EMF Induced Between Parallel Conductors
Consider first that the two wires (of length \$\ell\$) are placed very close to each other such that the distance R is very small (ostensibly zero). Consider also that the driven wire has self-inductance (L) and that this inductance has a reactance that develops a volt drop due to the current flowing: -



Because the two wires (driven and undriven) are very close, we can assume the field generated by the driven wire is wholly shared by the undriven wire and so we get 100% transformer action and, whatever voltage is across the length (\$\ell\$) of the driven wire will be (in fact has to be) across the receiving wire.

The voltage is \$I\cdot\omega L\$ and that voltage rapidly gets smaller as R increases. But, what is the inductance of a wire: -



However, you can use an online calculator for that to get the answer if you don't want a math/algebra solution: -



So that gives you the induced voltage when R is zero. To calculate the voltage when R is non-zero requires understanding of how the coupled flux falls away with an increasing distance R.

This is probably best understood by thinking about how the flux density falls away with distance from a long straight conductor (based on Biot Savart).

Flux density, B = \$\dfrac{\mu_0\cdot I}{2\pi d}\$

Where I is the current and d is the distance from the conductor. If this is integrated from d=0 to d=infinity, you get the total flux and this of course is used to calculate wire inductance as indicated at the top of the answer.

If you then integrate only as far as distance R you get the flux that isn't coupled to the receiving wire. The difference between the two is the flux that couples to the receiving wire and, as a proportion of the total flux that is the coupling factor, k: -



So, you calculate k (based on distance R) and that multiples by the voltage across the driven inductance/wire to give the induced emf.