329lect15 - 15 Inductance solenoid, shorted coax 3 2 Given...

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15 Inductance — solenoid, shorted coax Given a current conducting path C ,themagnet icfux Ψ linking C can be expressed as a Function oF current I circulating around C . Ψ I I, E = - L dI dt V ( t )= L dI dt + - ± 3 ± 2 ± 1 0 1 2 3 ± 3 ± 2 ± 1 0 1 2 3 x z IF the Function is linear, i.e., iF we have a linear fux-current relation Ψ= LI, then constant L = Ψ I is termed the self-inductance 1 oF path C ,ane lementary inductor . Di±erentiating the fux-current relation with respect to time t ,and using the Fact that E = - d Ψ dt , we ²nd that the emF oF inductor L is simply E = - L dI dt , which is a voltage rise across the inductor in the direction oF cur- rent I (which, oF course makes L dI dt avo l taged ropinthesame direction, as used in circuit courses). 1 Amu tua linduc tance M 12 ,bycon t ra s
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For an inductor consisting of n -loops, the inductance L n Ψ I , and the emf E measured around n -loops is E = - d dt n Ψ= - L dI dt once again. Example 1: An n -turn coil has a resistance R =1Ω and inductance of 1 μ H. If it is conducting 3 A of current at t =0 ,determ ine I ( t ) for t> 0 . Solution: Current Fow in the resistive n -turn coil will be driven by emf E = - L dI dt matching the voltage drop RI .H en ce - L dI dt = RI dI dt +
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329lect15 - 15 Inductance solenoid, shorted coax 3 2 Given...

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