# Apparent paradox in no costants fields

1. Jun 1, 2015

### andrea96

if there is an uniform, infinitly extended,magnetic field that is changing with a law B(t), I can calculte the induced electric field using an arbitrary circle; then the induced circuitation on the circle is $$\epsilon _i=-\pi r^2 \frac{dB(t)}{dt}$$, and since the simmetry of the system ( infact each point on the circle isn't different from an other ) the electric field on the circle is $$\frac{r}{2} \frac{dB(t)}{dt}$$. But here there is a big problem: a different choise of r should not change the result ( again from the simmetry of the system ), but I have obteined an r-dependent result! I have a little explaination of this paradox, but I'd really like to know your answers...
Thanks!

2. Jun 1, 2015

### Chandra Prayaga

It is not clear why you think the result should be independent of r. The flux does depend on the area of the circle, which depends on r.

3. Jun 1, 2015

### andrea96

Excuse me I haven't explained very well: the circuitation certenly depend on r, because it is a "propety" of the circle... but the electric field cannot depend on r because it cannot depend on the choise of a particular circle instead of an other! In fact electric field is a "property" of the space and it should depend only on the point, but in this case, in each point I can build a lot of circle with differents r obtaining different result of the component of the electric field along the tangential direction.

4. Jun 1, 2015

### rumborak

Each point of the circle experiences the change in B, and thus will contribute to the overall emf. The more circumference of the circle, the more emf. The formula says exactly what one would expect.

5. Jun 1, 2015

### willem2

But what is the direction on the electric field? any point lies on many different circles with tangents in all possible direction. Symmetry seems to require that the electric field is zero everywhere. The only way I can see out of this, is that uniform, infinitely extended fields aren't possible. They would have to come from infinitely large current loops.

6. Jun 1, 2015

### Staff: Mentor

The electric field does not depend on r, but the induced EMF does. Those are two different things. The EMF is the integral of E around the circle, which scales as r, just like you found.

7. Jun 1, 2015

### rumborak

I see what willem and andrea are saying though, and it is a very interesting question. Why/how how does the wire segment know what direction the resulting electric field should point to? The wire segment is unaware of the rest of the circle after all.
My best guess is that it's a "net force" type of effect, where maybe neighboring line segments influence it.

I feel one should be able to construct Faraday's Law from the ground up through the Lorentz force. But that one already is only dependent on B, not the change of B, so i don't see how to construct the law. Apparently this is something that Einstein pondered and got him towards Special Relativity, so we're not pondering something stupid here.

EDIT: I think willem's suggestion of "there can not be a completely uniform field" might be the answer though. Magnetic fields are circular, so the field lines can not possibly be exactly parallel. When the field changes in time, different parts of the wire segment will experience different fields. The electrons in the wire will thus experience different Lorentz forces, which then likely results in the net physical force (and resulting electrical force) along the wire segment.

Last edited: Jun 1, 2015
8. Jun 1, 2015

### stedwards

I'm going to back up a little. $\oint E \cdot l = -\frac{d}{dt} \int\int B \cdot dA$ .

For your circular path, $2\pi r E = -\frac{d}{dt}2\pi r^2 B$ .

You have a quantity you called circulation. This the the induced voltage around the loop. The EMF
I'll use the symbol V. $V=2 \pi r E$
or
$V=\frac{d}{dt}2\pi r^{2} B$

$V=\frac{d\Phi}{dt}$
$\Phi$ is the magnetic flux. It's just B added up over an area.
The voltage increases with $r^2$ because the quantity of Magnetic flux in the loop also increases with r.

If the magnetic flux were independent of r, and all located at near r=0, the induced voltage would be independent of the path. This may be the invariant you were thinking of.

This why the voltage out of a power transformer depends on the number of secondary turns, but not their shape and size to any great degree, because most of the magnetic flux is in the iron core.

Does this help? I could be addressing the wrong issue.

Last edited: Jun 1, 2015
9. Jun 1, 2015

### andresB

I have the same feeling.

Last edited: Jun 2, 2015
10. Jun 2, 2015

### willem2

There's more than one solution to maxwells equations:∇.E = 0 and ∇×E = -dB/dt

$$E = (r - r_0) × \frac {\delta B}{\delta t}$$ is solution for every r_0,

You can also combine two or more solutions, if E1 and E2 are solutions, so is $c_1 E_1 + c_2 E_2$ as long as $c_1+c_2 = 1$

11. Jun 2, 2015

### andrea96

Perfect! This is the paradox: I can obtain infinity different results for each direction of the electric field. You have said that the explaination of this paradox is the impossibility to creat that magnetic field: this could be an interesting answer, but I'd like to inivite you to think that it's impossible to build also an infinity plane or linear distribution of charge, but in these case the electric field is well definite $$E=\frac{\sigma}{2\epsilon}$$ or $$E=\frac{\lambda}{2\pi \epsilon r}$$. I'd like to show you also that a similar paradox came out if I want calculate the electric field in an infinity and uniform 3-dimentional distribution of charge: the field should be 0 since the simmetry, but calculating it using an arbitray Gaussian surface, I can obtain any value of the electric field chooising any Gaussian surface. And you can see as other two similar paradoxes exist: infinity 3-dimentional current distribution and infinity uniforno-costant electric field. There is a common reason in all these four paradoxes. Instead, this "problem" there isn't in the cases of infinity uniform plan and linear distribution of charge...

12. Jun 2, 2015

### Staff: Mentor

I don't know how your confusing EMF around a loop with the E field at a point constitutes a paradox.

13. Jun 2, 2015

### andrea96

Maybe I have explained vary bad: I used the EMF to calculate the E field, the result about EMF isn't a paradox (it obviously depend on r), but the result of E filed constitutes a paradox. I try to explain better: in the space there is a uniform infinitely extended B field that is changing in time. This variation of magnetic field due an E field. To calculate the electric field in a certain point I calculate the electric circuitation on a circle (the circle must pass on the point in wich I want to calculate the E filed) and exploiting the simmetry of the system ( from the infinity extention of the B field, the inducted E field has to be the same in any point on the circle ) I can calculate the E field : $$E=\frac {EMF}{2\pi r}=\frac {-\pi r^2\frac {dB (t)}{dt}}{2\pi r}=-\frac {r}{2} \frac {dB (t)}{dt}$$.
From the result is evident that the E field depend on the particular choise of the circle. Now choosing an other circle passing in the point where I'm calculating the electric field with an other radius r', I obtain a different result for the E field : $$E=-\frac {r`}{2} \frac {dB (t)}{dt}$$. And this is a paradox: I have obtained two different values for the E field in a certain point, but the field should be univocally determinated...
I hope that my leatest explaination was better...

Last edited: Jun 2, 2015
14. Jun 2, 2015

### rumborak

If I understand his andrea's point correctly, and it's what I was getting to also, is how can the local field be dependent on the rest of the circle?
And yes, I know of course that the Maxwell/Farady equation say exactly that. But it still begs the question how this "holistic" result can arise when in the end, it's just singular electrons in a wire experiencing their local magnetic field.

Last edited: Jun 2, 2015
15. Jun 2, 2015

### vanhees71

I've not read the entire thread. Of course, such a field doesn't exist in nature, because it would have infinite energy, but it can of course exist locally approximately. So suppose in some finite region of space we have a homogeneous time-dependent magnetic field $\vec{B}=\vec{B}(t)$. Then according to the Faraday Law we have
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \dot{\vec{B}}(t).$$
That's a constant wrt. to spatial variables. So we look for a solution of this seen as a differential equation. The solution is determined only up to a gradient field
$$\vec{E}=\frac{1}{2c} \vec{x} \times \dot{\vec{B}}(t)-\vec{\nabla} \Phi(t,\vec{x}).$$
So $\vec{E}$ won't be homogeneous. Of course, you need more information to calculate the electric field completely.

16. Jun 2, 2015

### nasu

This (second line, left hand side) will be true only if the electric field will the tangent to that circle in all points of the circle and have the same magnitude at all points.
So you are assuming already some geometry of the field.

More general, the integral around the circle will give you info just about the tangential component of the field.
You cannot find the entire field from that unless the field happens to be tangential to some circle.

17. Jun 2, 2015

### andresB

I don't know what I'm doing wrong but in this case I'm getting that Ampere's law is not compatible with Faraday's law.

rot B= dE/dt, but

rot B =0, since the magnetic field is uniform and only depend on time. so the electric field doesn't depend on time.

But, on the other hand

Rot E= - dB/dt

and the right hand side can depend on time since B(t) is arbitrary, but then rot E will depend on time in contradiction with the previous result.

18. Jun 2, 2015

### Staff: Mentor

This calculation is incorrect. See Vanhees post for the correct E field. You can plug it back in and see that it satisfies Maxwell's equations, whereas yours does not. However, ...

Even if your expression were correct, it would still not constitute a paradox. You have simply incompletely specified the boundary conditions so you have not determined a unique solution to Maxwell's equations. Not only is there a valid solution to Maxwell's equations which satisfies your conditions, there are an infinite number of such solutions. Having multiple solutions does not constitute a paradox.

Last edited: Jun 2, 2015
19. Jun 2, 2015

### andrea96

Thanks for your answer! However my expression for E doesn't satisfy Maxwell equation because, as I said in one of my initials post, it isn't the vectorial electric field, but just his component along the tangential direction. However I understand what you are saying about the unicity of the solution, but if I specify that the magnetic field is generated by a very big coil? What happens?

20. Jun 2, 2015

### Staff: Mentor

I think that is correct then for lops centered on the origin and 0 gradient field. It may be that there is some relationship between the center of your loop and the gradient.

Assuming that you also specify that there are no other sources and some appropriate boundary conditions (absorbing, reflecting, zero at infinity, or whatever) then you could find a unique solution. In the center of the coil it should look like Vanhees' solution with 0 gradient field in coordinates with the origin at the center of the loop.