Mutual-Inductance of Rogowski Coil (Toroid of Circular Cross-section)

[Steve_C]

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

 

[mark726]

Thank you for your post and for sharing your previous attempt at calculating the mutual inductance of a circular cross-section toroid. Your approach seems to be on the right track, but there are a few things that need to be corrected in your solution.

Firstly, your integral for the magnetic field B is only valid for a point on the axis of the toroid. To calculate the magnetic field at any point within the toroid, you will need to use the Biot-Savart law, which takes into account the contributions from all the current-carrying loops in the toroid. This will result in a more complicated integral, but it is necessary for an accurate calculation of mutual inductance.

Secondly, your approach for calculating the flux through the toroid is not entirely correct. The flux through a surface is given by the dot product of the magnetic field and the area vector of the surface. In this case, the area vector is perpendicular to the surface, and the magnetic field will be perpendicular to the area vector at each point on the surface. Therefore, the dot product of the two will be the magnitude of the magnetic field multiplied by the area of the surface, which is a constant value for a toroid with a circular cross-section. This means that the flux through the toroid will simply be the magnetic field multiplied by the cross-sectional area of the toroid.

Finally, your equation for self-inductance is not quite correct. The self-inductance of a toroid is given by L = μN^2A/l, where μ is the permeability of the toroid, N is the number of turns, A is the cross-sectional area, and l is the length of the toroid. In this case, A would be equal to πa^2, since the cross-section is a circle with a radius of a. Therefore, the correct equation for self-inductance would be L = μπN^2a^2/l.

To calculate mutual inductance, you will need to use the equation M = L2/L1, where L1 and L2 are the self-inductances of the two toroids. In this case, you would use the equation above to calculate L1 and L2, and then plug those values into the equation for mutual inductance.

I hope this helps clarify some of the issues with your solution. Please let me know if you have any further questions