A solenoid is wound with a single layer of insulated copper wire of diameter 2.800 mm and has a diameter 5.000 cm and is 1.500 m long. Assume that the adjacent wires touch and that the insulation thickness negligible. what is the inductance per meter(H/m) for the solenoid near its center.
To find the inductance per meter (H/m) for the solenoid near its center, we can use the formula:
L = (μ₀ * N² * A) / l
Where:
L is the inductance
μ₀ is the permeability of free space (4π × 10^⁻7 H/m)
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid
First, let's find the number of turns, N:
The solenoid is wound with a single layer of insulated copper wire. The diameter of the wire is given as 2.800 mm, which means the radius of the wire is 1.400 mm (0.0014 m). We can calculate the number of turns per meter, n:
n = 1 / (2 * π * r)
Where:
n is the number of turns per meter
r is the radius of the wire
n = 1 / (2 * π * 0.0014)
Next, let's calculate the cross-sectional area of the solenoid, A:
The diameter of the solenoid is given as 5.000 cm, so the radius of the solenoid is 2.500 cm (0.025 m). We can calculate the cross-sectional area, A:
A = π * r²
Now, let's calculate the inductance, L:
L = (μ₀ * N² * A) / l
Plug in the values we have calculated to find L.