Chapter 28Inductance;Magnetic Energy Storage
Chapter 28 Inductance; Magnetic Energy Storage
Self inductanceEMFElectric currentmagnetic field(changing)(changing)LPhenomenon of self-inductionMagnetic flux p αc current INΦβ= LIL is self inductance of the coil. Unit: Henry (H)
Self inductance Magnetic flux ΦB ∝ current I L Electric current magnetic field EMF (changing) (changing) Phenomenon of self-induction N LI = B L is self inductance of the coil. Unit: Henry (H)
EMFof inductorA coil with significant L: inductordΦdlREMF induced in a inductor:6二dtdt1) L shows the electromagnetic inertia of a coil2) L depends on: geometry & ferromagnetics3) Inductor in AC circuit → reactance /impedance
EMF of inductor 1) L shows the electromagnetic inertia of a coil A coil with significant L: inductor EMF induced in a inductor: B d dI N L dt dt = − = − 2) L depends on: geometry & ferromagnetics 3) Inductor in AC circuit → reactance /impedance
Determine inductance7For a long solenoid:B= μonln=NIl1Total magnetic flux:Nβ = N-μonl.S = μon'I .SI = μon?I.VL = μon?VSelf inductance:How to avoid inductance for a resistor?
Determine inductance For a long solenoid: B nI = 0 Total magnetic flux: S N N nI S = B 0 2 0 = n I Sl 2 0 = n I V Self inductance: 2 L n V = 0 How to avoid inductance for a resistor?
InductanceofcoaxialcableExamplel: Determine the inductance per unitlength of a coaxial cable with thin conductorsSolution: Magnetic field ?B= LolRAmpere's law:R2元r福RuoTn2元R2元1R2ΦμoBIn11R2元
Inductance of coaxial cable Example1: Determine the inductance per unit length of a coaxial cable with thin conductors. Solution: Magnetic field ? R R1 2 I I 0 2 I B r Ampere’s law: = 0 1 2 B I dr r = 0 2 1 ln 2 I R R = 0 2 1 ln 2 B R L I R = =