Suppose you wish to make a solenoid whose self-inductance is 3.0 mH. The inductor is to have a cross-sectional area of 2.0*10^-3 m2 and a length of 0.052 m. How many turns of wire are needed?
Use the formula for inductance L that you will find here:
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In your case, you know L, l (the length), and area pi a^2, so solve for N
To find the number of turns of wire needed for the solenoid, we can use the formula for self-inductance:
L = (μ₀ * N² * A) / l
Where:
L is the self-inductance in Henrys (H)
μ₀ is the permeability of free space, approximately equal to 4π * 10^-7 H/m
N is the number of turns
A is the cross-sectional area in square meters (m²)
l is the length of the solenoid in meters (m)
In this case, we are given:
L = 3.0 mH = 3.0 x 10^-3 H
A = 2.0 x 10^-3 m²
l = 0.052 m
Rearranging the formula, we can solve for N:
N = √((L * l) / (μ₀ * A))
Now, substituting the values into the formula:
N = √((3.0 x 10^-3 * 0.052) / (4π * 10^-7 * 2.0 x 10^-3))
Simplifying the expression:
N = √((0.000156) / (8π x 10^-10))
Calculating further:
N ≈ √19.5 x 10^9 ≈ 1397 turns (rounded to the nearest whole number)
Therefore, approximately 1397 turns of wire are needed for the solenoid.