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As the title suggests, whenever there is a moving wire in a magnetic field of constant magnitude, an induced voltage is produced. Now, due to the separation of charges, which is a result of the magnetic force acting on the charges within the wire, an induced electric field is produced. The induced electric field starts on the positive charges and ends on the negative ones and is in the opposite direction of the induced current. Does this mean that it is a conservative electric field? We have been told that the induced electric field generated due to changing magnetic flux in, let us say, a loop is non-conservative since it forms closed loops, and thus has no beginning nor end. Also, it is in the same direction of the induced current that opposes the change in magnetic flux (changing magnetic field, changing surface area of the loop, etc). These characteristics or properties are the exact opposite of those of the induced electric field in a moving wire. Any help is appreciated! one more question, does non-electrostatic mean non-conservative exactly, or does it add more to the meaning?

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Is the induced electric field generated due to the separation of charges in a moving wire in a magnetic field non-conservative?

The induced electric field generated by non-moving charges is always conservative.

$$\nabla \times \vec{E_1} = 0$$

We have been told that the induced electric field generated due to changing magnetic flux in, let us say, a loop is non-conservative since it forms closed loops

The induced electric field generated by a changing magnetic flux is never conservative. It is divergence-free.

$$\nabla \cdot \vec{E_2} = 0$$

These characteristics or properties are the exact opposite of those of the induced electric field in a moving wire.

The charges in a moving wire are (most likely) moving.

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  • $\begingroup$ Thanks you so much! It is all very clear now :). $\endgroup$
    – MOMC
    May 13 at 23:50

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