Final answer:
The magnetic field strength inside an iron nail wrapped with an insulated copper wire carrying 0.80 A current, considering its ferromagnetic properties that increase the field by a factor of 120, is 1.608 tesla.
Step-by-step explanation:
To calculate the magnetic field strength inside the iron nail when it is wrapped by a copper wire with current, we must know the coil's turn density (number of turns per unit length), the current passing through the coil, and the enhancing effect of the iron core's ferromagnetic properties.
The formula to calculate the magnetic field (B) inside a solenoid is B = μ × n × I, where μ is the permeability of the core material, n is the number of turns per unit length and I is the current. The permeability of free space (μ0) is 4π × 10-7 T·m/A. However, due to the ferromagnetic properties of the iron core, the magnetic field will be increased by a factor which is given in the problem.
First, we calculate n, which is the number of turns per unit length (240 turns over 1.8 cm, or 0.018 m). n = 240 turns / 0.018 m = 13,333.33 turns/m.
The field without the iron core would be calculated using the permeability of free space as: B0 = μ0 × n × I = 4π × 10-7 T·m/A × 13,333.33/m × 0.80 A ≈ 0.0134 T. After considering the increase due to the ferromagnetic properties by a factor of 120, the magnetic field inside the nail is:
B = 120 × 0.0134 T = 1.608 T.
Therefore, the magnetic field strength inside the nail, considering its ferromagnetic properties, is 1.608 tesla.