2 answers
(click or scroll down)
Suppose you wish to make a solenoid whose self-inductance is 1.6 mH. The inductor is to have a cross-sectional area of 1.3 x 10-3 m2 and a length of 0.065 m. How many turns of wire are needed?
bobpursley
answered
11 years ago
11 years ago
Explain Bot
answered
11 months ago
11 months ago
To find the number of turns of wire needed for a solenoid, you can use the formula:
N = sqrt((L * A) / ฮผโ)
Where:
N is the number of turns of wire
L is the length of the solenoid
A is the cross-sectional area of the solenoid
ฮผโ is the permeability of free space (constant value)
Given:
L = 0.065 m
A = 1.3 x 10^(-3) m^2
ฮผโ = 4ฯ x 10^(-7) Tยทm/A
Let's substitute these values into the formula:
N = sqrt((0.065 * 1.3 x 10^(-3)) / (4ฯ x 10^(-7)))
Now we can simplify the expression in the square root:
N = sqrt(8.45 x 10^(-6) / (4ฯ x 10^(-7)))
N = sqrt(8.45 / (4ฯ x 10^(-7 + 7)))
N = sqrt(8.45 / (4ฯ))
N โ sqrt(0.670)
Finally, we can calculate the number of turns N:
N โ 0.819
Therefore, approximately 0.819 turns of wire are needed for the solenoid.