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According to Faraday's law a changing flux through a loop will create an Electric-field curling around that loop. And if the loop is a conducting wire the current looping will create an opposing magnetic field as below(Lenz):

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But what if there is no wire and no current but just the circulating/looping Electric-filed around an imaginary loop? Would that Electric field generate an opposing magnetic filed as well?

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  • \$\begingroup\$ Your diagram has a failing which I have outlined in my answer to your previous question. The rising B field doesn't have any net change - yes the field de-intensifies through the coil because the secondary flux cancels the primary flux on the inside but, outside the coil, the secondary flux adds to the primary flux thus no net change in primary B field. \$\endgroup\$ – Andy aka Jul 7 '17 at 12:24
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Edit: WARNING: Some of my descriptions may be false.

Induced current creates a magnetic field that is opposite to the one that created it. This is the reason why the magnetic field magnitude decreases when it flies trough the conductor. It will decrease according to how easily the eddy currents can be induced inside a material. If the material is non-conductive, eddy currents will not be created and hence the magnetic field will just "fly" trough.

If you used a super conductor, because it has no resistance, it will create electrical current based on the whole magnetic field strength. Because of this, this current will produce exactly the same magnitude of the opposite magnetic field which will cancel out the magnetic field. This is also the reason why magnetic fields can't penetrate the super conductor and is a perfect shield. It also gives rise to the "levitation effect".

To the second part of your question: The changing magnetic field will indeed create an electromotive force around that invisible loop, but if no current is flowing, opposing magnetic field WILL NOT be created. This is the reason why the magnetic field isn't being "consumed" but will radiate outwards indefinetely or untill getting into contact with some electricaly charged particles.

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  • \$\begingroup\$ Imagine flux is increasing at a constant rate the electric field will appear constant circulating around the loop(no conductor just space and an imaginary loop). BUT will that constant circulating electric filed generate an opposing magnetic field. I.e electric field here is looping instead of a current is looping. So what I understand is electric field will be looping/curling but it will not generate an opposing magnetic field since electric field looping is not the same thing current looping. Is that correct? \$\endgroup\$ – atmnt Jul 7 '17 at 10:01
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    \$\begingroup\$ That is correct. Current creates magnetic field. Electric field by itself can not create a magnetic field unless it is acting upon charged particles and those then create the magnetic field that is the opposite in direction of the one that created the movement of those particles. \$\endgroup\$ – MaDrung Jul 7 '17 at 10:07
  • \$\begingroup\$ But what if the magnetic flux increases with a changing rate?? Then the electric field curling will be changing as well. In that case will that generate an opposing magnetic field? \$\endgroup\$ – atmnt Jul 7 '17 at 10:09
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    \$\begingroup\$ Note: a magnetic field can also be generated by a changing electric field (no conduction needed here): that component of current is called displacement current (it was Maxwells's achievement to understand that) \$\endgroup\$ – Curd Jul 7 '17 at 10:58
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    \$\begingroup\$ @Madrung: what about curl H = dD/dt? No charged particles here, just empty space, where a time dependent displacement field alone gives rise to a magnetic field. \$\endgroup\$ – Bart Jul 7 '17 at 11:24

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