Suppose you have a conducting circular wire loop and a magnet moving towards each other. They move along the $z$ direction with nonrelativistic constant speed $v$. Let the $B$ field of the magnet in its reference frame be parallel to the $z$ direction: $$\vec{B} = (0,0,-z),$$ so the strength of $\vec{B}$ linearly decreases along $z$. Let the surface normal of the area enclosed by the wire loop be pointing in the $z$ direction too. Now here is my problem: If I'd like to calculate the fields and forces acting upon charges in the wire, I get different results.

In the magnet's reference frame, there is a static magnetic field $\vec{B} = (0,0,-z)$, and the wire loop is moving with a constant velocity $\vec{v} = (0,0,1)$ towards the magnet. In this case there is no $E$ field, so the resulting force on charges is $\vec{F} = q\vec{v}\times\vec{B}$. Since $\vec{v}$ and $\vec{B}$ are parallel, the cross product is 0, so there is no force.

If I look at the wire's reference frame, there is a time-varying magnetic field producing an electric field. Here's where I don't quite get it. The electric field will be $$\vec{E}' = \vec{v}\times\vec{B}$$ according to Wikipedia (gamma is aproximately 1). In this case $$\begin{align} \vec{E}' &= 0 \\ \vec{v} &= 0 \\ \vec{F} &= q(\vec{E} + \vec{v} \times \vec{B}) = 0 \end{align}$$ just like in the other reference frame.

But if I try to use Faraday's equation instead, the wire loop sees a time-varying magnetic field, with a non-zero surface integral, so there has to be a non-conservative, non-zero $E$ field acting upon the charges.

I've tried this with different magnetic fields and every time my calculations fail miserably. I'm missing something somewhere, but I don't know why.

  • $\begingroup$ BTW, good presentation for a homework problem. $\endgroup$ Nov 1, 2015 at 11:12
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    $\begingroup$ thank you, @Daniel, I've edited the tags, so it now contains "homework-and-exercises" :) I'm new here, so I didn't know this tag exists. $\endgroup$
    – David
    Nov 1, 2015 at 11:58
  • $\begingroup$ The magnetic flux reduces the further you go from the magnet, and vice versa. So in either case, there will be an induced electric field acting on the loop, causing a current, which then causes the loop to experience a net Lorentz force. That's what's important, the flux change at the loop due to the relative motion of the magnet and the loop, which is equivalent in either reference frame. $\endgroup$ Nov 1, 2015 at 12:13
  • $\begingroup$ I.E., you can only use Faraday's Law, as a linked flux change is involved, which you have ignored and taken into account only pure Lorentz forces. $\endgroup$ Nov 1, 2015 at 12:21
  • $\begingroup$ Also, FYI, magnetic flux intensity drops off as $1/r^3$ along the axis of the magnetic dipole. $\endgroup$ Nov 1, 2015 at 12:23

1 Answer 1


You're missing the fact that your situation as you've defined it is physically impossible. Remember that one of Maxwell's equations is $\vec{\nabla}\cdot\vec{B} = 0$, but your magnetic field configuration $\vec{B} \propto (0, 0, -z)$ doesn't satisfy this. Given that, it's not surprising that Maxwell's other equations give inconsistent results.

Incidentally, $(0, 0, -z)$ is not the magnetic field of a point magnet. You would need some extended, continuous distribution of magnetic charge to get that field. If you did assume that magnetic monopoles exist and you can have them distributed in a way that produces this field configuration, you'd have a "fluid" of magnetic monopoles in motion in the wire's reference frame, which would account for the electric field.

  • $\begingroup$ Thank you very much!! I tried to simplify this problem so much, that I forgot to check wether this is even possible or not. I will try to formulate this problem in a real magnetic field and do the calculations. I hope that I don't get contradicting solutions. $\endgroup$
    – David
    Nov 1, 2015 at 13:38
  • $\begingroup$ You won't, if you do it right. I've seen a lot of people make this mistake (including myself), and it always works out cleanly once you make sure all the proper constraints are satisfied. $\endgroup$
    – David Z
    Nov 1, 2015 at 13:49
  • $\begingroup$ It is really good to hear, I am not the only one making these kind of mistakes. I will be more careful in the future and take into account all the Maxwell equations when trying to create a problem. $\endgroup$
    – David
    Nov 1, 2015 at 13:54

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