
              i  current.
    Φ  Magnetic flux through one turn.
SO …
          N
       L
            i
          N
       i
           L
           1 2 1 2 N 1
       W  Li  i          N i
           2       2    i   2
            0 NiA
       
               l
           1   0 NiA    1      2 2 A
       W  Ni              0 N i
                                2

           2  l  2 0               l
       1      A
W       N i 2     2 2

   2 0
              0
              l
From before:
      0 Ni
B
       l
       1   A   1 2
W      Bl   2 2
                  B V (volume)
   2 0    l 2 0
or
   W   1 2                ENERGY IN THE
u       B                 FIELD TOO!
   V 2 0
IMPORTANT CONCLUSION
 A region of space that contains either a magnetic
  or an electric field contains electromagnetic
  energy.
 The energy density of either is proportional to the
  square of the field strength.