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Unformatted text preview: 1 Important fact: Magnetic Flux Φ B is proportional to the current making the Φ B All our equations for Bfields show that B α I 2 ˆ 4 r r l d I B d × = v v π µ thru I l d B µ = ⋅ ∫ v v BiotSavart Ampere Inductance I B B ∝ ∝ Φ Flux I B ∝ Φ Flux Assumes leaving everything else the same. i If we double the current I , we will double the magnetic flux through any surface. Inductance 2 SelfInductance (L) of a coil of wire LI B ≡ Φ I L B Φ ≡ This equation defines selfinductance. Note that since Φ B α I, L must be independent of the current I. L has units [L] = [Tesla meter 2 ]/[Amperes] New unit for inductance = [Henry]. Selfinductance An inductor is just a coil of wire. The Magnetic Flux created by the coil, through the coil itself: ∫ ⋅ = Φ A d B B v v This is quite hard to calculate for a single loop. Earlier we calculated B at the center, but it varies over the area. Inductors (coil of wire) 3 Consider a simpler case of a solenoid I L N nI B inside   µ µ = = v Recall that the Bfield inside a solenoid is uniform! N = number of loops L = length of solenoid * Be careful with symbol L ! Inductors I L N nI B inside   µ µ = = v ∫ = = ⋅ = Φ A nI N BA N A d B N B ) ( µ v v L An NnA I L B 2 µ µ = = Φ = SelfInductance Length Inductors 4 Coil 1 Coil 2 I I Inductor 1 consists of a single loop of wire. Inductor 2 is identical to 1 except it has two loops on top of each other. How do the selfinductances of the two loops compare? A) L 2 = 2 L 1 B) L 2 > 2L 1 C) L 2 < 2L 1 Answer: L2 > 2L1 , in fact L2 = 4 L1. The self inductance L increases by 4. L = Φ /i. If we keep i fixed, but double the number N of turns, Φ increases by 4 . Total flux. N doubles, but Φ 1 = BA also doubles because when we double the number...
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This note was uploaded on 12/08/2010 for the course PHYSC 1220 taught by Professor Feiguin during the Fall '10 term at Univeristy of Wyoming Laramie.
 Fall '10
 Feiguin

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