Energy transfer with electromagnetic waves

Question

As we know that the electrical energy is transferred via electromagnetic waves from the source to the load. My question is that, even if there is transfer of energy through the electromagnetic waves, then how these waves get to know that where the load is actually connected, i.e. at which position or where they have to actually go?

And, if an electromagnetic wave involves only magnetic field and electric field, then why, if any load brought towards the closed circuit doesn't starts to work? For an example, if a bulb is connected to a circuit and it's glowing and then, if we bring an another bulb towards it, then it should also start glowing without any connections because electromagnetic waves are still present in the surroundings which are transferring the energy to the bulb which was already in the closed circuit, then why electromagnetic waves only target the bulb which is connected to the circuit?

Answer 1

When we say that electromagnetic waves transfer energy, we mean that the electromagnetic field has energy stored in it - just like the particles of waves on whater have kinetic and potential energy.

Classical view
Let us consider a classical radio emitter: oscillations of current in the antenna produce electromagnetic field,a nd some of the energy of these oscillations is lost to the field - so we need a constant power supply to sustain these oscillations. The electromagnetic field propagates in all directions. Suppose now we have a receiver at some point - electromagnetic field induces current in this receiver, which is transformed into a signal, e.g., by a loudspeaker. The electromagnetic field around the antenna loses some energy to the oscillations in the receiver, but it has no effect on the field elsewhere.

Quantum view
From the quantum viewpoint the intensity of the EM field is the number of photons emitted. Each photon has energy $\hbar\omega$. Emitter creates photons and receiver absorbs them. If emitter emits $n$ photons, it furnishes the field with energy $n\hbar\omega$, whereas the receievr absorbs $m$ photons, i.e., energy $m\hbar\omega$, leaving $(n-m)\hbar\omega$ in the field ($n-m$ photons).

Impedance matching
Just because a circuit is surrounded by an EM field, it does not mean that there will be much energy transferred between them. The efficient condition for coupling an emitter or receievr to the field is knwon as impedance matching. Antenna is par excellence the best-known device used to achieve this goal - a classical antenna has length equal to half-wavelength of the emitted radiation. If it were too short, the field would not really feel the variation of the current in space. If it were too long, the effect of the positive and negative current on the field would cancel out.

For an example, if a bulb is connected to a circuit and it's glowing and then, if we bring an another bulb towards it, then it should also start glowing without any connections because electromagnetic waves are still present in the surroundings which are transferring the energy to the bulb which was already in the closed circuit, then why electromagnetic waves only target the bulb which is connected to the circuit?

The bulb connected to a circuit is coupled to it efficiently, whereas the other is not. The situation is however somewhat different here than in the radio transmission, since the first bulb is directly driven by the current, rather than via EM waves propagating in space.

Answer 2

I assume that this question is not asking about radio transmission, but about some recent Youtube videos that explained that even in DC (or low frequencies like 50-60 Hz) circuits with wires connecting all the elements, the energy is actually carried in the fields around the wires rather than the wires themselves.

As we know that the electrical energy is transferred via electromagnetic waves from the source to the load.

As pointed out in the comments, this would be better worded as saying energy is transferred via the electromagnetic field, rather than by waves.

how these waves get to know that where the load is actually connected, i.e. at which position or where they have to actually go?

The wires guide the fields.

Mathematically the wires put boundary conditions on the equations from which we determine the fields. More simply, the arrangement of the wires controls where the fields go, and so they control where the energy is transferred from and to.

if any load brought towards the closed circuit doesn't starts to work? For an example, if a bulb is connected to a circuit and it's glowing and then, if we bring an another bulb towards it, then it should also start glowing without any connections because electromagnetic waves are still present in the surroundings

As pointed out in the comments, this can work. I've never tried it myself, but if you bring an old-school fluorescent tube near high voltage (10's or 100's of kV) power transmission lines, the tube can light up.

In the case of smaller bulbs near lower-voltage lines, the metal (and to a lesser extent, other materials) of the bulb also affects the fields around it. Metals in particular effectively collapse the electric fields around them, because the electrons in the metal rearrange to create their own field that opposes the external field. Again, mathematically we'd say the metal creates a boundary condition on the field.

Only if the bulb is very large (like the fluorescent tube) or the field is very strong (like around the high voltage transmission line) do we see significant amounts of power transferred to the bulb floating in space near the other electric circuit.