A question about Faraday’s law of induction for a closed path
The integral form of Faraday's law is \$\oint\vec{E}\cdot dl = -d/dt\oint\vec{B}\cdot d\vec{s}\$

Knowing the magnetic flux through your loop, you can calculate the voltage around the loop.

The continuity equation \$\nabla\cdot\vec{J} = -d\rho/dt\$ tells you that your current density is constant around your loop, since there is no charge being added or taken away.

Ohm's law says \$\vec{J} = \sigma \vec{E}\$

You would integrate over a distance to turn this into the more common form of Ohm's law \$I = V/R\$.

In your example \$\vec{J}\$ is constant, so if \$\sigma\$ decreases in a portion of the loop, \$\vec{E}\$ must increase. This is why the 3kOhm resistor has 3V across it while the 1kOhm only has 1V across it.