A coil consists of N=10^5 turns of wire wrapped uniformly around a plastic torus. The inside radius of the torus is r0=0.15 m and the outer radius is r1=0.16 m. Thus, each winding has a diameter d=0.01 m. What is the self-inductance (in H) of this coil? Work in the approximation d≪r0

To find the self-inductance of the coil, we need to use the formula for the self-inductance of a solenoid, which can be derived by considering the magnetic field produced by each turn of wire.

The formula for the self-inductance of a solenoid is given by:

L = μ₀N²A / ℓ

Where:
L is the self-inductance
μ₀ is the permeability of free space (constant value)
N is the number of turns of wire
A is the cross-sectional area of the solenoid
ℓ is the length of the solenoid

In this case, our coil consists of N = 10^5 turns of wire wrapped uniformly around a plastic torus. The diameter of each winding (d) is given as 0.01 m. However, we are given the approximation that d ≪ r₀, where r₀ is the inside radius of the torus.

Since d is much smaller than r₀, we can assume that the diameter of each winding can be neglected in the calculation of the cross-sectional area and length of the solenoid.

To calculate the cross-sectional area (A), we need to find the difference between the areas of the outer and inner circles of the torus:

A = π(r₁² - r₀²)

Next, we need to calculate the length of the solenoid (ℓ), which is the circumference of the torus:

ℓ = 2πr₀

Finally, we can substitute these values into the formula for self-inductance:

L = μ₀N²A / ℓ

Now let's calculate the self-inductance of the coil:

μ₀ is a constant value, approximately 4π × 10⁻⁷ H/m.

N = 10^5 (given)

r₀ = 0.15 m (given)

r₁ = 0.16 m (given)

d = 0.01 m (given)

First, calculate the cross-sectional area:

A = π(0.16² - 0.15²) = π(0.0256 - 0.0225) = 0.0031 m²

Next, calculate the length of the solenoid:

ℓ = 2π(0.15) = 0.9425 m

Now substitute the given values into the formula for self-inductance:

L = (4π × 10⁻⁷ H/m)(10^5)²(0.0031 m²) / 0.9425 m

Simplifying the expression:

L ≈ 1.298 x 10⁻² H

Therefore, the self-inductance of this coil is approximately 0.01298 H.