Concept Questions

•Calculate flux downwards – we get EMF clockwise•Flux is decreasing•Derivative of flux is negative•EMF is positive clockwise•Current will flow clockwise

A wire, initially carrying no current, has a radius that starts decreasing at t = 0. As it shrinks, which way does current begin to flow in the loop?A) Clockwise B) Counter-clockwise C) No currentD) Insufficient information

Bd

dt

E

The current that flows will then create a magnetic field, which inside the loop, will A) Strengthen it B) Weaken itC) No change D) Insufficient information

•Right hand rule – B-field downwards•Reinforces magnetic field•Tries to keep the B-flux constant

Concept Question

•As magnet falls, some places have magnetic fields that diminish•Current appears, replacing magnetic field•This acts like a magnet, pulling it back up•At bottom end, current appears to oppose change•This repels the magnet, slowing it down

What happens as I drop the magnet into the copper tube?A) Falls as usual B) Falls slowerC) Falls faster D) Floats constantE) Pops back up and out

•Current is only caused by motion of magnet•If motion stops, resistance stops current

•If motion is small, opposition will be small•It doesn’t stop, it goes slowly

N

S

S

N

S

N

What if we used a superconductor?

Concept Question

What is Kirchoff’s law for the loop shown?A) + L (dI /dt) = 0 B) – L (dI /dt) = 0 C) None of the aboveD) I don’t know Kirchoff’s law for switches

0 EdI

Ldt

dI

dt LE I t

LE +

–

L

I

•The voltage change for an inductor is L (dI/dt)•Negative if with the current•Positive if against the current

Concept Question

L

= 12 V

+–

In the steady state, with the switch closed, how much current flows through R2? How much current flows through R2 the moment after we open the switch?A) 0 A B) 6 A C) 3 AD) 2 A E) None of the above

R1 =

2

•In the steady state, the inductor is like a wire•Both ends of R2 are at the same potential: no current through R2

•The remaining structure had current I = /R1 = 6 A running through it

6 A6 A

I = 6 A

R2 =

4

•Now open the switch – what happens?•Inductors resist changes in current, so the current instantaneously is unchanged in inductor•It must pass through R2

I = /R1 = 6 A

6 A

L

= 10 V+–

The circuit at right is in a steady state. What will the voltmeter read as soon as the switch is opened?A) 0.l V B) 1 V C) 10 VD) 100 V E) 1000 V

R1 =

10

•The current remains constant at 1 A•It must pass through resistor R2

•The voltage is given by V = IR

R2 =

1 k

•Note that inductors can produce very high voltages•Inductance causes sparks to jump when you turn a switch off

I =

1 A

1 A 1000 V IR

1000 VV

+–

Loop has unin-tended inductance

V

Concept Question

A capacitor with charge on it has energy U = Q2/2C, but Q is constantly changing. Where does the energy go?A) It is lost in the resistance of the wireB) It is stored as kinetic energy of the electronsC) It is stored in the inductorD) Hollywood!

Concept Question

C

L

Q

I

0 cosQ Q t

1

CL

dQI

dt 0 sinQ t

2

20 cos2C

QU t

C

212LU LI 2 2 21

02 sinLQ t

2

20 sin2L

QU t

C

20

2C L

QU U

C

Energy sloshes back and forth

•Let’s find the energy in the capacitor and the inductor

If the voltage from a source looks like the graph below, about what voltage should it be labeled?A)0 V B) 170 V C) 120 V D) 85 VE) It should be labeled some other way

Concept Question

•Average voltage is zero, but that doesn’t tell us anything•Maximum voltage 170 V is an overstatement•Power is usually proportional to voltage squared

Concept Question

A 60 W light bulb is plugged into a standard outlet (Vrms = 120 V). What is the resistance of the bulb?A) 15 B) 30 C) 60 D) 120 E) 240

2rmsI RP

2rmsV

R

pP

2120 V

60 W

240 R

2rms2

VR

R

Capacitors and Resistors Combined

•Capacitors and resistors both limit the current – they both have impedance•Resistors: same impedance at all frequencies•Capacitors: more impedance at low frequencies

1CX C

max max

max max C

I V R

I V X

Concept QuestionThe circuit at right might be designed to:(A)Let low frequencies through, but block high frequencies(B) Let high frequencies through, but block low frequencies(C) Let small currents through, but not big currents(D)Let big currents through, but not small currents

Impedance TableResistor Capacitor Inductor

Impedance R

Phase 0

VectorDirection

right down up

1CX

C LX L

12 1

2

Inductors are good for(A)Blocking low frequencies (B) Blocking high frequencies (C) Blocking large currents(D)Blocking small currents

max max

max max

max max

C

L

I V R

I V X

I V X

Concept Question1.4 k

2.0 F

60 Hz170 V

?

In the mystery box at right, we can put a 2.0 F capacitor, a 4.0 H inductor, or both (in series). Which one will cause the greatest current to flow through the circuit?A) The capacitor B) The inductor C) bothD) Insufficient information

L= 4.0 H

1.3 k

1.5 kC

L

X

X

•We want to minimize impedance•Make the vector sum as short as possible•Recall, capacitors point down, inductors up•The sum is shorter than either separately

1.4 k

1.5

k

1.3 k

Concept QuestionRC

f

Vmax

L2rms2

RV

Z P

22L CZ R X X

How will XL and XC compare at the frequency where the maximum power is delivered to the resistor?A) XL > XC B) XL < XC C) XL = XC D) Insufficient information

•Resonance happens when XL = XC.•This makes Z the smallest•It happens only at one frequency•Same frequency we got for LC circuit

1CX

C LX L

00

1L

C

0

1

LC

Concept Question

RCL

•In the example we just did, we found only some frequencies get through

1 RC R L

f

Vmax = 5 V

What happens if this is impossible to meet, because 1/RC > R/L?A) The inequality gets reversed, R/L < < 1/RCB) Pretty much everything gets blockedC) Only a very narrow frequency range gets through

Concept Question

When the voltage shown in blue was passed through two components in series, the current shown in red resulted. What two components might they be?A) Capacitor and InductorB) Inductor and ResistorC) Capacitor and Resistor

1tan L CX X

R

•The phase shift represents how the timing of the current compares to the timing of the voltage•When it is positive, the current lags the voltage

•It rises/falls/peaks later•When it is negative, the current leads the voltage

•It rises/falls/peaks earlier

Voltage

Current

Concept Question

1 2

1 2N N

E EN1 =

5000 N2 =

?

V

1 = 10 k

V

V2 =

120

VA transformer has 10,000 V AC going into it, and it is supposed to produce 120 V AC, suitable for household use. If the primary winding has 5,000 turns, how many should the secondary have?A) 120 B) 240 C) 60D) None of the above

2

10,000 V 120 V

5,000 N 2 V

2

120 V

2 VN 60

A wave has an electric field given by What does the magnetic field look like?

A) B)

C) D)

0ˆ sinE

kz tc

B i 0ˆ sinE

kz tc

B k

0ˆ sinE

kz tc

B i 0ˆ sinE

kz tc

B k

0ˆ sinE kz t E j

•The magnitude of the wave is B0 = E0 / c•The wave is traveling in the z-direction, because of sin(kz - t).•The wave must be perpendicular to the E-field, so perpendicular to j•The wave must be perpendicular to direction of motion, to k•It must be in either +i direction or –i direction•If in +i direction, then E B would be in direction j i = - k, wrong•So it had better be in the –i direction

Concept Question

0 0E cB

Concept Questionc f Which of the

following waves has the highest speed in vacuum?A) InfraredB) OrangeC) GreenD) It’s a tieE) Not enough info

Radio WavesMicrowavesInfraredVisibleUltravioletX-raysGamma Rays

Incr

easi

ngf I

ncre

asin

gRed

OrangeYellowGreenBlue

Violet

Cross-Section•To calculate the power falling on an object, all that matters is the light that hits it

•Example, a rectangle parallel to the light feels no pressure•Ask yourself: what area does the light see?

•This is called the cross sectionS P = F P=

If light of intensity S hits an absorbing sphere of radius a, what is the force on that sphere?A) a2S/c B) 2a2S/c C) 4a2S/c

P S c

•As viewed from any side, a sphere looks like a circle of radius a•The cross section for a sphere, then, is a2

Equations for Test 3

End of material for Test 3

Impedance:Transformers:

Speed of Light

Frequency, Wavelength

2 f

1f T

dIL

dtE

1 H 1 V s/A Units:

Inductors:

Power and Pressure

I V P

max maxI V Z2rmsRIP

1 2

1 2N N

E E

1 1 2 2I V I V

83.00 10 m/sc 2k

S P =

F P=

f k c

2F P=

Bd

dt

E

Faraday’s Law