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What is the direction of E?

Discussion in 'Electronic Design' started by Chris Carlen, Sep 18, 2003.

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  1. Chris Carlen

    Chris Carlen Guest


    I am trying to understand the reasons behind the signs in Faraday's Law.

    Suppose there is a single loop of wire (N=1) on a horizontal surface
    (like the paper) through which a magnetic flux density B is increasing
    in the upward direction. The loop is almost closed, but a resistor load
    is connected across the small gap in the loop.

    We will define the surface normal ds to be up, so that the contour C of
    the loop, is counterclockwise. We also define V_tr_emf (the induced
    voltage across the gap in the stationary loop by the changing magnetic
    field) to be positive where when following the contour we first
    encounter the gap, and negative when we encounter the other side of the

    Since B is increasing up and given the definitions established, it is
    apparent that (assuming a uniform B always perpendicular to the
    surface,) the quantity N*d(flux)/dt is negative, and so the real voltage
    measured by a voltmeter at the gap would be opposite the definition of
    V_tr_emf, so:

    V_tr_emf = -N*d(flux)/dt , which is the definition of Faraday's Law as
    given in my EM text. Things are looking good so far.

    +--------+ | |
    | | | loop: |
    | | | B increasing |
    / (-) +-------+ 1 out of paper |
    \ (+) toward you, | i current points
    / R V_tr_emf | | down in accordance
    \ (-) | | with Lenz' Law
    / (+) +-------+ 2 ds points | V
    | | | out of paper |
    | | | toward you |
    +--------+ | |

    C---> contour is CCW

    Here's where problems arise:

    We know that the E field in the resistor is with the direction of
    current flow. But the current flow in the loop is against the direction
    of the actual polarity of V_tr_emf. Remember that the *real physically
    observed* V_tr_emf has the opposite sign of the *defined* V_tr_emf.

    What is the direction of E in the loop (E_loop)?

    I would think that E_loop must be in agreement with the real physical
    value of V_tr_emf, which would make E point opposite the direction of
    current flow in the loop. This is also consistent with the fact that
    the loop is a generator, so the magnetic field is importing energy to
    the charge carriers, thus, the current must be moving *against* E_loop
    in the loop, while in the external circuit current is moving *with* E
    through the resistor, where the current is dissipating energy.

    But if that is so then the contour integral around C from point 2 to
    point 1 of E_loop is positive.

    Thus we have:

    integral_C(E_loop . dl_vec) > 0


    N*d(flux)/dt > 0

    Thus we have the opposite sign on one or the other terms compared to
    what is known in Faraday's Law (the full relation in MAxwell's equations:)

    integral_C(E_loop . dl_vec) = - N*d(flux)/dt

    On the other hand, if we say that the E_loop field is in accordance with
    the defined direction of V_tr_emf, then the quantity

    integral_C(E_loop . dl_vec) < 0

    which agrees with what we know about the other side of Faraday's Law.

    So the question is which way does E_loop really point?

    I expect it to point opposite the direction of i_loop, as charge must be
    gaining potential as it traverses the loop. This is also consistent
    with Kirchoff's Law, or:

    integral_C_closed(E . dl_vec) = 0 around the entire circuit, which can
    only be true if E_loop is against the current in the loop, while E is
    with the current in the resistor.

    But with this interpretation, the math of Faraday's Law has the wrong sign.

    Where am I going wrong?


    Christopher R. Carlen
    Principal Laser/Optical Technologist
    Sandia National Laboratories CA USA
    -- NOTE: Remove "BOGUS" from email address to reply.
  2. Genome

    Genome Guest

    By thinking too hard about things......

    There are two types of electrons.

    Those who do what the others are doing and those who do what they are being
    told to do by others.

    Those who do what the others are doing live on capacitors. Those who are
    told what to do by others live in inductors.

    Now you know what's going on, perhaps you can see where the negative sign
    comes from.

    I think I'll copyright that one.

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