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

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 .....
View Full Document

{[ snackBarMessage ]}

Page1 / 47

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

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online