We can generalize the concept of motional emf for a conductor with any shape, moving in any magnetic field, uniform or not (assuming that the magnetic field at S each point does not vary with time). For an element d l of the conductor, the contri- bution dE to the emf is the magnitude dl multiplied by the component of v X B (the magnetic force per unit charge) parallel to d l ; that is, d E = (v X B)dlcostheta For any closed conducting loop, the total emf is the integral.

"This expression looks very different from our original statement of Faraday’s law which stated that emf = -dmagnetic flux/dt. In fact, though, the two statements are equivalent’'

What I don’t understand is that according to faradays law if there is no change in magnetic flux in the closed loop then the induced current is 0. However suppose that we have a closed conducting loop moving through a uniform magnetic field. Faraday’s Law tell us that the induced emf is 0 since there is no change in flux. However according to the first statement there should be a motional emf created as the loop moves through a uniform magnetic field. This is contradictory anyone knows what went wrong ?


1 Answer 1


If the field is uniform and the velocity is constant then we can pull $\mathbf v\times\mathbf B$ out of the integral.

$$\mathcal E=\oint_C(\mathbf v\times\mathbf B)\,\cdot\text d\mathbf l=(\mathbf v\times\mathbf B)\cdot\oint_C\text d\mathbf l$$

Since $\oint_C\text d\mathbf l=0$, $\mathcal E=0$, so there is no contradiction.

  • $\begingroup$ Doesn’t seem to work, if the loop is moving to the right and dl is upwards [] (suppose this square is the loop) then v X dl is vdl out of the page. Could it be instead that dl in the right part of the loop is + while in the left its negative so they cancel ? For your answer v x dl is almost never 0 or am I doing something wrong ? $\endgroup$
    – qubitz
    Commented Jun 10, 2021 at 0:54
  • $\begingroup$ @qubitz Sorry there is an easier way if v is also constant. Let me edit $\endgroup$ Commented Jun 10, 2021 at 0:57
  • $\begingroup$ @qubitz See my edit $\endgroup$ Commented Jun 10, 2021 at 1:04
  • 1
    $\begingroup$ Imagine a square conducitngi loop moving parallel to one of its sides. There is no emf generated by the sides. The front and back conductors both push charges to the same side, with the same emf, so in terms of the loop they c ancel and emf around the loop is zero. Similar argument, but using integrals, if the loop is not square. $\endgroup$
    – Peter
    Commented Jun 10, 2021 at 1:05

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