So, i have one question about how the induced electric fields will be either conservative or not. So, first of all, we got Faraday-Lenz's Law: $$\epsilon = - \frac{d\phi}{dt}$$

This only refers to the flux of the magnetic field, and since the flux is an integral of $\vec{B}·\vec{n}dA$, i can just change the area to get an induced difference of potential.

And now, we got one of Maxwell's equation (that derives from Faraday's Law) : $$ \nabla \times E = - \frac{\partial B}{\partial t} $$ Here is my question: From Maxwell's equation, one can see that you will have a curl different from $0$ if the magnetic field $B$ changes along the time. But, in the first case, one can deduce from Faraday's law that you can induce and electric field in some wire loop just by changing the area, without changing the field. So, my question is basically: Will the electric field induced by a constant field $B$ in a wire loop (just changing the area along time) be conservative, (since $\frac{\partial B}{\partial t} = 0 $, then there will be no curl) or the electric field induced will be one of those exceptions in vector calculus, where the field has a curl that equals to zero, but it is not conservative? Thanks.


1 Answer 1


The second equation is often known as the local version of Faraday's Law. It would be helpful to put in the coordinates.

$$\nabla \times E(x,y,z,t)= -\frac{\partial B(x,y,z,t)}{\partial t} $$ When you have an expanding loop, the coordinates of a point of the loop are not fixed, but instead change over time. To apply the equation, you have a changing coordinate system, and so you have to apply a Lorentz transformation. The math is messy, but you can see the induced field appear when you substitute a Lorentz transformation into the equation.

  • $\begingroup$ But, in a point (x,y,z) won't B remain the same along time? $\frac{\partial B}{\partial t} $ would still be $0$. $\endgroup$ Nov 9, 2016 at 13:54
  • 1
    $\begingroup$ @Dovah-king, for a fixed point, yes. But when you have an expanding loop, the points you care about have a moving coordinate system. $\endgroup$
    – Paul
    Nov 9, 2016 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.