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Equation for Induction of Electric Current in Magnetic Field

  1. Jan 16, 2013 #1
    I am looking for something that should be relatively simple to find, but as such, I have not been successful. I know that an electric current can be created by moving through a magnetic field. I have found equations that indicate this can be based on the velocity and the length of said object moving through the field. However, I have not been able to identify a single equation to suit my needs.

    Basically - I am looking for an equation to calculate the electric current (and voltage? in order to calculate the power...) of a loop or coil of conducting material (wire) moving through a magnetic field (say, the Earth's?).

    Any help will be much appreciated.

    Thank you.
  2. jcsd
  3. Jan 16, 2013 #2

    Philip Wood

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    Depends on whether or not you are looking for understanding, or just for a formula to use for a particular case.

    For the latter: the e.m.f. ('voltage'), ε, generated in a straight wire of length L moving at speed v at right angles to its own length and to a magnetic field of flux density B is ε = BLv.
    If the wire is part of a circuit of resistance R, the current, I, will be I = BLv / R. But the rest of the circuit must stay stationary while the wire of length L moves. For example, the wire may slide along stationary rails, making contact with them at each end of the wire. The rails connect to the rest of the circuit. If the whole of the circuit, including the wire, moves all together, there will be no emf and no current.

    To understand why there is no emf in a circuit which moves without turning through a uniform field requires a proper understanding of electromagnetic induction. You need, in particular, to understand what is meant by flux linkage and cutting of flux.
    Last edited: Jan 16, 2013
  4. Jan 16, 2013 #3
    Alright, that makes sense. Is there any way to generate a continuous EMF by moving a loop/coil (for greater length, L) through the Earth's magnetic field, B, at some velocity, v?
  5. Jan 16, 2013 #4


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    It is possible, but completely unfeasible. The Earths magnetic field is very weak and on our everyday scale it is very uniform.
  6. Jan 16, 2013 #5

    Philip Wood

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    Amplifying Drakkith's response, translating a whole circuit (that is moving it without turning it) through a uniform field will produce zero emf and zero current.

    The Earth's field, over a large scale, is not uniform, and if you moved your circuit fast enough over a large enough distance for the non-uniformity to show up (i.e. a distance comparable with the Earth's radius), then there would be an emf, but this is almost certainly impracticable.

    What IS feasible is to rotate a coil about its diameter in the Earth's field, making connections to an external circuit by means of slip rings. This is, of course, none other than an a.c. generator (with a very weak magnetic field!).
  7. Jan 16, 2013 #6
  8. Jan 16, 2013 #7


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    nice explanation, Philip. I hadn't really thought about this before. So the idea is that the loop is rigid, and is not being rotated, so the velocity at all parts of the loop is the same, and the magnetic field is uniform. And the equation for the emf due to the magnetic field is:
    [tex]\oint (\vec{v} \wedge \vec{B}) \ \cdot d \vec{L} [/tex]
    But since the integrand is a constant vector, then we can simply write it as the gradient of some scalar field, for example if:
    [tex]\vec{v} \wedge \vec{B} = q \hat{z} [/tex]
    (where q is just a number, not charge or anything) then we can write it as the gradient of the scalar field [itex]\phi=qz[/itex] And since we can write it in this form, the emf is:
    [tex]\oint \nabla (\phi) \ \cdot d \vec{L} [/tex]
    But luckily for us, this is the definition of:
    [tex]\oint d \phi [/tex]
    And this must equal zero because it is around a closed loop. Therefore the emf in this situation must equal zero. (Sorry if this is using too much vector calculus). I am interested in the explanation using cutting of flux and flux linkage. Because I never really got comfortable with those concepts.
  9. Jan 16, 2013 #8

    Philip Wood

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    I love your vector calculus derivations. You certainly know how to wield these tools!

    In my earlier posts I used the principle that the emf is equal to the rate of change of flux linking the loop. [The nice thing about this it that it works for both the cases of a circuit moving in a magnetic field (when the charge carriers in the circuit experience a magnetic force q(vxB) and for a stationary circuit sitting in a changing B, when the charge carriers experience an electric force qE.] In the case of a circuit translating in a uniform field, the flux linked with the circuit never changes (if you like, new lines of flux replace old lines), so the emf is zero.

    It is, of course possible to show the equivalence of the emf line integrals in both cases (moving circuit and stationary circuit+changing B) to d[itex]\Phi[/itex]/dt. I used to be able to do it without looking it up!
  10. Jan 16, 2013 #9
    The induced current & voltage will always be related per Ohm's law, i.e. emf = I*R. But the induced voltage "emf" is related to the **net flux** per emf = B*l*v, where B is the net flux density.

    This net flux density is the external field density Bext, combined with any flux density generated by the induced current I, which is Bint (internal). Lenz' law states that Bint has a polarity opposite to Bext.

    If the closed path resistance R is high, greater than XL for the loop, then the induced emf is Bext*l*v, & the Bint (internal flux density) can be ignored as it is way smaller than Bext. A small induced current generates a small B field which is negligible compared with Bext.

    But as R decreases, the internal B field starts increasing, cancelling the external B. The net result is that the induced emf is less than in the open loop case. But emf/I = R in all cases. I just wanted to convey that there is no inherent "voltage regulation" here. As the loop resistance R decreases, so does emf.

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