# lect3 - Lecture 3 EE743 Lecture 3 EE743 Magnetic Circuits...

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Unformatted text preview: Lecture 3 - EE743 Lecture 3 - EE743 Magnetic Circuits Professor: Ali Keyhani Professor: Ali Keyhani 2 Magnetic Circuits ■ Consider the following coil Φ Φ Φ Area = A i N i V l g l i H B A B BA μ = Φ = = Φ 3 Magnetic Circuits ■ Recall that Ampere’s Law states that the line integral of the field intensity “ H H ” about a closed path is equal to the net current enclosed within this closed path of integration, that is, Ni l H l H Ni dl H dl H Ni dl H g g i i a b g b a i enclosed = + ∫ = ∫ + ∫ = ⋅ 4 Magnetic Circuits ■ Consider a cross section of the magnetic circuit depicted before. ■ The flux density “ B ” is related to the field intensity “H ” by B B H H i i ds ds A A i = 2 m Weber H B μ 5 Magnetic Circuits ■ The permeability of the medium is usually represented by μ . ■ The surface integral of the flux density is equal to the flux Φ , as shown below = ∫ ⋅ = Φ amperes Joules Weber s d B A 6 Magnetic Circuits ■ If an uniform flux density is assumed, the following relationships can be obtained, ■ Substituting for H and B, we will have the following expression Φ = Φ = Φ = ⋅ = Φ ⋅ = Φ i g i g g g g i i i A A A B A B Ni l B l B g g g i i i = + μ μ 7 Magnetic Circuits m H material tic ferromagne the of type the on depending Ni A l A l o rg o rg g ri o ri i g g g i i i 7 10 4 1 4000 500- × = = = → = = = Φ + Φ π μ μ μ μ μ μ μ μ μ μ μ 8 Magnetic Circuits ■ Let ■ The above equations represents the circuit depicted below, ( 29 ( 29 g i g i g g g g i i i i Ni Ni A l A l ℜ + ℜ = Φ = Φ ℜ + ℜ = ℜ = ℜ μ μ 9 Magnetic Circuits ) ( ) ( 2 2 g i g i N i i N i N L ℜ + ℜ = ℜ + ℜ = Φ = ) ( 2 g i N L ℜ + ℜ = g g g g i i i i A l A l μ μ = ℜ = ℜ 10 Magnetic Circuits ■ Example: Example: Assume l g =10-3 m, l i =0.06x 10- 6 m, A i = A g = 0.1 m 2 , μ ri =500 , μ rg =1, μ =4 π x 10-7 H/m H H g i g i 1 10 95 . 7 1 10 96 .....
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lect3 - Lecture 3 EE743 Lecture 3 EE743 Magnetic Circuits...

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