Derive an expression for the self inductance of a narrow air-cored tor

1.	Derive an expression for the self inductance of a narrow air-cored toroid of circular cross section.
Answer» Consider a narrow air-cored toroid of circular cross section of radius r, central radius R and number of turns N. So that, assuming r << R, the magnetic field in the toroidal cavity is considered to be uniform, equal to B = \(\cfrac{μ_0 NI}{2πR}\) = µ₀nI ………….. (1) where n = \(\cfrac{N}{2πR}\) is the number of turns of the wire 2nR per unit length. The area of cross section, A = πr². The magnetic flux through one turn is Φ_m = BA = µ₀nIA ………… (2) Hence, the self inductance of the toroid, L = \(\cfrac{NΦ_m}I\) = (2πRn) µ nA = µ₀2πRn²A = µ₀n² V …………… (3) = \(\cfrac{μ_0N^2r^2}{2R}\) ………….. (4) where V = 2πRA is the volume of the toroidal cavity. Equation (3) or (4) gives the required expression.

Derive an expression for the self inductance of a narrow air-cored toroid of circular cross section.

Answer»

Consider a narrow air-cored toroid of circular cross section of radius r, central radius R and number of turns N. So that, assuming r << R, the magnetic field in the toroidal cavity is considered to be uniform, equal to

B = \(\cfrac{μ_0 NI}{2πR}\) = µ₀nI ………….. (1)

where n = \(\cfrac{N}{2πR}\) is the number of turns of the wire 2nR per unit length. The area of cross section, A = πr².

The magnetic flux through one turn is

Φ_m = BA = µ₀nIA ………… (2)

Hence, the self inductance of the toroid,

L = \(\cfrac{NΦ_m}I\) = (2πRn) µ nA = µ₀2πRn²A = µ₀n² V …………… (3)

= \(\cfrac{μ_0N^2r^2}{2R}\) ………….. (4)

where V = 2πRA is the volume of the toroidal cavity. Equation (3) or (4) gives the required expression.

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