Final answer:
The self-inductance (L) of the solenoid can be calculated using the formula (L = (μ₀ * N² * A) / l). Plugging in the values for the given solenoid with a length of 50.0 cm, diameter of 10.0 cm, and 1000 loops, we find that the self-inductance is 9.42 mH (Option a).
Step-by-step explanation:
The self-inductance (L) of a solenoid can be calculated using the formula:
L = (μ₀ * N² * A) / l
Where:
μ₀ = permeability of free space (4π × 10⁻⁷ T·m/A)
N = number of loops in the solenoid
A = cross-sectional area of the solenoid
l = length of the solenoid
For the given solenoid with a length of 50.0 cm (0.5 m), diameter of 10.0 cm (0.1 m), and 1000 loops:
A = π * (d/2)² = (3.14) * (0.1/2)² = 0.00785 m²
Plugging in the values, we get:
L = (4π × 10⁻⁷ T·m/A) * (1000²) * 0.00785 m² / 0.5 m = 9.42 * 10⁻³ H = 9.42 mH
Therefore, the self-inductance of the solenoid is 9.42 mH (Option a).