What are the conditions for magnetic field and electric fields to be closed?

Question

Assume we have a statement:

The electric field $E$ cannot form closed loops

I am trying to prove/disprove this claim.

The conditions for a closed loop is:

$$\oint_CE\cdot dl=0$$

Using stokes theorem:

$$\iint_S(\nabla\times E)\cdot\hat n\;dS=\oint_CE\cdot dl\quad\rightarrow\quad\nabla\times E=0$$

So my solution is that the electric field forms closed loops when we have a conservative field.

However in the solution it says that:

The electric field $E$ cannot form closed loops

• False
• If $\nabla\cdot E=0$ everywhere and $\nabla\times E=-\dfrac{\partial B}{\partial t}\ne0$ the field lines will be closed loop (i.e. this is non-conservative field)

Its says electric field forms closed loops when we have a non-conservative field.

Can anyone point out where I went wrong?

My question is why must there be a non-conservative field for there to be closed loop.

Answer 1

The condition $$\nabla \times \mathbf E=0$$ is true only if the electric field does not change with time, as in electrostatic situations.

In the case where the electric field does change with time, we have the condition $$\nabla\times\mathbf E=-\dfrac{\partial\mathbf B}{\partial t}$$ In this case the dynamic nature of the electric field makes it non-conservative. This also means that the path integral $$\oint_C\mathbf E\cdot d\mathbf l\ne 0$$ but instead equals emf. We can show this by using Stoke's theorem so that $$\oint_C\mathbf E\cdot d\mathbf l =\iint_S (\nabla\times\mathbf E) \cdot d\mathbf S=-\frac{\partial}{\partial t}\iint_S\mathbf B\cdot d\mathbf S=-\frac{d\Phi}{dt}=V_i $$ where $\Phi$ is magnetic flux and $V$ is induced emf.

So my solution is that the electric field forms closed loops when we have a conservative field.

If electric field lines formed closed loops, then this would mean that these lines would originate and then terminate at the same point which is not what we see. We do see this for magnetic fields, and the condition $$\nabla \cdot \mathbf B=0$$ is always satisfied for magnetic fields, but not always for electric fields. What we find is that electric field lines are always directed outward from a positive charge, and converge inward to a negative charge.

The condition $$\nabla \cdot\mathbf E=0$$ is true except where there are sources. That is, for a field without a source, then the divergence of that field vanishes. When there is a source, then we have the condition $$\nabla \cdot\mathbf E=\frac{\rho}{\epsilon_0}$$ where $\rho$ is the charge density.

Answer 2

The condition $$\oint_CE\cdot dl=0$$ is not the condition for a closed loop. The reason is that the way you use that condition implies that $C$ is an arbitrarily chosen closed loop. If it was coinciding with a field line, you would be using the conclusion (all the field lines are closed) to prove it. In general, to make contact with $\nabla\times E$ one has to use arbitrary closed curves $C$, even not corresponding to field lines.

A counteraxample may help to understand the above point. Let's consider the 2D vector field $$ {\bf F}= -y {\bf \hat x} +x {\bf \hat y}. $$ Its line fields are all the circles of arbitrary radius $R$ with center at the origin ($x^2+y^2=R^2$). However, $\nabla\times {\bf F}=2 \neq0$ and $$\oint_CE\cdot dl=2 \pi R \neq0.$$

Instead, the correct condition for closed loops is the condition that there are no sinks or sources. I.e. $$ \oint_{\Sigma}{\bf F}\cdot d{\bf S}=0$$ for all the closed surfaces $\Sigma$. Corresponding to the condition of a divergenless (solenoidal) field $$ \nabla \cdot{\bf F}=0 $$ and to the absence of any charged surface. Under such conditions, if the field would be conservative, it would vanish everywhere. Therefore, the only possibility to have a non-zero field is a solenoidal, non-conservative field with at least one closed path such that $$\oint_CE\cdot dl\neq0$$

Answer 3

Sorry for my poor english. My native language is french.

Usually, we define a field line avoiding singular points where the field is zero. (Indeed, this last case poses particular problems: several lines of field can arrive there)

If the field is not zero along the path, it cannot change direction along the field line and therefore the circulation on a field line cannot be zero.

If a field line forms a closed path, we would therefore have the integral of the field on a closed path which would be non-zero: the field cannot therefore be conservative.

Answer 4

The equation you have in the question is Faraday's Law for a non-varying magnetic field. However if you have varying field then that can produce a non-conservative field. This is the general equation (Faraday's Law) with the right-hand side equal to zero for electro or magnetostatics.

$$\int \vec E \cdot \vec {dl}=-\frac{\partial \vec B}{\partial t} $$

Edit after the question was rephrased: $\\$

If we have electrostatics and magnetostatics meaning no time varying fields then the electric field cannot form closed loop since that would mean that the field lines start and end on the same charge and we know that cannot be the case. But for a non-conservative field one that is produce by having a time varying magnetic field which can generate an electromotive force EMF the fields can start and end on the same point.

Answer 5

There’s definitely some confusion here - if a field line forms a closed loop, then by definition the electric field is tangential to the loop at every point of the loop, which means that $\oint \mathbf E\cdot d\mathbf r\neq 0$.

In a conservative field, closed loop integrals of that type always vanish; as a result, if any field lines form closed loops, then the field must be non-conservative. The converse is not necessarily true, and I would imagine that finding the precise conditions under which field lines close on themselves would be quite difficult.