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Faraday's law is given by:

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

On the right hand side of the equation, we have a quantity representing how $B$ changes over time. On the left hand side of the equation we have a quantity representing the spatial variation of $E$ (the curl of $E$).

The most common interpretation is that, we can determine the spatial variation of the induced electric field, if we know how $B$ changes over time. That is, the right hand side is producing the results obtained from left hand side of the equation.

Suppose instead however, I set up an electric field such that $\nabla \times E$ is non 0, will this induce a time-varying magnetic field given by the above equation?

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up vote 1 down vote accepted

I think your last question stated in the following way might clear up your confusion. If I setup an electric field $\mathbf E(t, \mathbf x)$ such that $\nabla\times\mathbf E(t_0,\mathbf x_0)\neq 0$ at some spacetime point $(t_0, \mathbf x_0)$, then Faraday's law tells me that whatever $\mathbf B$-field there is around, it must necessarily satisfy $$ \frac{\partial\mathbf B}{\partial t}(t_0, \mathbf x_0)\neq 0 $$ This does not mean that $\mathbf B(t_0, \mathbf x_0)\neq 0$, the magnetic field could vanish for the instant $t_0$, but since its time derivative must me nonzero, it means that in the next instant the magnetic field must, in fact, be nonzero. So you can be guaranteed that if you have managed to setup an electric field with non-vanishing curl at some instant, then quite soon there will have to be a nonzero magnetic field around.

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Thanks Josh. I think you meant to say "if you have managed to setup an electric field with NON- vanishing curl" in your last paragraph. – Mew Feb 1 '13 at 8:10
Yeah I did sorry! (I should really re-read answers after posting them :)) – joshphysics Feb 1 '13 at 8:13

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