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## Homework Statement

Calculate Mutual-inductance of a circular cross section toroid.

circular cross-section radius :a

toroid mean radius :R

Previous attempt for self inductance :https://www.physicsforums.com/showthread.php?t=537149

## Homework Equations

B=μNI/(2pi(R+y)) (Cartesian coordinates)

flux=∫∫B.dA

dA=dxdy, x from 0 to √(a^2 -y^2), y from (-a to a) (using symmetry, only integrate over half of x)

## The Attempt at a Solution

Discarding the constants for 'unclutteredness' we have

∫∫(1/(R+y))dx dy

integrating with respect to x the integral becomes

∫(√(a^2 -y^2))/(R+y) dy

using a trig substitution of y=asin(θ); the boundaries (y=a and -a) become (pi/2 and -pi/2)

the integral then simplifies to

∫(acos(θ)^2)/(R+asin(θ)) dθ

which is equal to ln(R+asin(θ))

solving with the boundaries, it give ln((R+a)/(R-a)) which is essentially ln(B/A) with B the most outer radius, A the most inner radius of the toroid.

from, the self inductance can be equated as L=(flux)*N/I

The solution seems strange, but it may be my own perspective that is blurred because I have been struggling with this problem for about two weeks.

any comment would be appreciated

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