The solar cells will provide some current when illuminated. This can be applied to charge the battery. It doesn't matter if something else is connected to the battery at the same time that is drawing current, you still get to use the power from the solar cells effectively. In fact, it's more efficient to power the end load directly than to first store the energy in a battery and later have the battery power the load. The point of the battery is to be able to use the power when it's needed, not just when the solar cells can provide it.
For example, let's say the solar cells can produce 1A in full sunlight, and this 1A can be applied to charging the battery. Now let's say a load is turned on that draws 400mA. The 1A from the solar cell will be split so that 400mA drives the load, leaving 600mA left to charge the battery. This is perfectly fine. If the load were more than the solar cell current, this would still work. Let's say the load takes 1.5A. 1A of that would come straight from the solar cell, and 500mA from the battery.
In other words, there is nothing wrong with the solar cell current relieving some of the load on the battery as apposed to charging the battery and then having the battery supply the load current. As I said before, that is actually more efficient since there is some loss in storing and the retreiving energy from the battery.
Knowing when the battery is full and what to do about it is a separate topic. Some battery types are quite fussy about how they are charged. For those you will need a active circuit to monitor the battery voltage and possibly temperature to decide what current to charge it with or what voltage to hold it at.
Lead-acid batteries are pretty forgiving in this area and don't have a problem with continuous charging even when full. The simplest solution is to arrange and size the solar cells such that they can only produce the maximum float voltage at the maximum current the battery can take when full. Then you can just connect the solar cells accross the battery and be done with it. This will charge the battery much more slowly than it is capable of when it is low, but is a simple setup guaranteed not to damage it.
If the solar panel size or cost is important, you need to use the energy from it more efficiently. In that case I'd use a switching power supply driven by a microcontroller that takes into account what current the solar cells can put out, the battery voltage, and possibly the battery temperature to decide what current, if any, to charge it with. This gets complicated. You have to read the battery datasheet carefully and implement the charging regime accordingly.
The dual of Faraday's Law is Ampere's Law but, while Faraday's Law is fundamental to the physics of an inductor, Ampere's Law is not fundamental to the physics of a capacitor.
Now, it is true that, in circuit theory, the capacitor and inductor are duals:
$$i_C = C\frac{dv_C}{dt} \leftrightarrow v_L = L \frac{di_L}{dt}$$
However, we have to be more careful outside the context of circuit theory.
In physics, the fundamental relationship
$$Q = CV$$
clearly requires the existence of electric charge and an electric scalar potential due to a conservative electric field. This equation relates electric charge and electric scalar potential.
The closest we can get to a dual of this is
$$\Phi = LI $$
which relates magnetic flux and electric current. But magnetic flux is not the dual of electric charge.
The missing ingredient here is the hypothetical magnetic charge (magnetic monopole) which is the dual of electric charge.* Were magnetic charge \$Q_m\$ (measured in webers) to exist, it would be a source or sink of a conservative magnetic field (measured in amperes per meter) and there would be an associated scalar magnetic potential (measured in amperes).
We could thus relate magnetic charge and magnetic scalar potential with a magnetic "capacitance" measured in henrys.
Further, we could relate electric flux to magnetic current (measured in volts) with an electric "inductance" measured in farads.
To summarize, while electric flux and magnetic flux are duals, and changing magnetic flux is fundamental to the physics of an inductor, changing electric flux is not fundamental to the physics of a capacitor. Indeed, it is the electric field itself, not the electric flux, that is fundamental.
*Assuming magnetic charge exists, Maxwell's equations become
$$\nabla \cdot \vec D = \rho_e$$
$$\nabla \cdot \vec B = \rho_m$$
$$\nabla \times \vec E = - (\vec J_m + \frac{\partial \vec B}{\partial t})$$
$$\nabla \times \vec H = \vec J_e + \frac{\partial \vec D}{\partial t}$$
Best Answer
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.
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.