Change in current for a moving sliding bar?
Would the motion of the slider change the current that is flowing with it?


Not directly. If the motion of the conductor in the magnetic fields of the rest of the circuit causes a voltage in it, then this voltage could cause the current in the circuit inductance to change.


  At \$ t=0\$ current is \$ I_i\$ while it's at rest, when it beings to move is it still \$ I\$ = \$ I_i\$? 


At that instant yes, current cannot change discontinuously in a circuit with inductance.


  Change in current from possible change in \$ R\$ due to the motion?


I presume by \$R\$ you mean the contact resistance of the slider, or perhaps the effective resistance of the wide bar due to the path that the current takes through it, its resistance will be lower to a mid point than an edge point. In either case yes. A change in resistance will create a change of voltage due to the current flowing, which will cause a change of current in the circuit inductances.


  Or change in current from a possible induced \$ -\epsilon\$ due to the motion? 


To induce a voltage, the conductor movement has to 'cut' lines of magnetic field. Another way to describe the same thing is for the total flux enclosed by the loop of current (including its power supply) to change. The field of L1 is so arranged that movement of L2 does not cut it. 

Whether movement of L2 cuts any flux depends on how the rest of the circuit is arranged. If power is supplied by conductors going off in the x direction to a power supply in the plane of the paper, then that loop will be changed in area by the movement of L2, and so will induce a voltage. That is, the field from those conductors and the power supply is in the z direction at L2, so is cut by the x movement of L2. If the power supply is displaced only in the z direction (towards the viewer) the loop area is not changed by the movement, and there will be no induced voltage.


  Even though I modelled the magnetic field of the two components as two separate elements and can't find them affecting one another. 


Yes, that's correct. But the circuit (there's always a circuit) isn't complete in your diagram yet.