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EEE340Lecture 221 Note that the correspondences between fields and circuits are: But and I are governed by Ampere’s circuital law in 90 o. According to (6.23), aligns with is 90 o with Therefore and I are 90 o apart. The magnetic flux through a given area S, bounded by contour C is (6.24, 6.25)

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EEE340Lecture 222 6-4: Biot-Savart Law The differential magnetic due to a differential segment of current where For a closed path C’ of current I In contrast, the electric field (6.33) (6.32)

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EEE340Lecture 223 Example 6-4: Find of a straight wire of length 2L carrying current I. Solution Employing (6.32) where R P(x,y,0) dz’ (0,0,z’) 0 -L L

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EEE340Lecture 224 To integrate, we employ variable transformation Hence Finally, (6-35)

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EEE340Lecture 225 Example 6-5 Find the magnetic flux density at the center of a square loop of side w, carrying current I Solution Using the result from the previous example, Question1. Can you compute the B at the center of a rectangular loop? Question2. Can you compute the B at a point within a rectangular loop?

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EEE340Lecture 226 Example 6-6: Loop wire of radius b carrying current I. Find the -field at the symmetry axis. Loop antenna (dynamic). Solution: Now z R b

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EEE340Lecture 227 Note that is not a constant. The component is cancelled out during the integration.

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EEE340Lecture 228 6-5: Magnetic Dipole A circular current loop of radius b, carrying current I makes a magnetic dipole. For R>>b, the magnetic vector potential at a point P(r, , /2) is where is the magnetic moment, Using we have the B-field (6.45) (6.48)

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EEE340Lecture 229 Equations (6.45) and (6.48) the dual of the electric dipole. where is the electric dipole moment. (3.53b) (3.54)

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EEE340Lecture 2210 6-6: Magnetization and Equivalent current Density The magnetization vector In contrast to the polarization vector The equivalent volume current density and surface current density These two equations are the dual of (A/m) (6.55) (C/m 2 ) (3.79) (6.62) (6.63) (3.89) (3.88)

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