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