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Nesterboy [21]
3 years ago
11

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?
Physics
1 answer:
monitta3 years ago
5 0

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:

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