Generating%20Currents

1
Generating Currents

• Consider the following circuit a bar moves on
  two rails that are connected at one end. The
  whole setup has a magnetic field that goes
  through it.
• B Ä Ä
• v
• Ä Ä

2
Generating Currents

• If the bar is moving to the right, and the bar
  contains electrons that are free to move (as in
  metal), then the electrons are also moving to the
  right. There is a magnetic force on the
  electrons pushing them down with magnitude
  Fmagnetic q v B.
• B Ä Ä
• Fmag on e v
• Ä Ä

3
Generating Currents

• Thus, negative electrons should pile up at the
  bottom of the moving bar, leaving a net positive
  charge at the top of the bar. But this should
  act just like a battery!
• B Ä Ä
• Fmag on e v
• Ä - Ä

4
Generating Currents

• To find the voltage of this battery, we note
  that as the charges pile up at the ends of the
  rod, an Electric field will be set up. The
  electrons will continue to pile up until the
  Electric Force
• (Fel qE) balances the Magnetic Force
  (FqvB).
• B Ä Ä
• Fmag on e v
• Felec on e ?
• Ä - Ä

5
Generating a Voltage

• We now have, at equilibrium SF 0, or
• Felec on e Fmag on e , or qE qvB , or
• E vB .
• We know that the electric field is related to
  voltage by E -DV / Ds , (here Ds L, the
  length of the bar). Thus we have for the
  voltage DV v B L .

6
Generating Power

• We have generated a voltage, and now to generate
  electric power (PIV) we need to have that
  voltage drive a current. Since we have completed
  the circuit by connecting the ends of the rails,
  we will have a complete circuit - and so we will
  get a current depending on the resistance in the
  rails (VIR).

7
Conservation of Energy

• We have made an electric generator that can
  generate electrical energy. But according to the
  Law of Conservation of Energy, we can only
  convert energy from one form into another.
• In the case of the electric generator, where
  does this energy come from?

8
Generating Currents

• Note that as electrons flow clockwise around the
  circuit, this acts the same as a current of
  positive charges going counterclockwise, as
  indicated on the diagram below. Note that a
  current flows up the bar.
• B Ä I Ä
• I Fmag on e ? v
• Felec on e ? I
• I Ä - Ä

9
Generating Currents

• Is there a magnetic force on this current due to
  its flowing through a magnetic field? YES!
• Note that the direction of the force on this
  current is to the left. This will act to slow
  the bar down. In effect, this apparatus converts
  the kinetic energy of the bar into electric
  energy!
• B Ä I Ä
• I Fmag on e ? v
• Felec on e ? I Fmag on I
• I Ä - Ä

10
Generalizing

• We have from the previous apparatus
• DV v B L .
• We note that v Dw/Dt where w is the width of
  the circuit (distance from end of rails to bar),
  so DV (Dw/Dt) B L .
• Can we take the D /Dt and apply it to all the
  variables DV D(B L w) / Dt ?
• From experiment, YES!

11
Faradays Law

• We also note that wL A (area of circuit). We
  can also have N number of loops, so we finally
  get DV D(N B A) / Dt . This is called
  Faradays Law.
• When we consider direction as well, we see that
  the magnetic field, B, has to cut through the
  area, A. If we assign a direction to A that is
  perpendicular to the surface, we get an even more
  general form
• DV D (N B A cos(qBA) / Dt .

12
Faradays Law

• If the magnetic field is not constant over the
  whole area, we need to break the area into pieces
  and then sum up the pieces. This leads to
  Faradays Law as
• DV d ? B ? dA / dt ,
• where we have used the dot product to represent
  the cos(qBA) .

13
Magnetic Flux

• DV d ? B ? dA / dt
• The quantity in brackets ? B ? dA is called the
  magnetic flux with units of Webers. It is a
  measure of how much magnetic field cuts through a
  certain area. Faradays Law says there is a
  voltage produced whenever this magnetic flux
  changes in time.
• Webers Volt-sec Tm2 .

14
Magnetic Field

• DV d ? B ? dA / dt
• Because of its importance in magnetic flux, the
  magnetic field, B, is sometimes called the
  magnetic flux density.
• Although we already have a unit for B, the Tesla,
  because of Bs importance to inductance and its
  relation to magnetic flux density, it also has an
  equivalent unit called the Weber/m2 Tesla .

15
Lenzs Law

• DV D (N B A cos(qBA) / Dt or
• DV d/dt ? B ? dA
• The above formula is for determining the amount
  of voltage generated. But what is the direction
  of that voltage (what direction will it try to
  drive a current)?
• The answer is Lenzs Law the direction of the
  induced voltage will tend to induce a current to
  oppose the change in magnetic field through the
  area.

16
Example 1

• Using Lenzs Law, which direction should the
  induced current flow when the bar is moving to
  the right as indicated below?
• B Ä Ä
• v
• Ä Ä

17
Example 1 (cont.)

• Since the area through which the magnetic field
  is going is increasing, the magnetic flux is
  increasing. According to Lenzs Law, the induced
  current will flow in a direction so as to create
  a magnetic field that will oppose the increase.
  Here it means the current will create a field
  going out of the loop and so using the RHR the
  current must flow counterclockwise!
• B Ä ? Ä
• I ? Binduced ? v
• ? I
• Ä ? Ä

18
Example 1 (cont.)

• This is the same direction we determined
  previously!
• B Ä ? Ä
• I ? Binduced ? v
• ? I
• Ä ? Ä

19
Example 2

• Consider the situation below in which the North
  pole of a magnet is brought closer to the center
  of a loop of wire.
• Which way will the induced current be in the
  wire?
• FRONT VIEW VIEW from LEFT
• S N ? B B ?
• v ?

20
Example 2 (cont.)

• Since the magnetic field through the loop is
  increasing, the flux is increasing. By Lenzs
  Law, the induced current will try to create a
  magnetic field opposing the increase, and the
  current direction is determined by the RHR.
• FRONT VIEW VIEW from LEFT
• ? Iin ? Iin
• S N ? B B ?
• v ? ? Bin Bin?
• ? ?

21
Example 2 (cont.)

• If the North pole of the magnet is moved away
  from the coil, then the external field will
  decrease, and Lenzs Law will say that the
  induced current should create a magnetic field to
  replace the decreasing external magnetic field.
• FRONT VIEW VIEW from LEFT
• ? Iin Iin ?
• S N ? B B ?
• ? v ? Bin Bin?
• ? ?

22
Lenzs Law

• In figuring out the direction of the induced
  current using Lenzs Law, there are four steps
• Determine the direction of the external magnetic
  field, Bexternal. Recall that for a magnet the
  magnetic field goes out of a North pole and into
  a South pole for a current creating the field,
  use the right hand rule.
• Determine whether that external field, Bexternal,
  is increasing, decreasing, or constant.

23
Lenzs Law

• Determine the direction of the induced field,
  Binduceda) if the external field, Bexternal,
  is increasing, the induced field, Binduced, will
  be in the opposite directionb) if the external
  field, Bexternal, is decreasing, the induced
  field, Binduced, will be in the same direction
• c) if the field is not changing, then there
  will be no induced field and so no induced
  current.
• 4. Using the right hand rule, determine the
  direction the induced current would have to
  flow to create the induced field, Binduced.

24
Lenzs Law

• The Computer Homework assignment, Vol. 4 3,
  deals with Lenzs Law and will give you practice
  with this as well.

25
Transformers and Inductors

• DV D (N B A cos(qBA) / Dt .
• We have already seen how changing the area of a
  circuit in a magnetic field generates a voltage.
• We have also seen how changing the magnetic field
  strength through the circuit can generate a
  voltage - this is the basis of an inductor and a
  transformer.

26
Transformer

• Well consider an inductor in a little bit. But
  now consider the situation in the figure below.
  Note that circuit 1 is not electrically connected
  to circuit 2. But it is connected magnetically,
  since the magnetic field will flow through the
  iron square ring.
• I1 ?
• B1? ? B1
• 1 2

27
Transformer (cont.)

• If I1 changes in circuit 1, then B1 changes. But
  since the iron ring carries B1 through circuit 2,
  a voltage and current will be induced in circuit
  2. The current in circuit 2 depends on how the
  current in circuit 1 changes (as well as the
  number of loops in both circuit 1 and circuit 2).
• I1 ?
• B1? ? B1
• 1 2

28
Transformer (cont.)

• The voltage in circuit 2 depends on the change
• in the current in circuit 1
• V2 M12 dI1/dt
• M12 is called the mutual inductance, and is
• measured in Henries, where
• henry volt / amp/sec volt-sec / amp.
• I1 ?
• B1? ? B1
• 1 2

29
Mutual Inductance

• Like capacitance and resistance, mutual
  inductance depends NOT on voltage and/or current,
  but on the geometry and materials. In this case,
  the mutual inductance will depend on the number
  of turns in both circuits 1 and 2 as well as the
  geometries (solenoids) used in circuits 1 and 2.
  It also depends on the magnetic properties of the
  material connecting the two circuits.

30
Electric Generators

• Finally, we could change the direction of the
  area in relation to the field - this is the basis
  for the most common kind of generator. This
  generator looks just like the electric motor,
  except we put in rotational motion and get
  current instead of putting in current and getting
  rotational motion!

31
Electric Generator

• When we use the crank to turn the area inside the
  magnetic field, we are changing the magnetic
  flux, BA, through the area. By Lens Law this
  should generate a voltage and current!
• r
• N L S
• pole B ? pole
• w
• crank

32
Electric Generator

• (two complete rings
• rather than one split ring)
• N S
• crank (turn at frequency, f)

33
Electric Generators

• DV d/dt (N B A cos(qBA) .
• When we change the angle, qBA, with respect to
  time qBAwt , we get the following relation DV
  N B A w sin(qBA) , or
• VAC Vo sin(wt) where Vo NBAw, and where
  w dqBA/dt 2pf .
• This kind of voltage is an alternating voltage
  (AC voltage) since the sine function alternates
  between positive and negative.

34
Electric Generators

• You should be able to DESIGN your own generator
  based on the voltage (V) and frequency (f) you
  want out of the generator.
• Design means you have choices for some of the
  parameters and use the relation to solve for the
  last parameter to make the equation work VAC
  Vo sin(wt) where Vo NBAwwhere w 2pf.

35
AC Circuits

• VAC Vo sin(wt)
• For this kind of AC voltage, we can determine the
  amplitude of the voltage (Vo NBAw) . But since
  the average of sine is zero, how do we treat the
  average?
• What is usually important is the power delivered
  by the electric circuit. From PIV we see that
  both the current and the voltage are important.

36
AC Circuits

• From Ohms Law, we have V IR, where R is a
  constant that depends on the material and
  geometry of the materials used to conduct the
  current. Thus, I V/R (NBAw/R) sin(wt)
  Io sin(wt) , where Io NBAw/R Vo/R .
• We see that an alternating voltage makes an
  alternating current!
• From this we see that the electric power is
• P(t) I(t)V(t) IoVo sin2(wt) .

37
AC Power

• P(t) I(t)V(t) IoVo sin2(wt)
• Note that the Power involves the square of the
  sine function, and so the Power oscillates but is
  always positive.
• But what we are usually interested in is the
  average power. From the calculus, we find that
  the average of sin2(q) ½. Thus
• Pavg ½IoVo .

38
Average of sin2(q)

• Avg of sin2(q) S sin2(qi) / S (1)
• This is approximate, so we could extend the
  summation to an integral
• Avg of sin2(q) ? sin2(q) dq / ? (1) dq , where
  the limits of integration go from 0 to 2p for
  both the numerator and the denominator.

39
Another way

• Notation well use lt gt to indicate average.
• A tricky way to get this without integrating is
• to note that sin2(q) cos2(q) 1 for all qs,
  so ltsin2(q) cos2(q)gt lt1gt 1 . But
  ltsin2(q) cos2(q)gt ltsin2(q)gt ltcos2(q)gt.
• Note that ltsin(q)sin(q)gt ? ltsin(q)gtltsin(q)gt.
• Sine and cosine differ only by the starting
  point, so ltsin2(q)gt ltcos2(q)gt over a complete
  oscillation. Hence, average of ltsin2(q)gt ½ !

40
RMS Voltage and Current

• In order to work with AC circuits just as we did
  with DC circuits, we create a voltage and current
  called rms (root mean square).
• rms square the values, take the average of the
  squares (mean), and then take the square root of
  that average ltsin(q)2gt½ ½½.
• Vrms Vo (½)1/2 and Irms Io (½)1/2
  so that we have
• Pavg ½ IoVo IrmsVrms, and Vrms IrmsR
  which are just like the formulas for DC
  circuits!

41
Eddy Currents

• Another consequence of Faradays Law is the
  existence of eddy currents. Well look at a
  demonstration in class to see these eddy currents
  at work.

42
Eddy Currents in a spinning conducting disk

• Note the Magnetic force on the electrons moving
  with the disk, and note how Lenzs Law applies
  here. Recall that the electron flow is opposite
  that of normal current flow.
• electron flow
  electron flow
• ? v ?Binduced? ? Binduced
  v ?
• excess e- ? Fmag on e-
  Fmag on e-? excess e-
• deficit of e-
• ? Bexternal ? ?
• ? ? ?

43
Eddy Currents in a spinning conducting disk

• On the left (right) side, electrons are now
  moving to the left (right) in a magnetic field
  directed into the page, so there is a force on
  the electrons directed up (down) which is
  opposite the velocity of the disk and so will
  slow down the disk.
• electron flow
  electron flow
• ? v ?Binduced? ? Binduced
  v ?
• excess e- ? Fmag on e-
  Fmag on e-? excess e-
• deficit of e-
• ? Bexternal ? ?
• ? ? ?

44
Eddy Currents

• Since eddy currents will tend to slow a rotating
  conductor down (turning kinetic energy into heat)
  as it goes in and out of magnetic fields, it will
  be useful to see if we can figure out how to
  reduce those eddy currents so we can increase the
  efficiency of both motors and generators.
• In the example in class of the rotating disk,
  one way is to make slits in the disk so the eddy
  currents cant go very far.

45
Extension of Amperes Law

• Recall from Part 3 Amperes Law
• ?closed loop B ? dL moIencircled
• Consider a circuit with a capacitor that is being
  charged. There is a current flowing to the
  capacitor, and so there will be a magnetic field
  that encircles the wire. However, there is no
  current flowing across the capacitor.
• Will there be a magnetic field encircling the
  charging capacitor?

46
Amperes Law (extended)

• Gauss Law for Electric Fields
• ?? E ? dA Qenclosed/eo
• Recall also that I dq/dt. These two can be
  combined to give for a charging capacitor
• Ieffective eo d/dt ?? E ? dA .
• With this, Amperes Law can be extended
• ?closed loop B ? dL moIencircled moeo d/dt
  ?? E ? dA

47
Maxwells Equations

• We now have the four main laws for electric and
  magnetic fields. Together they are called
  Maxwells Equations. The four laws are

48
Maxwells Equations

• Gauss Law for Electric fields
• ??closed area E ? dA Qenclosed / eo
• Gauss Law for Magnetic fields
• ??closed area B ? dA 0
• Amperes Law
• ?closed loop B ? dL moIencircled moeo d/dt
  ?? E ? dA
• Faradays Law
• ?closed loop E ? dL -DV -d/dt ? B ? dA
  .