Faradays law of induction - where does the energy for the current in second loop come from?
In case 1, when you closed the switch, there will be a time period where the current ramps up to the steady state value. This is due to the loop inductance. It stores energy in the magnetic field. When you open the switch there will be a spark across the contacts (case 1) that dissipates the energy stored.

In case 2, at the instant the switch is closed, the current flowing will be different to case 1. This is because transformer action induces a secondary voltage and current will flow in loop 2 via the ammeter. So now you have a short period of time where energy is delivered into loop 2. This adds an extra current demand on loop 1 when the switch closes and is in addition to the ramping-up current seen in case 1.


  Is there any formula for the energy drawn from the battery when a
  second loop is present ?


In case 1 with the resistance in loop 1 at zero then, when the switch closes there is a rising current (di/dt) = V.L i.e. a rearrangement of Faraday's law. This is altered into an exponentially rising current that ultimately limits at V/R when the loop has resistance.

The formula for this is: -



This current creates a proportional magnetic field and some fraction of that field couples with loop 2 so, this generates a voltage in loop 2: -

\$V = N\dfrac{d\Phi}{dt}\$ where \$\Phi\$ is the coupled magnetic flux and N= 1 in your example.

That induced voltage drives current through the effective secondary inductance, the loop resistance and the ammeter. All in series.

The voltage and current (in loop 2) are taking energy from loop 1 in addition to the inductive energy held in the magnetic field for case 1.