Two closely wound circular coils have the same number of turns , but one has twice the radius of the other . What is the ratio of self inductances of the two coils ?
The self-inductance of a coil is given by the formula:
L = (μ₀ * N² * A) / l,
where L is the self-inductance, μ₀ is the permeability of free space (constant), N is the number of turns, A is the cross-sectional area of the coil, and l is the length of the coil.
Let's assume the smaller coil has a radius 'r' and the larger coil has a radius '2r' (twice the radius of the smaller coil). Since the number of turns is the same for both coils, let's denote it as 'N' for simplicity.
For the smaller coil:
A₁ = π * r², (cross-sectional area)
l₁ = 2πr. (length)
For the larger coil:
A₂ = π * (2r)² = 4πr², (cross-sectional area)
l₂ = 2π(2r) = 4πr. (length)
Now ,calculate the self-inductance for each coil:
L₁ = (μ₀ * N² * A₁) / l₁ = (μ₀ * N² * π * r²) / (2πr) = (μ₀ * N² * r) / 2,
L₂ = (μ₀ * N² * A₂) / l₂ = (μ₀ * N² * 4πr²) / (4πr) = (μ₀ * N² * r²) / r = μ₀ * N² * r.
To find the ratio of the self-inductances, we can divide L₂ by L₁:
(L₂ / L₁) = [(μ₀ * N² * r) / 2] / (μ₀ * N² * r)
= (μ₀ * N² * r) / (2 * μ₀ * N² * r)
= 1 / 2.
Therefore, the ratio of the self-inductances of the two coils is 1:2 (or 1/2).
