You need to produce a solenoid that has an inductance of 2.51 μH . You construct the solenoid by uniformly winding 1.17 m of thin wire around a tube. How long, in centimeters, should the tube be?
L=2.51•10⁻⁶ H
x=1.17 m
z=?
L=μ₀N²A/z
N=x/2πr => r= x/2πN =>
A=πr²=πx²/4π²N²
L=μ₀N²A/z= μ₀N²πx²/4π²N²z=
= μ₀x²/4z
z= μ₀x²/4L= 4π•10⁻⁷/4•2.51•10⁻⁶=0.12 m
Thanks Elena. I was really stuck on the N=x/2πr. I didn't know that was a property.
But I believe there is an error in the math
z should be z= μ₀x²/4πL = .054 m
Thanks again
To find the length of the tube needed, we can use the formula for inductance of a solenoid:
L = (μ₀ * N² * A) / l
Where:
L = Inductance of the solenoid (2.51 μH)
μ₀ = Permeability of free space (4π x 10^(-7) T·m/A)
N = Number of turns (unknown)
A = Cross-sectional area (unknown)
l = Length of the solenoid (1.17 m)
We need to find the length of the tube, which is equivalent to the length of the solenoid (l). Let's rearrange the formula to solve for l:
l = (μ₀ * N² * A) / L
Rearranging further:
l = (L * N² * A) / μ₀
Now, we can convert the given inductance (2.51 μH) to Henries:
L = 2.51 x 10^(-6) H
Substituting the values into the formula:
l = (2.51 x 10^(-6) * N² * A) / (4π x 10^(-7))
Now, we need to find the relationship between the length of the tube (l) and the number of turns (N) and the cross-sectional area (A). The length of wire used to create the solenoid will be equal to the length of the solenoid (l):
l = 1.17 m
Since the wire is uniformly wound around the tube, the length of wire used will be equal to the circumference of the tube, multiplied by the number of turns (N):
l = 2π * r * N
where r is the radius of the tube.
Rearranging:
N = l / (2π * r)
Now, we need to relate the cross-sectional area (A) to the radius (r). The cross-sectional area of the solenoid will be equal to the area of the circular face of the tube:
A = π * r²
Substituting the value of A into the equation for l:
N = l / (2π * sqrt(A/π))
Simplifying:
N = l / (2 * sqrt(A))
Now, substitute the expression for N back into the formula for l:
l = (2.51 x 10^(-6) * (l / (2 * sqrt(A)))² * A) / (4π x 10^(-7))
Now, we can solve for A:
Multiply both sides by (2 * sqrt(A))²:
l * (2 * sqrt(A))² = (2.51 x 10^(-6) * A) / (4π x 10^(-7))
Simplify the left side:
4 * A * l² = (2.51 x 10^(-6) * A) / (4π x 10^(-7))
Divide both sides by A:
4 * l² = (2.51 x 10^(-6)) / (4π x 10^(-7))
Simplify the right side:
4 * l² = (2.51 / (4π)) x (10^(-6) / 10^(-7))
Simplify further:
4 * l² ≈ 0.20 x 10
Divide both sides by 4:
l² ≈ 5 x 10^(-2)
Take the square root of both sides:
l ≈ sqrt(5 x 10^(-2))
l ≈ sqrt(5) x sqrt(10^(-2))
l ≈ 0.707 x 10^(-1)
l ≈ 0.071 cm
Therefore, the length of the tube should be approximately 0.071 cm.
To determine the length of the tube needed for the solenoid, we can utilize the formula for inductance of a solenoid:
L = (μ₀ * N² * A) / l
Where:
L = Inductance
μ₀ = Permeability of free space (constant)
N = Number of turns
A = Cross-sectional area of the solenoid
l = Length of the solenoid
We are given:
L = 2.51 μH = 2.51 x 10^(-6) H
N = Number of turns = 1
l = Length of the solenoid = 1.17 m
The question asks for the length of the tube in centimeters, so we'll find the length of the tube in meters first and then convert it to centimeters.
Rearranging the formula for l, we have:
l = (μ₀ * N² * A) / L
We need to find A, the cross-sectional area of the solenoid. The cross-sectional area of a solenoid is given by:
A = π * r²
Where r is the radius of the solenoid.
From the information provided, the wire is uniformly wound around the tube. This implies that the length of the wire is equal to the length of the solenoid.
Therefore, the length of the wire (l_wire) = 1.17 m.
The length of the wire is the same as the circumference of the tube, which is given by:
2π * r = l_wire
Solving for the radius (r):
r = l_wire / (2π)
Now we can calculate the radius of the solenoid.
Next, substitute the values for N, A, and L into the formula for l:
l = (μ₀ * N² * A) / L
Finally, convert the value of l from meters to centimeters.
By following these steps, you should be able to find the length of the tube in centimeters.