A solenoid is 10.0 cm long, has a radius of 0.500 cm and has 8000 equally spaced loops. What angular frequency needs to drive this inductor so that it has a reactance of 6.000e+3 ohms?
To determine the angular frequency needed to drive the inductor, we first need to calculate its inductance using the given information.
The inductance of a solenoid can be calculated using the following formula:
L = (μ₀ * N² * A) / l
Where:
L is the inductance
μ₀ is the permeability of free space (4π * 10⁻⁷ T m/A)
N is the number of turns (loops)
A is the cross-sectional area of the solenoid
l is the length of the solenoid
Given:
N = 8000 loops
A = π * r² (where r is the radius of the solenoid)
l = 10.0 cm = 0.100 m
Let's calculate the inductance:
A = π * (0.500 cm)² = 0.785 cm² = 7.85e-5 m²
L = (4π * 10⁻⁷ T m/A) * (8000 loops)² * (7.85e-5 m²) / 0.100 m
L ≈ 2.480e-3 H
Now that we have the inductance, we can calculate the angular frequency (ω) using the reactance (X) and the formula:
X = ω * L
Where:
X is the reactance
L is the inductance
ω is the angular frequency
Given:
X = 6.000e+3 ohms (Ω)
Let's calculate the angular frequency:
ω = X / L
ω = (6.000e+3 Ω) / (2.480e-3 H)
ω ≈ 2.419e+6 rad/s
Therefore, the angular frequency needed to drive this inductor so that it has a reactance of 6.000e+3 ohms is approximately 2.419e+6 rad/s.