A pot core inductor, similar to the one in Figure 1.28 on page 47 of live text,

Question

A pot core inductor, similar to the one in Figure 1.28 on page 47 of live text, is shown below in cross section. This core is cylindrical with a single air gap of length g. the center leg is a cylinder, while the remainder is a cylindrical shell. the core is constructed in two parts, with the coil inserted before joining the two core sections. Core dimensions: R1=1.50cm, R2= 4.00 cm, R3 = 4.50 cm, h = 0.75 cm, I = 2.5 cm. Core permeability. mu mu Air gap length g = 1.00 mm. Determine the reluctance ot the magnetic circuit (neglecting the core reluctance and neglecting fringing at the air gap). If the coil has N = 250 turns, calculate the coil inductance

Explanation / Answer

Magnetic reluctance is analogous to electrical resistance, and is in units of At/W (Ampere turns per weber). There is an equivalent of Ohms law for magnetic circuits, where: Reluctance = Magneto Motive Force / magnetic flux Where: Reluctance is in units of ampere turns per weber. MMF is in ampere turns. magnetic flux is in webers.. Reluctance is proportional to LengthOfPath / Permeability x CrossSectionalArea. The permeability of air is 1, the permeability of magnetic materials up to a few thousand. Using realistic figures of 0.04m path and area of 0.0001m^2... Air, 40 Iron, 0.16 Therefore air is ~250 time the magnetic reluctance of iron. In the electrical case air is totally non conductive (except when it becomes ionised - breaks down). The resistance ratio between air and a metallic conductor approaches infinity. This may help you..thank you very much

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