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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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chamberlainuket

Answer:

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

Explanation:

The self-inductance L of the toroid is mu(0)N^2a^2/2R.

Calculation of the self-inductance L:

Since

The toroidal inductor has a circular cross-section of radius a. The toroid has N turns and radius R.

The toroid is narrow ( a≪R )

So,

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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