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15 January, 10:32

A toroidal inductor has a circular cross-section of radius a a. The toroid has N turns and radius R. The toroid is narrow (a≪R), so the magnetic field inside the toroid can be considered to be uniform in magnitude. What is the self-inductance L of the toroid?

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  1. 15 January, 10:36
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    L=N/I*magnetic flux

    =N/I*BA=N*mu (0) NIA/2pi*R=mu (0) N^2/2pi*R) * Pi*a^2

    =mu (0) N^2a^2/2R
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