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Question about Faraday's law of induction

  1. Dec 20, 2015 #1
    I was studying the Maxwell equation for Faraday's law:

    ∇×E = -(∂B/∂t)

    I then did some math and noticed that the electric field is a conservative vector field, because
    ∇×E= <0,0,0>

    Since this is the case, based on the above Maxwell equation this would set the time derivative of the magnetic field equal to 0 as well (meaning that the magnetic field does not change with respect to time).

    If the magnetic field remains constant with respect to time, then what information exactly is supposed to be taken from this equation? Initially, I thought it was supposed to tell you about the electric field that is induced by changing magnetic fields, but the fact that the electric field is conservative seems to disagree with that thought (unless the formula for computing the electric field varies from situation to situation).
     
  2. jcsd
  3. Dec 20, 2015 #2
    Hi, space_time. I'm curious about your math to arrive at ∇×E= <0,0,0>.
     
  4. Dec 20, 2015 #3
    E= (KQ/|r|2) * (r/|r|) where r is the position vector <x,y,z>

    Transforming r into Cartesian coordinates, this equation turns into:

    E= (KQ/(x2 + y2 + z2)^(3/2)) * <x,y,z>

    Taking the curl of this yields <0,0,0>.

    Do you think I made some kind of arithmetic mistake?
     
  5. Dec 20, 2015 #4
    No, looks fine. The definition of the electric field vector field you began with is valid only for electrostatics, though.

    It's more generally true that the flux through a surface relates to the enclosed charge, which allows for the field to curl.
     
  6. Dec 23, 2015 #5
    This equation relates the production of an electric field to the rate of change of magnetic field intensity. E produced by a changing B is not conservative. If the curl E is found to be zero you know it is not due to a changing magnetic field intensity since for a conservative field div E =0.

    in general E = - ∂A/∂t - V where A is the vector potential due to current densities and V is the scalar potential due to static charges.
     
  7. Dec 24, 2015 #6

    Philip Wood

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    Gold Member

    Just to emphasise what's already been said…

    Up to the 1830's it was thought that the electric field was conservative (curl E zero everywhere, as we say now), essentially because E was thought to be the sum of fields due to point charges obeying Coulomb's law.

    In the 1830's, Faraday (and, I think, Henry) discovered that a magnet thrust into a stationary coil induced a voltage. Since the electrons in the coil weren't moving initially (at least not in a co-ordinated way) it must be an electric field (not a magnetic field per se) that urged them, producing the voltage. This is is the phenomenon summed up by curl E = -dB/dt. [Sorry about lack of partials.]

    Electric fields arising in this way (when magnetic fields change) are not subject to the restriction curl E = 0.
     
    Last edited: Dec 24, 2015
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