Figure 22.52 shows four different ways to connect the copper wire. Based on the analysis we have just carried out, involving identifying whether or not there is a battery-like emf in a loop, what is the brightness of both bulbs in circuits $\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}$and $\mathbf{4}$?

The solenoid works on the principle of electromagnetism. When an electric current passes through the coil, a magnetic field is created. When a metal core is placed inside the coil, the magnetic flux lines are focused on the core. This increases the inductance of the coil compared to an air core.

The two bulbs are connected in series around a solenoid with a varying magnetic field. An alternating current in the solenoid provides a time-varying magnetic flux through the circuit to produce an emf to light the bulb.

Consider that each bulb has a resistance $R$. By applying Ohm’s law, the current through the bulb is,

$I=\frac{emf}{2R}$

From the given figure, the copper wire connected in first three cases makes the circuit into parallel combination of the bulbs. In case 4, the connected copper wire remains the two bulbs in series. So, in the series combination, same current will flow in two bulbs, the glow of the two bulbs is same. But, the intensity of bulb is less when you compared with the intensity of bulbs in other three cases.

In the first case, the connected copper wire covers one bulb and varying magnetic field. Due to the varying magnetic field, there exists an induced emf across the lower bulb, so it glows bright.

In second case, the connected copper wire covers upper bulb and varying magnetic field. Due to the varying magnetic field, there exists an induced emf across it, so it glows bright. The other bulb doesn’t include in this, so it becomes dark.

In the third case, the connected copper wire covers upper bulb and varying magnetic field. Due to the varying magnetic field, there exists an induced emf across it, so it glows bright.