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Registered Member #3343
Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Can somenone inform how to calculate the reluctance of a ferrite rod in free space?
It should be noted that the magnetic circuit has 2 reluctances: one provided by the ferrite rod itself Rrod, and other provided by the air gap (big) located between the ends of the rod.
So far wath Im gessing,
The total reluctance is Rrod + Rair
Rrod = Lrod/(Uini x 4x pi x 0.0000001 x Arod)
Rrod is the reluctance of the ferrite rod alone [AE/Wb] Rair is the reluctance of the space between the two extreme ends of the rod, [AE/Wb] Lrod is the lenght of ferrite rod [m], Uini is the initial permissivity (relative) of the ferrite Arod is the cross section area of ferrite [m^2]
So far so easy.
Rair = reluctance of the magnetic circuit by air between the extreme ends of the rod. Here Is the issue I ned the help!
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Hi Newton.
R_air can be calculated from the effective length/area of the flux return path in air, using U_air = 1. The tricky part is estimating the air length and area, both of which depend on the length and location of the electrical coil that provides H in the magnetic circuit you want to analyze.
I'd start by drawing a longitudinal section of the rod and coil, roughly to scale,. Then sketch a set of loops indicating my guess to the B field orientation and density in the neighborhood.
You could also:
- continue searching on the Internet for known solutions or formulas. A good starting place might be catalogs of companies that make ferrite cores, including rod shapes, and work backwards from their inductance formulas.
- wind a test coil and measure its inductance. I think that for moderate excitation at low frequencies, R_rod will be << R_air. (but R_air depends on flux geometry that is strongly influenced by the rod).
Registered Member #834
Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644
The approach is interesting, but I would expect a significantly larger effective area for the return path and smaller effective length, because the field lines exit the core not only at the extremities, but along all its length.
Registered Member #540
Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969
The field lines aren't fully circular and exit the end of the core at different points. The field of an open bar technically extends out to infinity but it drops off in, I would think, the same manner as an electrostatic dipole.
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A magnetic circuit: ferrite + air
