Answer:
L = (μ₀ × N² × A) / (2π × r)
Explanation:
To determine the self-inductance of a toroidal coil with N turns of wire wound on an air frame with mean radius r, we can use the following formula:
L = (μ₀ × N² × A) / (2π × r)
Where:
L is the self-inductance of the coil (in henries, H)
μ₀ is the permeability of free space (4π × 10⁻⁷ H/m)
N is the number of turns in the coil
A is the cross-sectional area of the toroid (in m²)
r is the mean radius of the toroid (in m)
Here's the step-by-step derivation:
Consider a toroidal coil with N turns of wire wound on an air frame with mean radius r.
The magnetic field inside the toroid is uniform and confined within the coil. The field outside the toroid is negligible.
The magnetic flux density (B) inside the toroid is given by:
B = (μ₀ × N × I) / (2π × r)
where I is the current flowing through the coil (in amperes, A).
The total magnetic flux (Φ) through the coil is the product of the magnetic flux density and the cross-sectional area (A) of the toroid:
Φ = B × A = (μ₀ × N × I × A) / (2π × r)
The self-inductance (L) of the coil is defined as the ratio of the total magnetic flux to the current:
L = Φ / I = (μ₀ × N × A) / (2π × r)
Simplifying the equation, we get:
L = (μ₀ × N² × A) / (2π × r)
Therefore, the self-inductance of a toroidal coil with N turns of wire wound on an air frame with mean radius r is given by:
L = (μ₀ × N² × A) / (2π × r)
Note that this equation assumes an ideal toroid with a uniform magnetic field and no leakage flux. In practice, the actual self-inductance may deviate slightly from this theoretical value due to factors such as the finite thickness of the wire and the presence of small gaps between the turns.