Introduction:
In order to achieve maximum power transfer from the ideal transformer to the 2 Ω resistor, the turns ratio between the primary and secondary windings must be determined. The turns ratio plays a crucial role in matching the impedance between the source and the load, ensuring maximum power transfer.

Understanding the problem:
The problem states that the secondary winding of the ideal transformer has 40 turns, and we need to find the number of turns in the primary winding for maximum power transfer to the 2 Ω resistor.

Key formula:
The turns ratio (N) between the primary and secondary windings of an ideal transformer can be calculated using the formula:

N = N_primary / N_secondary

Where N_primary is the number of turns in the primary winding and N_secondary is the number of turns in the secondary winding.

Calculating the turns ratio:
Given that the number of turns in the secondary winding (N_secondary) is 40, we can substitute this value into the formula to calculate the turns ratio:

N = N_primary / 40

Achieving maximum power transfer:
For maximum power transfer to the 2 Ω resistor, the impedance seen by the source (which is the primary winding in this case) should be equal to the complex conjugate of the load impedance.

Since the load impedance is purely resistive (2 Ω), the impedance seen by the source (primary winding) should also be purely resistive and equal to 2 Ω.

Impedance calculation:
The impedance seen by the source can be calculated using the turns ratio (N) and the load impedance (Z_load) as follows:

Z_source = (N^2) * Z_load

Given that the load impedance (Z_load) is 2 Ω and the turns ratio (N) is N_primary / 40, we can substitute these values into the equation:

2 Ω = ((N_primary / 40)^2) * 2 Ω

Simplifying the equation:

1 = (N_primary^2) / (40^2)

40^2 = N_primary^2

N_primary = 40

Conclusion:
Therefore, the number of turns in the primary winding for maximum power transfer to the 2 Ω resistor is 40.