Final Exam Study Guid Physic 51

7/23/2019 Final Exam Study Guid Physic 51

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Faraday’s Law of nduction

ε = −dΦB

dt ; ΦB = B A sin(ωt); Φ = ωt;

dΦB

dt = d

dtBA cos Φ; I = ε

R

If B// A then ΦB = BA; If B⊥ A then ΦB =BA cos 900 = 0; If other then ΦB = BA cosΦ

Flux through theurns of coil

ε = −N dΦB

dt ; A = πr2; dA = Lvdt;

ε = B Lv; F = I L × B; P = F v; ε =NBAω

Careful with the sign (decrease -; increase +)

Lenz’s Law The direction of any magnetic induction; UseRHR

Motion emf v L⊥ B)

ε = vBL; F = qv × B; F = |q | vB;v = ωL; dε = vBdL = ωBLdL; V ab =V a − V b = EL = vBL = ε

Motion emf

closed conduc-ion loop)

ε =

(vX B)d l; d = (v × B)d l; dε =

vBdl; ε =

vBdr; v = ωr

Faraday’s LawStationary inte-ration path)

Ed l = − dΦb

dt = ε; ( E is around the

loop then

Ed l = EC = E 2πr)

Displacementurrent in the

egion (betweenlates or parallellates)

iD = dΦE

dt ; q = C v; C = 0A

d ; v = Ed;

q = Cv = EA = ΦE ; ΦE = EA;

jD =

iD

A =

dE

dt ; iC =

dq

dt =

dΦE

dt ;

dE

dt =dT dt

(P A

); R = ρLA

; E 0 = ρJ C = ρI C A

; dE dt

=ωE 0

Total Electricaleld

E = E c + E n

Maxwell’s equa-ions in emptypace (no charger conductionurrent)

Bd l = 0µ0 d

dt

Ed A;

Ed l =

− ddt

Bd A; ic = 0; Qencl = 0; ΦE =

Ed A; ΦB =

Bd A

nterms of theorces

F = q ( E + v × B)



Mutually in-uced emf

ε2 = −M di1dt

; ε1 = −M di2dt

Mutual induc-ance

M = N 2ΦB2

i1= N 1ΦB1

i2; B1 = µ0n1i1 =

µ0N 1i1l

; ΦB2 = B1A = µ0N 1i1A

l ; M =

N 2N 1µ0i1Ai1l

; n1 = N 1l

elf-Inductance L = N ΦB

i ; ΦB = BA = µ0NiA

2πr

elf-Inductancemf

ε = −Ldidt

; N dΦB

i = L di

dt

Energy stored in

n inductor

U = L I

0 idi = 1

2 LI 2; L = 2U I 2

Magnetic energyensity in vac-um

u = B2

2µ0; u = U

V = U

2πrA = 1

2 µ0N 2I 2

(2πr)2

Magnetic energyensity in a ma-erial

u = B2

2µ; µ = K

mµ

0

Time constantor an R-Lircuit

τ = LR

; i = εR

(1 − e−(RL

)t) = I 0e−(RL

)t;L = −Rt

ln(1− iR

ε ); ε = I R

Fraction of theriginal energyonstant

U = U 0e−2(RL

)t; U = 12

Li2 =12 LI 0e−2(R

L)t; U 0 = 1

2 LI 20 ;

How long after closing the switch will the current(or energy) through the inductor reach one-halfof its maximum value. (energy → u = U max

2 =⇒

1−e−t

τ = 1√ 2

; current→ i = imax

2 =⇒ 1−e−

t

τ = 12

)



Angular fre-uency of oscil-

ation in an L-Circuit

ω =

1LC

= 2πf ; f = ω2π

; T = 1f

;

q = Q cos(ωt + Φ); i = −ωQ sin ωt;Q = Cε;

The charge is maximum at t = 0 so q = Q cos ωtwhere ω = 0

nductor-

CapacitorCircuit

Magnetic Energy= 12

Li2; Electric

Energy= q22C ;

12 Li2 + q2

2C = Q2

2C = 12 CV 2;

i = ±

1LC

Q2 − q 2; i = dq

dt; ω = 1

LC ;

q = Q cos(ωt + Φ)

Magnetic Energy = 0 when t = 0; q = Q when

i = 0; q = 0 when i = imax

Under-dampedL-R-C series

ircuit

ω

=

1LC − R2

4L2 ; R =

4LC

One-half the undamped frequency → ω

= 12 ω;

Potential across capacitor → q

C ; potential across

inductor → Ldidt

; Amplitude A(t) = A0e−R

2L t

A sinusoidaloltage

v = V cos ωt; ω = 2πf

A sinusoidalurrent (instan-aneous)

i = I cos ωt

Rectified averagealue of a sinu-oidal current

I rav = 2π

I

Root-mean-quare valuef a sinusoidalurrent

I rms = I √ 2

; I rms =

(i2)av

Root-mean-

quare valuef a sinusoidaloltage

V rms = V √ 2



Amplitude of oltage across resistor, acircuit

V R = I R; vR = iR = (IR)cos(ωt)

Amplitude of oltage across

n inductor, acircuit

V L = IX L; X L = ωL; vL = Ldidt

=L d

dt(I cos ωt) = −IωL sin(ωt); vL =

IωL cos

ωt + π

2

Amplitude of oltage across

capacitor, acircuit

V C = IX C ; X C = 1ωC

; q = I ω

sin(ωt);q = CvC ; vC = I

ωC sin(ωt); vC =

I ωC

cos

ωt − π2

Amplitude of oltage acrossn ac circuit (ineries)

V = IZ ; V = V 2R + (V L−

V C )2;

V = I

R2 + (X L −X C )2

mpedance of anL-R-C series cir-

uit

Z =

R2 + (X L −X C )2 = R2 + (ωL − 1

ωC )2

Phase angle of anL-R-C series cir-

uit

tanΦ = ωL−1

ωC

R = V L−V C V R

Average Powern a resistor

P av = 12

V I = V √ 2

I √ 2

= V rmsI rms;

V rms = I rmsR; P av = I 2rmsR

Power in a resis-or

P (t) = iV = i2R = I 2R cos2(ωt); V =iR

Φ = 0 → cos Φ = 1

Average Powern an Inductor

¶av = 0

Power in an In-uctor

P (t) = iv = iLdidt

=−IL sin(ωt) cos(ωt); v = L d

dtI cos(ωt)

Φ = ±900 → cos Φ = 0

Average Power

n a Capacitor

P av = 0 Φ =

±900

→cos Φ = 0



Average powernto a general acircuit

P av = 12 V I cos Φ = V rmsI rms cos Φ cos Φ is power factor

nstantaneousower (fromource)

P source = vi = V cos(ωt + Φ)I cos(ωt);P sourceave = IV

2 cosΦ = V rmsI rms cos Φ

The maximumnstantaneousower in resis-

ance

P = V I cos2(ωt); P max = V I = 2P av

L-R-C series cir-uit at resonace

X L = X C ; ω0L = 1ω0C

; ω0 = 1√ LC

;

I = V Z

; Z =

R2 + (X L −X C )2;

f 0 =

ω0

2π ; I rms =

V rms

Z ; V R−rms = I rmsR;V L−rms = I rmsX L; V C −rms = I rmsX C

max I → min Z; X L = X C at resonace; Z = R atresonance

Terminal voltagef transformerrimary andecondary

V 2V 1

= N 2N 1

; ε1 = −N 1dΦB

dt ; ε2 = −N 2

dΦB

dt ;

ε1ε2

= N 1N 2

; V 2 = V 1(N 2N 1

)

N 2 > N 1 → V 2 > V 1 → step-up; N 2 < N 1 → V 2 <V 1 → step-down

Current in trans-

ormer primarynd secondary

V 1N 1 = V 2N 2; I 2 = V 2R

; V 1I 1

= R

(N 2

N 1

)2;

P in = P out; N 1N 2

= I 2I 1

; Roff = R

(N 2

N 1)2

Find the ratio; V 2 is standard source; V 1 is operate

(open); To get from V 1 to V 2

eries Resistance I = V Rint+Rload

; P delivered to load =

I 2Rload = V 2Rload

(Rint+Rload)2; Rload = Rint

Filters and theLow-Pass

R = X C = 1ωC

; V out =V C cos(ωt + ΦC ) where ΦC =−π

2 ; V C = IX C = I

ωC ;V signal = V S cos(ωt + ΦS ) where

ΦS = tan−X C R

= tan− 1( −1

ωRC )

Transfer func-ion

V C V S

= 1

ωC

R2+ 1

ω2C 2

= 11+ω2R2C 2



Gauss’s Law forlectric field

Ed A = Qencl

0

Gauss’s Law formagnetism

Bd A = 0

Ampere’s Law

Bd l = µ0(ic + 0dΦE

dt )encl

Faraday’s Law

Ed l = −dΦB

dt

The speed of he waves fromhe wavelength-requency

v = c = λf ; c = 3× 108m/s

peed of electro-magnetic waves

n vacuum

c =

1µ00

Electromagneticwave in vacuum

E = cB;

Ed l = −Ea; dΦB

dt =

Bac; B = 0µ0cE ;

Bd l = µ00dΦE

dt ;

Bd l = Ba; dΦE

dt = Eac

1) Transverse, E ⊥ B, direction E × B; 2) E = cB, E ⊥ B are in phase; 3) In vacuum, definite and un-changing speed c; 4) Polarization, electromagneticwave require no medium

inusoidal Elec-

romagneticwaves

E y(x, t) = ˆ jE max cos(kx − ωt);

Bz(x, y) = k̂Bmax cos(kx − ωt)

Electromagneticwave in vacuum

E max = cBmax

peed of wave c = f λ

inusoidal elec-romagneticlane wave,ropagating in

x direction

E y(x, t) = E max cos(kx + ωt);Bz(x, t) = −Bmax cos(kx + ωt);λ = c

f ; ω = ck

Electromagneticwaves in differ-

nt materials

λ = cf

= vf

; v = c√ KK m

Remember to identify the wavelength of whichmaterials



Energy densityn electric and

magnetic fields

u = 12 0E 2 + 1

2µ0B2; B = E

c =√

0µ0E ;

u = 0E 2

Poynting vectorn vacuum (mag-itude/rate of

nergy flow)

S = 1µ0

E × B; S = 0cE 2; c = 1√ 0µ0

;

S = EBµ0

ntensity of a si-usoidal wave inacuum

I = S av = E maxBmax

2µ0= E 2max

2µ0c =

12

0µ0

E 2max = 12 0cE 2max; I = P

A; A =

2πr2

Flow rate of elec-romagnetic mo-

mentum

1A

dpdt

= S C

= EBµ0c

; dpdt

= EBµ0c2

= S c2

; S av =I

Radiation Pres-ure (averageorce per unitrea due to the

wave)

P rad = S avc

= I c

; wave reflected →P rad = 2S av

c = 2I

c ; F = P radA; F =

ma; m = ρV

Standing Electromagnetic wave → reflected wave

Wave functionsor the super-osition of two

waves

E y(x, t) = E max[cos(kx + ωt) −cos(kx − ωt)]; Bz(x, t) =Bmax[− cos(kx + ωt) − cos(kx − ωt)];cos(A±B) = cos A cos B ∓sin A sin B; E y(x, t) =−2E max sin kx sin ωt; Bz(x, t) =−2Bmax cos kx cos ωt

Nodal planes of E

x = 0, λ2 ,λ,...

Nodal planes of B

x = λ4 , 3λ

4 , 5λ4 , ...

Wavelengthsdepending on nalue)

λn = 2Ln

where n = 1, 2, 3, ..

Frequencies (de-ending on n

alue)

f n = cλn

= nc2L

where n = 1, 2, 3, ..

Final Exam Study Guid Physic 51

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