IV

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IV3 Energy of Magnetic Field
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Main Topics

• Transformers
• Energy of Magnetic Field
• Energy Density of Magnetic Field
• An RC Circuit
• An RL Circuit
• An RLC Circuit - Oscilations

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The Transformer I

• Transformer is a device with (usually) two or
  more coils sharing the same flux. The coil to
  which the input voltage is connected is called
  primary and the other(s) are secondary.
• Transformers are mostly used to convert voltages
  or to adjust (match) internal resistances.

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The Transformer II

• Let us illustrate the principle of functioning of
  a transformer on a simple type with two coils
  with N1 and N2 loops. We shall further suppose
  that there is negligible current in the secondary
  coil.
• Since both lops share the same magnetic flux, in
  each loop of each coil the same EMF ?1 is
  induced
• ?1 - d?/dt

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The Transformer III

• If we connect a voltage V1 to the primary coil,
  the magnetization in the core will grow until the
  counter EMF induced in this coil is equal to the
  input voltage
• V1 N1?1
• The voltage in the secondary coil is also
  proportional to its number of loops
• V2 N2?1

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The Transformer IV

• So voltages in both coils are proportional to
  their number of loops
• V1/N1 V2/N2
• It is more difficult to understand the case when
  the secondary coil is loaded and, of course, even
  much more difficult to design a good transformer.

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The Transformer V

• Suppose, our transformer has efficiency close to
  1. In this case, it is easy to show that currents
  are inversely proportional to the number of loops
  in each coil and resistances are proportional to
  their squares
• P V1I1 V2N1I1/N2 V2I2
• I1N1 I2N2
• R1/N12 R2/N22

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Energy of Magnetic Field I

• An inductance opposes changes in current. That
  means it is necessary to do work to reach a
  certain current in a coil. This work is
  transferred into the potential energy of magnetic
  field and the field starts to return it at the
  moment we want to decrease the current.
• If a current I flows through a coil and we want
  to increase it, we have to deliver power
  proportional to the rate of the change we want to
  reach.

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Energy of Magnetic Field II

• In other words we have to do work at a certain
  rate to move charges against the field of the
  induced EMF
• P I? ILdI/dt ?
• dW Pdt LIdI
• To find the work to reach current I, we
  integrate
• W LI2/2

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Energy Density of Magnetic Field I

• Similarly as in the case of a charged capacitor
  the energy here is distributed in the field, now
  of course magnetic.
• If the field is uniform, as in a solenoid, it is
  easy to find the density of energy
• We already know formulas for L and B
• L ?0N2A/L
• B ?0NI/L ? I BL/?0N

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Energy Density of Magnetic Field II

• W ½ ?0N2A/L (BL/?0N)2 ½ B2/?0 AL
• Since AL is the volume of the inside of the
  solenoid, where we expect (most of) the field ½
  B2/?0 can be attributed to the magnetic field
  energy density.
• Although we have found this formula for a special
  case of solenoid, it is valid more generally.

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RC, RL, LC and RLC Circuits

• Often not only static but also kinetic processes
  are important. So we have to find out how
  quantities depend on time when charging or
  discharging a capacitor or a coil. We shall se
  that circuits with LC will show a new effect
  un-dumped or dumped oscillations.

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RC Circuits I

• Lets have a capacitor C charged to a voltage Vc0
  and at time t 0 we start to discharge it by a
  resistor R.
• At any instant the capacitor can be considered as
  a power source and the Ohms law is valid
• I(t) Vc(t)/R
• This leads to differential equation.

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RC Circuits II

• All quantities Q, V and I decrease exponentially
  with the time-constant ?.
• Now, lets connect the same resistor and
  capacitor serially to a power supply V0. At any
  instant we find from the second Kirchhoffs law
• I(t)R Vc(t) V0
• a little more difficult differential equation.

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RC Circuits III

• Now Q and V exponentially saturate and I
  exponentially decreases as before. All with the
  time-constant ?.

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RL Circuits I

• A similar situation will be if we replace the
  capacitor in the previous circuit by a coil L.
• As the current grows in the circuit the sign of
  the induced EMF on the coil will be the same as
  was on the capacitor and we can again use the
  second Kirchhofs law
• RI(t) LdI/dt V0
• This is again a similar differential equation.

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RL Circuits II

• I starts from zero and exponentially saturates
  while the EMF on the coil (VL) starts from its
  maximal value, of course V0, and exponentially
  decreases.

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LC Circuits I

• Qualitatively new situation will be when we
  connect a charged capacitor C to a coil L.
• It can be expected that now the energy will
  change from the electric one to the magnetic one
  and back. We obtain un-dumped periodic movement.

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LC Circuits II

• This circuit is called an LC oscillator and it
  produces electromagnetic oscillations.
• We can use the second Kirchhoffs law
• -L dI/dt Vc 0
• This is again a differential equation but of
  different higher order.

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LC Circuits III

• What happens qualitatively
• If the capacitor is charged and there is zero
  current in the beginning it will tend to
  discharge through the coil. The discharge can
  only be by current in the circuit. The coil will,
  however prevent immediate discharge since it will
  allow only a certain rate of current growth.

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LC Circuits IV

• So current grows and the capacitor discharges.
  But since also the voltage on the capacitor
  decreases the grow slows down.
• At the point, the capacitor is discharged, the
  voltage on it and thereby the rate of the growth
  of the current are zero. But the current is in
  its maximum and the coil prevents it to drop
  instantly.

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LC Circuits V

• The EMF on the coil will now grow in the opposite
  direction to oppose the decrease of the current.
  But anyway the EMF as well as the decrease rate
  of the current grows. The capacitor will now be
  charged it the opposite polarity.
• At the moment the capacitor is fully charged the
  current is zero and everything repeats again

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LRC Circuits I

• If we add a resistor the oscillations will be
  dumped. The energy will be lost by thermal loses
  on the resistor.

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Homework

• No homework today!

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Things to read and learn

• Chapter 29 5, 6 30 1, 2

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The vector or cross product I

• Let ca.b
• Definition (components)

The magnitude c
Is the surface of a parallelepiped made by a,b.
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The vector or cross product II
The vector c is perpendicular to the plane made
by the vectors a and b and they have to form a
right-turning system.
?ijk 1 (even permutation), -1 (odd), 0 (eq.)

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The scalar or dot product

• Let ca.b
• Definition I. (components)

Definition II (projection of one vector
into the direction of the other one)

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Rotating Conductive Rod

• Torque on a piece dr which is in a distance r
  from the center of rotation of a conductive rod L
  with a current I in magnetic field B is

• The total torque is

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RC Circuit I

• We use definition of the current I dQ/dt and
  relation of the charge and voltage on a capacitor
  Vc Q(t)/C

• The minus sign reflects the fact that the
  capacitor is being discharged. This homogeneous
  differential equation can be easily solved by
  separating the variables.

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RC Circuit II

• We have defined a time-constant ? RC. We can
  integrate both sides of the equation

• The integration constant can be found from the
  boundary conditions Q0 CVc0

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RC Circuit III

• By dividing this by C and then by R we get the
  time dependence of the voltage on the capacitor
  and the current in the circuit.

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RC Circuit IV

• We again substitute for the current and the
  voltage and reorganize a little

• We get a similar equation for the charge on the
  capacitor but it doesnt have zero on the right
  side. We can solve it by solving first a
  homogeneous equation and then adding one
  particular solution e.g. Qk CV0 (final Q)

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RC Circuit V

• Since we have already solved the homogeneous
  equation in the previous case, we can write

The integration constant we again get from the
initial condition Q(0) 0 ? Q0 -CV0.
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RC Circuit VI

• By dividing this by C we get the time dependence
  of the voltage on the capacitor

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RC Circuit VII

• To get the current we have to calculate the time
  derivative of the charge

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RL Circuit I

• First we solve a homogeneous equation (with zero
  in the right) and add a particular solution Im
  V0/R (maximal current)

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RL Circuit II

• This can be solved by separation of the
  variables. Using the previous, defining the
  time-constant ? L/R and adding the particular
  solution, we get

• We apply the starting conditions I(0) 0 ? I0
  -Im and we get

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RL Circuit III

• The voltage on the coil we get from
• V LdI/dt

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LC Circuit I

• We use definition of the current I dQ/dt and
  relation of the charge and voltage on a capacitor
  Vc Q(t)/C

• We take into account that the capacitor is
  discharged by the current. This is homogeneous
  differential equation of the second order. We
  guess the solution.

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LC Circuit II

• Now we get parameters by substituting into the
  equation

• These are un-dumped oscillations.