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Homework Help: Help on quick question

  1. May 17, 2005 #1
    I was given this formula for magnetic field strength:

    H=vector product v *d(flux density vector)

    I wanted to know why it's dependent on velocity of charge ?

    And what is permittivity ? I'ts calibrated to vacuum I'm assuming, but what is there to resist charge ?
    thanks in advance

  2. jcsd
  3. May 17, 2005 #2


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    This will be an oversimplified view of what is going on, but hopefully it will give you the basic ideas.

    Basically, magnetic field strength depends on the velocity of the charge because stationary charges do not produce magnetic fields. You have probably learned that currents in wires produce magnetic fields that are proportional to the current. Current can be increased in two ways: either you can increase the number of charges moving at a certain rate, or you can increase the rate of motion of a number of charges, including just one charge.

    The whole subject of electric and magnetic fields is fairly complex. Whether a charge is stationary or moving depends on the frame of reference from which you are looking at the charge. In the rest frame of the charge, there is no magnetic field, but in any reference frame that sees the charge as moving there will be a magnetic field. Observers who are moving relative to one another see different electric and magnetic fields from the same charge.

    It is possible to have a magnetic field without having any local charges that you can observe in motion. All that is needed is an electric field that is varying in time. This is what happens with light waves. A light wave is the combination of a time varying electric field that is produced by some distant moving charges and a corresponding time varying magnetic field. The electric and magnetic fields propagate through space as electromagnetic waves at the speed we call the speed of light.

    The electric displacement field D is a convenient way to account for the complex situation that arises when an electric field is present in an overall neutral assembly of charges such as exists in all materials. The microscopic electric field inside any material is the result of any applied field plus the organization of all the protons and electrons that make up the material. The displacement field is a smoothed out field that represents an average over some small (on the scale of the size of the material) region within the material. Some materials become polarized when an electric field is applied, which results in a tendency of the internal charges to align so as to cancel the applied field. Such effects are better understood in terms of the D field rather than the microscopic E field.

    There is a similar consideration for magnetic fields inside of magnetic materials. The microscopic B field is very complex, and the smoothed H field is a more convenient way to look at things.

    In the simplest cases, D is proportional to E and H is proportional to B. Permittivity is the constant of proportionality between D and E, and permeability is the constant of proportionality between B and H. You will find the equations

    [tex] D = \epsilon E [/tex]


    [tex] B = \mu H [/tex]

    In any material the product the two proportionality constants is the reciprocal of the square of the speed of light in that material

    [tex] \epsilon \mu = \frac{1}{v^2} [/tex]

    In free space, the speed of light is c, and permittivity and permeability are given the subscript zero

    [tex] \epsilon_0 \mu_0 = \frac{1}{c^2} [/tex]

    The numerical values given to these parameters are different in different units, but the connection to the speed of light is the same.
    Last edited: May 17, 2005
  4. May 18, 2005 #3
    thanks for the reply.

    So why is [tex] \epsilon_0 \mu_0 [/tex] actually added to equations ?
    Is it to produce the correct numerical values ?

    And why is [tex] {c^2} = \frac{1}\epsilon_0 [/tex] ?

  5. May 18, 2005 #4


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    [tex] {c^2} \ne \frac{1}{\epsilon_0 } [/tex]

    [tex] {c^2} = \frac{1}{\epsilon_0 \mu_0} [/tex]

    [tex] \epsilon_0 \mu_0 [/tex] is not "added to" equations. These constants are part of the fundamental equations relating electric an magnetic fields. They are not arbitrarily added or taken away any more than m is added or taken away from F = ma or G is added or taken away from

    [tex] F = \frac{GMm}{r^2} [/tex]

    They are constants of proportionality. Their product is a proportionality constant in one of the four fundamental equations developed by Maxwell that serve as the foundation of electromagnetic theory. In free space

    [tex] \nabla \times B = \epsilon_0 \mu_0 \frac{\partial E}{\partial t} [/tex]

    Historically, the origin of electric and magnetic fields were recognized as coming from charge distributions, and currents in wires. The letters used to represent those quantities and the units for the physical quantities involved developed independently. Bringing them together, as Maxwell did, requires a proportionality constant. It happens that light travels at a particular speed in free space. When expressed in terms of the units that were developed for representing length and time that speed is some large number of meters per second. It is possible (and some theoretical physicists are fond of doing it) to define the speed of light to be 1 and develop a whole system of units surrounding that. I expect that is a pretty big leap from units that are familiar to you. In those units

    [tex] 1 = \frac{1}{4 \pi \epsilon_0} [/tex]

    [tex] 1 = \frac{\mu_0}{ 4 \pi} [/tex]

    so the product of the proportionality constants becomes

    [tex] \epsilon_0 \mu_0 = 1 [/tex]

    This does not really get rid of these quantities. It makes it unnecessary to write them, but at a price. The price is you need to redefine things like charge and current in units that you would not recognize.
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