Physics 202. Review Session for Exam3

Quizzes/Exercises: Determine Direction Of emf

  Indicate the direction of emf in the following cases: + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + |B| increases

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

B

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

B

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

|B| decreases

|B| decreases |B| increases

|B| decreases

path outside B

or

Faraday’s Law (Reminder)  The emf induced in a “circuit” is proportional to the

time rate of change of magnetic flux through the “circuit”.

  Notes:   “Circuit”: any closed path

 does not have to be real conducting circuit

  The path/circuit does not have to be circular, or even planar

A

B

θ

nominal direction of ε

Methods to Change Electric Flux

 Change of ΦB emf   To change ΦB:

  Change B emf produced by an induced E field   Change A motional emf   Change θ motional emf   Combination of above

dtBAd

dtd

uniform

)cos( θε −Φ

−= =B

B

Electric Generators

€

ΦB = Bdvt €

ΦB = B(wd − dvt)

€

vt

€

ε = RI; I =εR

; F ∝ I∝ε

Exercise: Calculate Inductance of a Solenoid

 show that for an ideal solenoid:

(see board)

Reminder: magnetic field inside the solenoid (Ch 28) €

L =µ0N

2Al

Area: A

# of turns: N

Energy in an Inductor   Energy stored in an inductor is U= ½ LI2

 This energy is stored in the form of magnetic field: energy density: uB = ½ B2/µ0 (recall: uE= ½ ε0E2)

 Compare: Inductor: energy stored U= ½ LI2 Capacitor: energy stored U= ½ C(ΔV)2

Resistor: no energy stored, (all energy converted to heat)

Basic Circuit Components Component Symbol Behavior in circuit

Ideal battery, emf ΔV=V+-V- =ε Resistor ΔV= -IR

Realistic Battery (Ideal) wire ΔV=0 (R=0, L=0, C=0) Capacitor ΔV=V- - V+ = - q/C, dq/dt =I Inductor ΔV= - LdI/dt

(Ideal) Switch L=0, C=0, R=0 (on), R=∞ (off) Transformer

Future Topics Diodes, Transistors,…

r ε

Turn on LR Circuit

Note: the time constant is τ=L/R

€

I =V0R(1− e

−t

L /R )

Turn off LR Circuit

Note: the time constant is τ=L/R

€

I = I0e−

tL /R

Charging a Capacitor in RC Circuit

Charging Note: τ≡RC is called time constant

Discharging a Capacitor in RC Circuit

discharging €

I(t) = −QRC

e−t /RC

Note the time constant τ=RC

€

q(t) /C + R dq(t)dt

= 0

Energy Stored in a LC Circuit

Reminder Energy stored in: Inductor: energy stored UL= ½ LI2 Capacitor: energy stored UC= ½ C(ΔV)2 = ½ Q2/C

Current And Voltages in a Series RLC Circuit

ΔVmax Sin(ωt +φ)

i= Imax Sin(ωt)

ΔvR=(ΔVR)max Sin(ωt) ΔvL=(ΔVL)max Sin(ωt + π/2) ΔvC=(ΔVC)max Sin(ωt - π/2)

22maxmax )( CL XXRIV −+=Δ

)(tan 1

RXX CL −= −φ

Quiz: Can the voltage amplitudes across each components , (ΔVR)max , (ΔVL)max , (ΔVC)max larger than the overall voltage amplitude ΔVmax ?

€

ΔV = ΔVmax sin(ωt + φ)

Summary of Impedances and Phases

Electromagnetic Waves   EM wave equations:

 Plane wave solutions: E= Emaxsin(kx-ωt+φ) B= Bmaxsin(kx-ωt+φ)  Properties:

  No medium is necessary.   E and B are normal to each other   E and B are in phase   Direction of wave is normal to both E and B (EM waves are transverse waves)   Speed of EM wave:   E/B = Emax/Bmax=c   Transverse wave: two polarizations possible

same φ, set to be 0

x

y

z

E

B c

We’

ll ha

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Momentum Carried By EM Waves

  EM waves: momentum = energy/c

 Radiation Pressure (P):

100% absorption

Δp = p P= S/c

100% reflection

Δp = 2p P= 2S/c €

P = u =Sc

Wavelength and Frequency  Because of the wave equation the wavelength of and

frequency of a EM wave in vacuum are related by:

 Example: Determine the wavelength of an EM wave of frequency 50 MHz in free space

€

λf = c = 3 ⋅108m /s

€

λ =cf

=3 ⋅108m /s50 MHz

=3 ⋅108m /s5 ⋅107s−1 = 6m

Example: Solar Energy  The average intensity of the EM radiation from the

Sun on Earth is S ~ 103 W/m2

  What is the average radiation pressure for 100% absorption:

  What is the force exerted by EM radiation by the Sun on a surface of 1 m2

€

P =Sc

=103W m2

3 ⋅108ms= 3.3 ⋅10−6 N m2

€

F = PA = 3.3 ⋅10−6 N m2 ⋅1m2 = 3.3 ⋅10−6N

Example: Solar Energy (cont)  The average intensity of the EM radiation from the

Sun on Earth is S ~ 103 W/m2

  What is the average radiation pressure for 50% absorption and 50% reflection:

€

P = PA + PR =SAc

+ 2 SRc

=S2c

+ 2S2c

=32Sc

€

P =32Sc

=32103W m2

3 ⋅108ms= 5 ⋅10−6 N m2