Chapter 30 Self Inductance, Inductors & DC Circuits Revisited

Chapter 30 Self Inductance, Inductors & DC Circuits Revisited

Self-Inductance and Inductors Self inductance determines the magnetic flux in a circuit due to the circuit s own current. B = LI Every circuit has some self inductance! Self-induced emf A circuit s self-inductance will give rise to an emf if the current is changing with time. E = d B = L di Inductor - a circuit device designed to have a particularly high self-inductance relative to the rest of the circuit. It is typically a solenoid. ) I ~E ~dl = L di

Example: Estimate the self inductance of a 2-cm long solenoid have 200 windings per cm and a radius of 0.3 cm. =µ 0 nin R 2 L = /I = µ 0 nn R 2 =(4 10 7 )(2 10 4 )(4 10 2 ) (0.003) 2 =0.3 mh

Clicker Question CT 33.29 What is the current through the resistor a very long time after the switch is closed? R = 20Ω A) 0 A B) 0.5 A C) 1 A D) 10 A E) other V = 10V L = 10H

Clicker Question CT 33.28 What is the current through the resistor immediately after the switch is closed? R = 20Ω A) 0 B) 0.5 A C) 1 A D) 10 A E) other V = 10V L = 10H

Clicker question What is the voltage at the top and the bo1om of the inductor immediately a8er switch is closed? R = 20Ω V = 10V L = 10H 1. V top = 10 V, V bo,om = 0 2. V top = 0, V bo,om = 0 3. V top = 10 V, V bo,om = 10 V 4. WTF (huh?)?! This seems wrong. Seems to depend on how I look at it

Faraday s Law of induction for the circuit: I ~E ~dl = L di Self-induced inductance causes line integral to be non-zero! E-field no longer conservative! V = 10V R = 20Ω L = 10H Thinking of circuits solely in terms of voltages (and using Kirchoff s Loop Law) is problematic since the electric field is X no longer conservative (it has curl) i V i 6= 0!

Where is the electric field? I E-field in battery E-field in resistor (if there is a current) That s it! No E-field anywhere else!!! ~E ~dl = L di E downward in battery Integral of E-field across resistor + integral of E-field across battery = induced emf. Note: It is not correct to think of a voltage difference existing across the inductor (as many books do) IR E b = L di

Solution to the Circuit: IR E b = L di I(t) = E b R [1 exp( Rt/L)] timescale to reach steady state: = L/R

Follow-up question Immediately after the switch is closed, what is the reading of the voltmeter V 2?

The switch has bin in position a for a long time. What happens when switch is switched to position B? I ~E ~dl = L di IR = L di I(t) = V bat R exp( Rt/L)

Why do you get a spark as you unplug your hair dryer while it is running?

Clicker Ques,on Suppose the two resistors are lightbulbs. The switch has been closed for a long time. When switch is opened, what happens to the 20- Ohm bulb? a) bulb instantly goes out (zero luminosity) b) luminosity of bulb momentarily stays the same and then fades c) Luminosity of bulb instantly increases and then fades

Energetics How much energy is dissipated by the resistor after the switch is in position b? I(t) = V bat exp( Rt/L) R P = I 2 (t)r = I 2 0 R exp[ 2Rt/L] E = Z 1 0 P (t) = I 2 0 R Z 1 0 exp[ 2Rt/L] E = L 2 I2 0 Where does the energy (converted to Ohmic dissipation) come from?

Energy of Magnetic Field The idea is that magnetic fields are a form of energy. Inside the solenoid, there is an energy density u B due to the B field. 1 2 LI2 = u B volume L = µ 0 nn R 2 = µ 0 nnv (V = R 2 d) 1 2 LI2 = 1 2 µ 0(nI) 2 V = 1 2 u B = B2 2µ 0 B 2 µ 0 Always true (general result is based on a much more formal derivation) V

Solar Flare

The LC Circuit Suppose switch closed at t = 0 with capacitor initially charged to Q. What is q(t)? q(t) =Q 0 cos(!t) I = dq L d2 q 2 + q C =0 m d2 x + kx =0 2 E dl = q C = LdI Looks like harmonic oscillator! (mass on spring) r 1 = LC

Energy in an LC circuit oscillates between electric and magnetic similar to mechanical harmonic oscillator (kinetic and potential)

-q +q LRC Circuit: With resistance in the circuit, LC oscillations damp I out. I E d = V C + IR = L di I I = dq L d2 q 2 + R dq + 1 C q =0 Similar to the damped harmonic oscillator: m d2 x 2 + bdx underdamped case: q(t) =q 0 exp( + kx =0 R/(2L)t) cos r 1 LC! R 2 4L 2 t

Back emf in motor circuit Circuits involving motors contain a rotating coil in a B field The rotating coil will generate an emf due to induction. This emf causes the current to be smaller than what it would be if it weren t rotating (Lenz s law). For this reason we call it a back emf.