hw5 - ECE 3025A Problem Set #5 Solutions 8.18. A wire of...

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ECE 3025A Problem Set #5 Solutions 8.18. A wire of 3-mm radius is made up of an inner material (0 <ρ< 2 mm) for which σ =10 7 S/m, and an outer material (2mm <ρ< 3mm) for which σ =4 × 10 7 S/m. If the wire carries a total current of 100 mA dc, determine H everywhere as a function of ρ . Since the materials have diFerent conductivities, the current densities within them will diFer. Electric ±eld, however is constant throughout. The current can be expressed as I = π ( . 002) 2 J 1 + π [( . 003) 2 ( . 002) 2 ] J 2 = π ! ( . 002) 2 σ 1 +[( . 003) 2 ( . 002) 2 ] σ 2 " E Solve for E to obtain E = 0 . 1 π [(4 × 10 6 )(10 7 )+(9 × 10 6 4 × 10 6 )(4 × 10 7 )] =1 . 33 × 10 4 V / m We next apply Ampere’s circuital law to a circular path of radius ρ , where ρ< 2mm: 2 πρH φ 1 = πρ 2 J 1 = πρ 2 σ 1 E H φ 1 = σ 1 2 = 663 A / m Next, for the region 2mm <ρ< 3mm, Ampere’s law becomes 2 πρH φ 2 = π [(4 × 10 6 )(10 7 )+( ρ 2 4 × 10 6 )(4 × 10 7 )] E H φ 2 =2 . 7 × 10 3 ρ 8 . 0 × 10 3 ρ A / m ²inally, for ρ> 3mm, the ±eld outside is that for a long wire: H φ 3 = I 2 πρ = 0 . 1 2 πρ = 1 . 6 × 10 2 ρ A / m 8.20. A solid conductor of circular cross-section with a radius of 5 mm has a conductivity that varies
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hw5 - ECE 3025A Problem Set #5 Solutions 8.18. A wire of...

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