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I have a few questions: 1) Is the induced emf from the inductor, induced on itself or another wire going through it? 2) How is the voltage able to jump from 0 to some other value after the circuit switch has been opened to disconnect the battery? 3) Finally, I thought that the magnetic field could do no work, so how does it store energy, doesn't that mean it is doing work on the charges in the inductor? I don't know if I worded that right, but I hope you understand what I am asking. this might be a stupid question but how is it possible for an inductor to have twice the voltage of a battery once it is disconnected from the battery?! where did this extra voltage come from?

Spring break 2013, more like one man one a lot of smart physics lectures and studying for physics. AND the right hand rule...... How are office hours going to be handled, what with spring break and all?

Seeing as it's pi day today, I'm trying to think of a clever and witty comment to make...but honestly all I can think about is how hungry I am and what I would give to just be eating some pie right now.

Electricity & Magnetism Lecture 18, Slide 1 Physics 212 Lecture 18

Today’s Concepts: A) Induction B) RL Circuits

Hour Exam 2 is October 31st Lectures 9-18

Electricity & Magnetism Lecture 18, Slide 2 From the Prelecture: Self Inductance

Wrap a wire into a coil to make an “inductor”…

dI e = -L dt

Electricity & Magnetism Lecture 18, Slide 4 What this really means:

emf induced across L tries to keep I constant.

dI = -L eL dt

L

current I Inductors prevent discontinuous current changes!

It’s like inertia!

Electricity & Magnetism Lecture 18, Slide 5 CheckPoint 2

Two solenoids are made with the same cross sectional area and total number of turns. Inductor B is twice as long as inductor A

2 2 LB = 0n r z

(1/2)2 2

1 LB = 2 LA Compare the inductance of the two solenoids

A) LA = 4 LB B) LA = 2 LB C) LA = LB D) LA = (1/2) LB E) LA = (1/4) LB

Electricity & Magnetism Lecture 18, Slide 6 How to think about RL circuits Episode 1:

When no current is flowing initially:

VL L  = I = 0 I = V/R R

L R L R

V VBATT BATT At t = 0: At t >> L/R:

I = 0 VL = 0 VL = VBATT VR = VBATT VR = 0 I = VBATT/R (L is like an open circuit) (L is like a wire)

Electricity & Magnetism Lecture 18, Slide 8 CheckPoint 2a In the circuit, the switch has been open for a long time, and the current is zero everywhere. I At time t = 0 the switch is closed.

What is the current I through the vertical resistor immediately after the switch is I = 0 closed? I L (+ is in the direction of the arrow)

A) I = V/R Before: IL = 0 B) I = V/2R C) I = 0 After: IL = 0 D) I = -V/2R E) I = -V/R I = + V/2R

Electricity & Magnetism Lecture 18, Slide 9 RL Circuit (Long Time)

What is the current I through the vertical resistor after the switch has been closed for a long time?

(+ is in the direction of the arrow)

A) I = V/R B) I = V/2R C) I = 0 - + D) I = -V/2R E) I = -V/R

+ -

KVR: After a long time in any static circuit: VL = 0 VL + IR = 0

Electricity & Magnetism Lecture 18, Slide 10 How to Think about RL Circuits Episode 2: V BATT When steady current is flowing initially: then switch is opened

VL L  = I = 0 R

R L R L R

I = V/R

At t = 0: At t >> L/R: I = V /R BATT I = 0 VR = IR VL = 0 VL = VR VR = 0

Electricity & Magnetism Lecture 18, Slide 11 CheckPoint 2b After a long time, the switch is opened, abruptly disconnecting the battery from the circuit. What is the current I through the vertical resistor immediately after the switch is opened?

(+ is in the direction of the arrow) circuit when switch A) I = V/R opened B) I = V/2R I = 0 C) L D) I = -V/2R E) I = -V/R IL = V/R R

Current through inductor cannot change DISCONTINUOUSLY

Electricity & Magnetism Lecture 18, Slide 12 Why is there Exponential Behavior?

- I + VL L  = R dI V = L L R V = IR dt + -

dI L + IR = 0 dt I(t) = I e-tR/ L = I e-t / 0 0 L where  = R

Electricity & Magnetism Lecture 18, Slide 13 I

L R

V V BATT L L  = R

Lecture: Prelecture:

Did we mess up? No: The resistance is simply twice as big in one case.

Electricity & Magnetism Lecture 18, Slide 14 CheckPoint 3a

After long time at 0, moved to 1 After long time at 0, moved to 2

After switch moved, which case has larger time constant? A) Case 1 B) Case 2 L L C) The same  =  = 1 2R 2 3R

Electricity & Magnetism Lecture 18, Slide 15 CheckPoint 3b

After long time at 0, moved to 1 After long time at 0, moved to 2

Immediately after switch moved, in which case is the voltage After switch moved: across the inductor larger? V A) Case 1 VL1 = 2R B) Case 2 R C) The same V V I = V = 3R Before switch moved: R L2 R Electricity & Magnetism Lecture 18, Slide 16 CheckPoint 3c

After long time at 0, moved to 1 After long time at 0, moved to 2

After switch moved for finite time, in which case is the current through the inductor larger? After awhile -t / A) Case 1 1 I1 = Ie B) Case 2

-t / 2 C) The same I2 = Ie

1 > 2 Immediately after: I1 = I2 Electricity & Magnetism Lecture 18, Slide 17 Calculation R The switch in the circuit shown has been 1 R2 open for a long time. At t = 0, the switch is closed. V L R3

What is dIL/dt, the time rate of change of the current through the inductor immediately after switch is closed

Conceptual Analysis Once switch is closed, currents will flow through this 2-loop circuit. KVR and KCR can be used to determine currents as a function of time. Strategic Analysis Determine currents immediately after switch is closed. Determine voltage across inductor immediately after switch is closed.

Determine dIL/dt immediately after switch is closed.

Electricity & Magnetism Lecture 18, Slide 18 Calculation R The switch in the circuit shown has been 1 R2 open for a long time. At t = 0, the switch is closed. V L I = 0 L R3

What is IL, the current in the inductor, immediately after the switch is closed?

A) IL = V/R1 up B) IL = V/R1 down C) IL = 0

INDUCTORS: Current cannot change discontinuously !

Current through inductor immediately after switch is closed is the same as the current through inductor immediately before switch is closed

Immediately before switch is closed: IL = 0 since no battery in loop

Electricity & Magnetism Lecture 18, Slide 19 Calculation R The switch in the circuit shown has been 1 R2 open for a long time. At t = 0, the switch is closed. V L R3

IL(t = 0 +) = 0

What is the magnitude of I2, the current in R2, immediately after the switch is closed? V V V VR R I = I = I = 2 3 A) 2 B) 2 C) 2 D) I2 = R R2 + R3 R + R + R 1 1 2 3 R2 + R3

We know I = 0 immediately after switch is closed R L 1 I R2

V V Immediately after switch is closed, circuit I = R R + R + R 3 looks like: 1 2 3

Electricity & Magnetism Lecture 18, Slide 20 Calculation

R I The switch in the circuit shown has been 1 R2 2 open for a long time. At t = 0, the switch is closed. V L R3

IL(t = 0 +) = 0 I2(t = 0 +) = V/(R1+ R2+ R3)

What is the magnitude of VL, the voltage across the inductor, immediately after the switch is closed?

R R R R R2 + R3 2 3 V = 0 V =V 2 3 V =V A) V L = V B) V L = V C) L D) L E) L R1(R2 + R3 ) R1 + R2 + R3 R1

Kirchhoff’s Voltage Law,

VL - I2 R2 - I2 R3 = 0 VL = I2 (R2 + R3) V VL = (R2 + R3 ) R1 + R2 + R3

Electricity & Magnetism Lecture 18, Slide 21 Calculation R The switch in the circuit shown has been 1 R2 open for a long time. At t = 0, the switch is closed. V L R3

What is dIL/dt, the time rate of change of the current through the inductor immediately after switch is closed VL(t = 0 +) = V(R2+ R3)/(R1+ R2+ R3)

d I V R + R d I d I V R + R d I V A) L = 2 3 B) L = 0 C) L = 2 3 D) L = dt L R1 dt dt L R1 + R2 + R3 dt L

The time rate of change of current through the inductor (dIL /dt) = VL /L

Electricity & Magnetism Lecture 18, Slide 22 Follow Up

R1 R The switch in the circuit shown has been closed 2 for a long time. V L R What is I2, the current through R2 ? 3 (Positive values indicate current flows to the right)

V V (R2R3) V I = + I = 0 I2 = - A) I 2 = + B) 2 C) 2 D) R2 + R3 R2 + R3 R1 + R2 + R3

After a long time, dI/dt = 0 Therefore, the voltage across L = 0

Therefore the voltage across R2 + R3 = 0 Therefore the current through R2 + R3 must be zero!

Electricity & Magnetism Lecture 18, Slide 23 Follow Up 2

R1 R I The switch in the circuit shown has been closed for a 2 2 long time at which point, the switch is opened. V L R3 What is I2, the current through R2 immediately after switch is opened ? (Positive values indicate current flows to the right)

V V V V I = + I = + I = - I = - A) 2 B) 2 C) I 2 = 0 D) 2 E) 2 R1 + R2 + R3 R1 R1 R1 + R2 + R3

Current through inductor immediately after switch is opened is the same as the current through inductor immediately before switch is opened

Immediately before switch is opened: IL = V/R1

Immediately after switch is opened: IL flows in right loop Therefore, IL = -V/R1

Electricity & Magnetism Lecture 18, Slide 24