I Faraday induction in constant B field, with non-conduction wires

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In a scenario with a constant magnetic field and a conducting loop that expands at a constant rate, the induced EMF and electric field can be calculated using E = Bv. However, if the wire is a perfect insulator, there would be no current, raising questions about the induced EMF and its effects on polarization in the dielectric material. The lack of current implies no induced magnetic field, which seems inconsistent with Maxwell's equations. Despite this, moving charges within a dielectric may still experience polarization due to the Lorentz force. Thus, even in a dielectric, an EMF can exist, leading to polarization effects.
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A standard textbook problem features a constant B field and a conducting loop that increases in area at constant rate.
It is easy to work out the induced EMF and the associated electric field magnitude and direction (CW or CCW). The magnitude of the E field
is E = B v where v is a velocity. The current in the loop is easy to work out and the rate of dissipation (I^2 R) compared with the external work/power needed to keep the loop area expanding at a constant rate.

My question is: What if the wire is actually a perfect insulator? Will there still be the same EMF and will that induce polarization in the dielectric
insulating material? There will be no current so, apparently, so no induced magnetic field thus no rate of change of magnetic field.
How is that consistent with the differential form of Maxwell's curl equations?
 
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When the wire is made larger the positive and negative charges are moved raadially and feel opposite tangential forces frrom Lorentz force. I would expect polarization in a polarizeable dielectric material
 
If the "wire" is a dielectric you'll also have the said EMF, which will polarize the wire.
 
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