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EEE340Lecture 201 5-5: Power Dissipation and Joule’s Law Circuits Power Fields Power density (5.53)

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EEE340Lecture 202 5-6: Boundary conditions at Conductor-to-Conductor Interface So that Therefore Equation (5.59) may also be derived from and (5.58) (5.59) (Kirchhoff’s current law)

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EEE340Lecture 203 Example 5-3: Two conductor media of 1 and 2 have a current J 1 with 1 to the norm. Find J 2 on the other side of the interface. Solution From (5.58), the tangential condition The normal condition Hence, (5.60) (5.61) (5.62) (5.63)

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EEE340Lecture 204 Chapter 6: Magnetostatic Fields 6-2: Fundamental Postulates of Magnetostatics Magnetic flux density B. or In contrast to electrostatic case, or There is no magnetic “charge,” as a result, magnetic flux lines are closed loops. or Ampere’s circuital law. You may still remember the right-hand rule: ( 6.6) (6.9) ( 6.7) (6.10)

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EEE340Lecture 205 Electro-static Magneto-static

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EEE340Lecture 206 Example 6-1: A long straight wire of radius b that carries a D.C. current I. Determine the magnetic flux density. Solution: Apply Ampere’s Law The D.C. current is uniformly distributed over the Cross-section. Inside the conductor where Hence

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EEE340Lecture 207 The total current enclosed by C 1 Therefore Outside the conductor The total current enclosed by C 2 is I. Therefore (6.11a) (6.11b)

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EEE340Lecture 208 The results are plotted here. Note that Ampere’s requires geometric symmetry. In Ampere’s law, we use contours in Circle or rectangular shape, while in Gauss’s law surfaces can be: Cylindrical Cartesian Spherical B b

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