The existance of Electric field

AI Thread Summary
The discussion centers on the existence of induced electric fields in relation to changing magnetic fields, specifically using Faraday's Law. It is established that when the magnetic field (B) changes, an induced electric field (E) should exist, but the calculations presented are questioned for their validity. The key point is that the induced electric field at a specific point (C) depends on the contributions from the entire magnetic flux in the surrounding area, not just the immediate vicinity. If the magnetic field is uniform and confined to a circular region, the induced electric field at the center may be zero, while a non-zero contribution can arise from other areas. The complexity of the problem arises when considering the extent of the magnetic field and how it influences the electric field distribution.
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Homework Statement


inductance-E_zps4bccb489.png

B is increasing with rate: dB/dt
when C is center of circle 1, Induced Electric Field is 0, but when i take another circle, in C, existing Induced Electric Field

Homework Equations


FaradaysLaww_zpsef8d3c97.png

I used this equation to prove

The Attempt at a Solution


Do Induced Electric Fields exist in C??
 
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Please show your working.

When the B field changes, there should be an accompanying electric field so that $$\nabla\times\vec E = -\frac{\partial \vec B}{\partial t}$$

If the loops you have drawn are conducting, then either of them should have an induced current.
 
what i mean is: in C, Existing of Induced Electric Fields
I have: E = (r/2).dB/dt( r is radius of circle 1)=> C doesn't have electric field
But when E' = (r'/2).dB/dt(r' is radius of circle 2) => C still have electric field( C belongs to circle2)
I think when B change, every where in space also have electric field, am i right
 
I think when B change, every where in space also have electric field, am i right
It is not always the case. The trouble is that your calculations make no sense.

You are trying to use Faradays Law:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html#c3

I do not see how you have done your calculations.
I do not see how you got your conclusions.

i.e.
I have: E = (r/2).dB/dt( r is radius of circle 1)=> C doesn't have electric field
If r>0, and |dB/dt|>0 then E≠0 so why do you say this means that there is no electric field?

Also - how did you use the fact that both B and E are vectors?

I think you need to be more careful in your definitions ... do you intend that the B field is defined throughout all space?
You also need to go through the calculation one step at a time.
 
Last edited:
induced-electric-fields_zpsddb3feb7.png

C have electric field, isn't it?
 
Oh I think I get you.
The calculation depends on where you put the center of the circle ...

But you've forgotten something - you are only calculating the contribution to the induced electric field due to the flux inside the circle.

In the first example you discovered that the flux at point C has no contribution to the field there.
In the second you found a non-zero contribution to the flux.
To get the total E field at C, you need to add the contributions from everywhere.

I don't think the calculation is actually valid for the B field you have in mind - but I can't tell for sure.
 
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If the uniform magnetic field is confined to a circular region, then the induced E will be zero at the center, C, of the region. You should be able to convince yourself that there is no contradiction with applying the integral form of Faraday's law to any closed path. (Note that for a circular path not centered at C, then you will not be able to pull E out of the integral.)

If the uniform B field extends to infinity in all directions, then there is no unique answer to the induced E field distribution. So, it's not a well-defined problem.

A similar situation occurs if you imagine all of infinite space filled with a uniform charge density ρ and try to find the E field produced by ρ using Gauss' law. There will be an infinite number of different E field distributions that satisfy Gauss' law.
 
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