Does electric field caused by time varying magnetic field form closed loops(electric field starts from a positive charge and ends at a negative charge)? and are they conservative or non-conservative fields?

  • 1
    $\begingroup$ Are you asking if this is always the case, or for an example where this is the case? $\endgroup$
    – DJBunk
    Feb 20, 2013 at 20:30
  • $\begingroup$ May be related to my answer to physics.stackexchange.com/q/5573/520 (or rather a comment under it). Short answer is "yes", but if you didn't believe me the first time there doesn't seem to be much point in repeating myself. $\endgroup$ Feb 20, 2013 at 21:45

1 Answer 1


The field lines certainly can form closed loops, for example in a transformer. But they don't have to either, an example being a plane EM wave.

For a specific example, consider the following long solenoid of radius $a$ and turn density $n$ with a time-varying current running through it: $I(t)$.

enter image description here

The solenoid is oriented along the x-axis, and current runs in the direction of $-x$ to $+x$. This current induces a magnetic field inside the solenoid:

$$\mathbf{B}=\mu_0 nI(t) \mathbf{i}$$

Now, by Faraday's Law:

$$\int \mathbf{E} \cdot \mathrm{d} \mathbf{s}=-\frac{\partial}{\partial t} \int \mathbf{B} \cdot \mathrm{d} \mathbf{A}$$

Which gives us a relationship the time derivative of magnetic flux through a loop and the electric field along that loop. Using a loop that is concentric with the solenoid gives:

$$2\pi r E=-\frac{\partial}{\partial t} \int \mu_0 n I(t) \mathrm{d}A=-\mu_0 n I'(t) \pi a^2$$

$$\Rightarrow E=-\frac{1}{2}\mu_0 n a^2~ \frac{I'(t)}{r}$$

where $I'(t)=\partial I / \partial t$, and we know $E$ is oriented along a closed loop.

  • $\begingroup$ In high school physics, it's given that electric field starts from a positive charge and ends at a negative charge, does the above mentioned field follow the same? $\endgroup$
    – vpp
    Feb 21, 2013 at 14:36
  • $\begingroup$ No. The electric field around the solenoid looks similar to the magnetic field around straight current. I.e. it looks similar to the field in: school-for-champions.com/science/images/… . $\endgroup$
    – Jold
    Feb 21, 2013 at 15:48
  • $\begingroup$ What is the difference between this field and the one caused by static charges?? $\endgroup$
    – vpp
    Feb 21, 2013 at 15:54
  • $\begingroup$ The way E fields are generated is given to us by Maxwell's equations, and the equation for finding the E field produced by a charge and the equation for finding the E field produced by a changing B field are different. $\endgroup$
    – Jold
    Feb 21, 2013 at 16:02
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    $\begingroup$ The electric field produced by a changing magnetic field is non-conservative. A conservative field is a field which can be written as the gradient of a potential function. In the case of the electric field produced by static charges, it can be written as $\mathbf{E}=-\nabla\phi$ where $\phi$ is the electric potential. When changing magnetic fields are introduced, it becomes $\mathbf{E}=-\nabla\phi-\partial \mathbf{A}/\partial t$ where $\mathbf{A}$ is the magnetic vector potential. Thus, it is no longer conservative. $\endgroup$
    – Jold
    Feb 22, 2013 at 5:24

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