Lecture 27: FRI 20 MAR - LSU

Lecture 27: FRI 20 MAR - LSU

Physics 2102 Jonathan Dowling Lecture 27: FRI 20 MAR Ch.30.79 Inductors & Inductance Nikolai Tesla Inductors: Solenoids Inductors are with respect to the magnetic field what capacitors are with respect to the electric field. They pack a lot of field in a small region. Also, the higher the current, the higher the magnetic field they produce. Capacitance how much potential for a given charge: Q=CV Inductance how much magnetic flux for a given current: =Li Using Faradays law: Tesla m 2 Units : [ L] = H (Henry)

Ampere Joseph Henry (1799-1878) Self-Inductance L of a Solenoid Solenoid of cross-sectional area A=r2, length l, total number of turns N, turns per unit length n Field inside solenoid: B = 0 n i Field outside = 0 B = NAB = NA 0 ni =Li L = inductance i

di E =L dt Example The current in a L=10H inductor is decreasing at a steady rate of i=5A/s. i If the current is as shown at some instant in time, what is the magnitude and direction of the induced EMF? (a) 50 V (b) 50 V Magnitude = (10 H)(5 A/s) = 50 V Current is decreasing Induced EMF must be in a direction that OPPOSES this

change. So, induced EMF must be in same direction as current The RL circuit Set up a single loop series circuit with a battery, a resistor, a solenoid and a switch. Describe what happens when the switch is closed. Key processes to understand: What happens JUST AFTER the switch is closed? What happens a LONG TIME after switch has been closed? What happens in between? Key insights: If a circuit is not broken, one cannot change the CURRENT in

an inductor instantaneously! If you wait long enough, the current in an RL circuit stops changing! 0, a capacitor acts like a wire; an inductor acts like a broken wire. a long time, a capacitor acts like a broken wire, and inductor acts like a RL circuits In an RC circuit, while charging, Q = CV and the loop rule mean: charge increases from 0 to CE current decreases from E/R to 0 voltage across capacitor increases from 0 to E In an RL circuit, while charging (rising current), emf = Ldi/dt and the loop rule mean:

magnetic field increases from 0 to B current increases from 0 to E/R voltage across inductor decreases from -E to 0 Example Immediately after the switch is closed, what is the potential difference across the inductor? (a) 0 V (b) 9 V (c) 0.9 V 10 W 9V 10 H

Immediately after the switch, current in circuit = 0. So, potential difference across the resistor = 0! So, the potential difference across the inductor = E = 9 V! Example Immediately after the switch is closed, what is the current i through the 10 resistor? (a) 0.375 A (b) 0.3 A (c) 0 40 W 3V 10 W 10 H Immediately after switch is closed, current through inductor = 0.

Hence, current through battery and through 10 resistor is i = (3 V)/(10 ) = 0.3 A Long after the switch has been closed, what is the current in the 40 resistor? (a) 0.375 A Long after switch is closed, potential (b) 0.3 A across inductor = 0. (c) 0.075 A Hence, current through 40 resistor i = (3 V)/(40 ) = 0.075 A Fluxing Up an Inductor How does the current in the circuit change with time? i di

iR + E L = 0 dt E t/ t i = (1 e ) R i(t) Fast/Small t E/R Slow/Large t Time constant of RL circuit: t = L/R t RL Circuit Movie QuickTime and a

Animation decompressor are needed to see this picture. Fluxing Down an Inductor The switch is at a for a long time, until the inductor is charged. Then, the switch is closed to b. i What is the current in the circuit? di LoopiR rule around + L =0 the new circuit: dt

i(t) Exponential defluxing E/R E Rt/ L E t/t i= e = e R R t Inductors & Energy Recall that capacitors store energy in an electric field Inductors store energy in a magnetic field.

di E =iR + L dt di 2 (iE ) =(i R ) + Li dt Power delivered by battery i P = iV = i2R 2 d Li 2 (iE ) =(i R ) + dt 2 = power dissipated by R

+ (d/dt) energy stored in L Inductors & Energy Li2 UB = 2 di P = Li dt Magnetic Potential Energy UB Stored in an Inductor. Magnetic Power Returned from Defluxing Inductor to Circuit. Example The switch has been in position a for a long

time. It is now moved to position b without breaking the circuit. What is the total energy dissipated by the resistor until the circuit reaches equilibrium? 10 W 9V 10 H When switch has been in position a for long time, current through inductor = (9V)/(10) = 0.9A. Energy stored in inductor = (0.5)(10H)(0.9A)2 = 4.05 J When inductor discharges through the resistor, all this stored energy is dissipated as heat = 4.05 J. E=120V, R1=10W, R2=20, R3=30, L=3H. What are i1 and i2 immediately after closing the switch?

What are i1 and i2 a long time after closing the switch? What are i1 and i2 1 second after closing the switch? What are i1 and i2 immediately after reopening the switch? What are i1 and i2 a long time after reopening the switch?

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