# How can a non-conservative field be a scalar multiple of a conservative field?

Okay so I was reading this from University Physics by Freeman and Young and on the topic of inductors as circuit element, they wrote that $\mathbf{E_c} + \mathbf{E_n} = 0$ which makes no sense to me

Here is an excerpt of the text and underlined the confusion with green

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The relationship $\vec E_c + \vec E_n = 0$ does not hold globally so the two fields are not scalar multiples.

Within, and only within, the (ideal) conductor that forms the inductor, the electric field must be zero.

Note that the text specifically says

so the total electric field ... within the coils must be zero.

UPDATE:

Consider the general $E$ field in terms of the scalar and vector potentials:

$\vec E = -\nabla V - \dfrac{\partial\vec A}{\partial t}$

The first term, the gradient of the scalar potential, is conservative (curl of the gradient is identically zero) so any non-conservative component must come from the second term so let's identify:

$\vec E_c = -\nabla V$

$\vec E_n = - \dfrac{\partial\vec A}{\partial t}$

Now, if these fields are scalar multiples, $\vec E_n$ must be conservative which implies that $\vec B = 0$.

But, in your problem, $\vec B$ is non-zero and time-varying so $\vec E_n$ non-zero and is not conservative and so the two fields are not scalar multiples.

However, it is clearly possible to impose the constraint $\vec E_c + \vec E_n = 0$ somewhere but not everywhere.

Consider the following two electric fields:

$\vec E_c = K \hat x$

$\vec E_n = -y \hat x$

Clearly, $\vec E_c$ is conservative and $\vec E_n$ is non-conservative.

However, it is also clear that $\vec E_c + \vec E_n = 0$ when $y = K$

And, there you have it, a simple example of a conservative field being cancelled somewhere but not everywhere by a non-conservative field.

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No, but even inside the inductor, how could this happen? I am confused as why that relationship should hold anywhere –  jip Aug 28 '12 at 23:44
@jak, how could what happen? What does "this" refer to? That a conservative field and a non-conservative field can exactly cancel within a limited region? –  Alfred Centauri Aug 28 '12 at 23:54
$\vec{E_c} + \vec{E_n} = 0 \implies \vec{E_c} = -\vec{E_n}$. How can multiplying a conservative field by -1 change the non-conservative behavior? –  jip Aug 29 '12 at 0:02
@jak, you're not, not multiplying a conservative field by -1 because the $\vec E_c$ and $\vec E_n$ are not proportional everywhere. They are proportional only within the conductor. $\vec E_c$ and $\vec E_n$ are not proportional outside the conductor so you are not multiplying the field by $-1$. –  Alfred Centauri Aug 29 '12 at 0:09
"They are proportional only within the conductor", so inside the conductor one is the multiple of another? –  jip Aug 29 '12 at 0:19

A field is conservative in an unbounded domain if its divergence is nul. In a bounded domain it is no longer the case, there is a surface integral which is non zero even if the divergence is nul.

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