What is the self-inductance of an air-core solenoid,1m long and 0.05m in diameter,if it has 1400 turns?(a)5.23mH(b)4.84mH(c)3.63mH(d)2.42mH
Do you have the formula for this?
5.23mh
To find the self-inductance of an air-core solenoid, we can use the formula:
L = (μ₀ * N² * A) / l
where:
L = self-inductance
μ₀ = permeability of free space (constant value of 4π * 10⁻⁷ H/m)
N = number of turns
A = cross-sectional area of the solenoid
l = length of the solenoid
First, let's calculate the cross-sectional area of the solenoid:
A = π * r²
where:
r = radius of the solenoid = diameter / 2
Given that the solenoid has a diameter of 0.05m, the radius will be 0.05 / 2 = 0.025m.
Therefore,
A = π * (0.025)²
Next, we substitute the given values into the formula to get the self-inductance:
L = (4π * 10⁻⁷ H/m * (1400)² * π * (0.025)²) / 1
Simplifying the equation:
L = (4 * 3.14 * 10⁻⁷ H/m * 1960000 * 0.000196)
L = 4.84 * 10⁻³ H
So, the self-inductance of the air-core solenoid is 4.84mH.
Therefore, the correct answer is (b) 4.84mH.