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Confused, velocity in motional EMF?

  1. Oct 4, 2014 #1
    Motional EMF is defined as E = -vBL
    Where B is the magnetic field in Tesla, L is the length in meters and ,v is the velocity.

    Lets say we have a distance(1m) where half of it(0.5m)the magnetic field is present. So, there is a wire that moves with constant acceleration throughout that whole distance. Is velocity equal to dx/dt? Where dx is the distance the field covers...? And dt is the time it takes for the wire to pass the field only correct? Does it make sense to say it's dx/dt with respect to the magnetic field? Change in the distance(in the magnetic field) and change in time(in the magnetic field) anything beyond the field does not matter or does not account to the velocity?

    Okay, what about using this formula: v = at , to calculate the velocity with respect to the time it takes for the wire to pass the field only not the whole path.

    Btw, when a wire moves in a constant uniform(or semi-uniform)magnetic field is that considered "change" in flux?
    Because -vBL = - dphi/dt?

    Edit: If the wire moved 0.5 meters in 100 milliseconds(before entering the field) the velocity is 0.5/0.100 = 5m/s... And then enters the field(0.5m wide) in another 100 milliseconds so the the velocity in motional emf is 0.5m/0.100 seconds or 1m/0.200 seconds?
    Last edited: Oct 4, 2014
  2. jcsd
  3. Oct 4, 2014 #2


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    I do not understand your sign convention nor how you deal with the vectors. The correct form of Faraday's Law in integral form for moving media reads
    [tex]\int_{\partial A} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{t}} \int_{A} \mathrm{d} \vec{f} \cdot \vec{B}.[/tex]
    Here [itex]\vec{v}(t,\vec{x})[/itex] is the velocity field of the medium, [itex]A[/itex] is the surface and [itex]\partial A[/itex] it's boundary with the usual orientation convention that the orientation of the boundary curve is related to the surface normal vectors according to the right-hand rule.
  4. Oct 4, 2014 #3

    I've not yet reached to that level, I'm baffled at your response and complexity... is there a simpler answer?
    I believe an example is required to help my understanding, and possibly clarify my point. Assume a magnetic field that covers an area of ( 0.5m x 0.5m) with magnetic field strength of 0.75Tesla. A conductor that is equal to that area is moved at constant acceleration of 10 m/s^2.

    I calculated the time s = 1/2 at^2 = 0.31 Seconds, so the velocity should be 0.5m/0.31 seconds = 1.58 m/s x 0.75T x 0.5m = -0.59Volts?
    Now let's change things a bit, instead of having a conductor equal to the area of the field, lets make it a bit bigger, 1m x 0.5m(L), still the velocity is equal to 0.5m/t ? t this time would depend on the larger conductor's acceleration being higher or lower than before.
  5. Oct 4, 2014 #4
    Is motional EMF the same as induced EMF where there is change in flux in both?
  6. Oct 4, 2014 #5
    One way for me to calculate the velocity could be using the acceleration and calculate the time.
    By using this formula v = at, it would give me the velocity and I can used it in the calculation of E = -vBL.
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