Self Inductance of a Toroidal Coil of Circular Cross-section

Let R be the mean radius of the toroid and N the total number of turns in the toroid. Let I be the current flowing in the windings. The lines of B for the toroid are concentric circles.

Applying Ampere’s law,

φ B.dl = μ x current enclosed by the parth.
The magnetic induction B is tangential around the circular path of radius R. Hence
        B 2πR = μ x NI
Or        B = μNI /2 πR

Let A be the area of cross-section of each turn of the toroid.
Total magnetic flux linked with the toroid is
        φ = BNA = μNI /2 πR . NA = μN²I /2 πR
But        φ = LI
. :        LI = μN²I /2πR
Or         L = μN²I /2πR



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