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Unformatted text preview: 1 Chapter 30: Inductance 2 Coil 1 Coil 2 A variable current I 1 flows in coil 1. I 1 then induces a current in coil 2. 1 21 2 I N ∝ Φ The flux ( Φ 21 ) through coil 2 due to coil 1 is §30.1: Mutual inductance 3 Writing this as an equality 1 21 2 MI N = Φ Where M is the mutual inductance. It depends only on constants and geometrical factors. The unit of inductance is the henry (1H = 1Vs/A). The induced emf in the coils will be dt dI M dt d N dt dI M dt d N 2 12 1 1 1 21 2 2 = Φ = = Φ = ε ε 4 Selfinductance is for when a current carrying coil induces an emf in itself. LI N = Φ The definition of selfinductance (L) is §30.2: Selfinductance and inductors 5 Example: The current in a 0.080 Henry solenoid increases from 20.0 mA to 160.0 mA in 7.0 s. Find the average emf in the solenoid during that time interval. ( ) V 10 6 . 1 s . 7 mA 20 mA 160 H 080 . 3 × =  = Δ Δ = Δ ΔΦ = t I L t N ε 6 Example: Consider a solenoid of length a that carries a current I . The solenoid has circular crosssection of area A . What is the selfinductance of the solenoid if it is filled with air? The flux linking the coils of the solenoid is I a A N NBA d N = = ⋅ = Φ ∫∫ 2 B μ A B The factor multiplying I is a collection of constants and geometrical factorsthis is the self inductance. a A N L 2 μ = 7 Example (30.45): Determine the selfinductance of the toroidal solenoid shown in the figure. r I N B π μ 2 = For a toroid: Determine the flux through a crosssection: = = = ⋅ = Φ ∫ ∫ ∫∫ ∫∫ = = = = a b Ih N dydx x I N dA r I N d h y y b x a x ln 2 2 2 B π μ π μ π μ A B 8 Example continued I a b h N = Φ ln 2 B π μ To find L, compare this result LI N = Φ to = a b h N L ln 2 2 π μ Find that 9 An inductor is essentially a coil of wire. An inductor is essentially a coil of wire....
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 Fall '07
 Fuchs
 Current, Inductance, Inductor, dt, Energy density, Kirchhoff

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