A coil consists of N=105 turns of wire wrapped uniformly around a plastic torus. The inside radius of the torus is r0=0.12 m and the outer radius is r1=0.14 m. Thus, each winding has a diameter d=0.02 m. What is the self-inductance (in H) of this coil? Work in the approximation d≪r0.
To calculate the self-inductance of a coil with the given parameters, we need to use the formula for the self-inductance of a solenoid:
L = (μ₀ * N² * A) / l
where:
- L is the self-inductance of the coil
- μ₀ is the vacuum permeability (equal to 4π × 10^-7 T*m/A)
- N is the number of turns
- A is the cross-sectional area of the coil
- l is the length of the coil
To start, let's calculate the length of the coil, l. Since the coil is formed by a wire wrapped uniformly around a torus, we can use the formula for the circumference of a circle to find the length of one winding:
C = 2π * r0
Since there are N turns, the total length of the coil is:
l = N * C
Next, we need to find the cross-sectional area of the coil, A. Since the windings are tightly wrapped around the torus, we can approximate the cross-section as a rectangle with a width equal to the diameter of one winding, d, and a height equal to the circumference of the torus.
First, let's calculate the circumference of the torus:
C₂ = 2π * (r₀ + r₁)
Next, let's calculate the height of the rectangle, which is the circumference of the torus:
h = C₂
Finally, we can calculate the cross-sectional area of the coil:
A = d * h
Now, we have all the values needed to calculate the self-inductance, L:
L = (μ₀ * N² * A) / l
Substitute the appropriate values into the formula and evaluate to find the self-inductance of the coil.