# Voltage across rod in time varying magnetic field

If a slim conductor of some length $$l$$ and diameter $$d\ll l$$ is placed in a magnetic field $$B$$, and the field is changed by $$\frac {dB}{dt}$$, what (if any) is the voltage $$V$$ induced across the ends of the conductor?

In my case of interest, the slim conductor is a wire, fixed in space, that is victim of interference by an adjacent current, where the $$\frac {dB}{dt}$$ is caused by $$I\ \sin(\omega t)$$ in a source wire.

I am particularly interested in calculating a specific case (given $$I_0$$, $$\omega$$, and $$r$$ the distance between the two wires), as well as the fundamental connections to Maxwell's laws, probably the Maxwell- Faraday equation.

I am familiar with Lenz's law , but in my case of interest there is no return path or "ground plane", and so the victim wire has no current loop, or EMF loop. I can't form a curl integral, and no area is determined, and thus no time varying flux. Nevertheless, I would expect the above "wire rod" case to be the Maxwellian foundation of, or at least a step towards the Lenz "loop around flux" case. (Or perhaps I am terribly on the wrong foot here.)

The closest I come to this problem is by the Lorentz force , as it also involves a rod, and it involves a EMF on a charge in motion in a magnetic field. In contrast, my question centers around a time-varying magnetic field, without motion.

To be clear, the rod is fixed in space, and I am interested in the voltage calculation, not the motion or forces. When $$B$$ is varying, a varying $$E$$-field (call it external) also appears along the wire. Therefore, I think, if your induced voltage produces an electric field inside the wire (call it internal) which is in the direction of the external varying $$E$$-field, the wire or rod accelerates along the $$E$$-fields, and in the meanwhile, it rotates about its center of mass because the internal $$E$$-field, due to your induced voltage, has accumulated the positive and negative charges, respectively, at each end of the rod, and thus the motion of these charges in the $$B$$-field produce a torque on the rod due to the Lorentz forces exerted on the rod's ends in the opposite directions.