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Initially, a point($P$) is defined in space, with a magnetic field($-B\hat{k}$) produced by a current element ($Idl$). If the wire begins to move in the $-v\hat{i}$ direction, would that produce an electric field around $P$ since the magnetic field is changing?

$$ \nabla \times E = -\frac{\delta B}{\delta t}$$

I ask this because, I've been pondering with Faraday's paradox, and remembered this experiment:

enter image description here

One of the steps states the following result:

The disc is held stationary while the magnet is spun on its axis. The result is that the galvanometer registers no current.

The reason(which was very insightful) for that is; rotating the magnet(and the dipoles within) does not cause the magnetic field to move.

Which made me consider the case of a current carrying elements. I know that system is not applying the Lorentz force(specifically the the magnetic force ($v\times B$)).

In addition, if the wire(or $Idl$) moves in any other direction would it still yield an non-conservative $E_{ns}$?

enter image description here

How is the current element and it's relation to the magnetic field produce by it, dynamically different than the rotating magnet with a stationary magnetic field produced by the magnet?


1 Answer 1


Firstly for a single current element the act of moving the path is just to transform the path (stretch y to the left). At P the magnetic field is into the page. The direction does not change unless the extrapolated path crosses P. Yes, there is an electric field set up due to the change in magnitude. It however does not make a closed loop as there are a continuum of points P. The curl is expreesed by an electric field gradient; being stronger closer to the path where the change in magnetic field is stronger.

  • $\begingroup$ Could be stretched or a rigid body that's moves fully, either way I agree with the induction of E due to the change of the magnetic field. However, could you clarify on the following: 1) Extrapolated path w. r. t the direction of B 2)The relevance of a closed loop and the continum of points P. Thanks $\endgroup$
    – Geodesic
    Commented Mar 28, 2019 at 10:23
  • $\begingroup$ In the diagram the current is on a line segment. As charges don't dissapear it either keeps on going somewhere (hence extrapolated) or builds up charge. $\endgroup$ Commented Mar 28, 2019 at 20:39
  • $\begingroup$ P can be chosen arbitrarily. Continuum because space is a continuum. $\endgroup$ Commented Mar 28, 2019 at 20:41
  • 1
    $\begingroup$ The closed loop is how you have drawn E. Curl is equivalent to integrating the dot product of a vector field with a loop around the point, but the field doesn't have to follow a circle. In this case the E field is roughly parallel with the stretched line segment, just stronger closer in. This gradient will make the integral on the near side of the circle bigger than the far side, so instead of cancel out it gives a non zero curl matching the change in time of the B field. $\endgroup$ Commented Mar 28, 2019 at 20:47

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