Electricity, Magnetism, Optic And Heat With Answer.docx

Chapter 1: Electric Charge, Forces and Fields Important Terms: Electric Charge = It is a fundamental property of matter. It is a fundamentally associated with atomic particles, the electron and the proton. Law of Charges or Charge-Force Law = “Like charges repel each other, and unlike charges attract each other” Coulomb = The SI unit of charge which was named after the French physicist, Charles A. Coulomb. Net Charge = It has an excess of either positive or negative charges. Conservation of Charge = The net charge of an isolated system remains constant. Conductors = They are groups of substances that have the ability to transmit electric charge. Valence electrons are loosely bound. Insulators = They are poor electrical conductors. Semiconductors = They are materials whose ability to transmit charge is much less than that of metals but much greater than that of insulators. Electroscope = A device that can be used to demonstrate the characteristics of electric charge. Electrostatic Charging = A process by which an insulator or an insulated conductor receives a net charge. Charging by Friction = The transfer of charge is due to the contact between the materials and depends on the nature of the materials. Charging by Contact or by Conduction = Refers to the flow of charge during the short period of time the electrons are in motion. Charging by Induction = Provide a path by which electrons can escape. Polarization = Separation of charge. Coulomb’s Law = Electric force is directly proportional to the product between two charges and inversely proportional to the square of the distance between the charges. Electric Field = A force per unit charge.

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Electric Lines of Force = The convenient way of graphically representing the electric field pattern in space through. Electric Dipole = It consist of two separate electric charges Gaussian Surface = Consider a single positive electric charge in an imaginary closed surface surrounding it. Gauss’s Law = The net number of electric field lines passing through an imaginary closed surface is proportional to the amount of net charge enclosed within that surface. Particles and electric charge Particles Electron(-) Proton(+) Neutron

Electric charge(C) -1.602x10-19 +1.602x10-19 0

Mass(kg) 9.109x10-31 1.672x10-27 1.674x10-27

Important Equations Quantization of Electric Charge Q = ne Where:

Q= charge(C or coulomb) n = integral charge (coulomb / electron or C/e) e= electron charge (e or electron)

Coulomb’s Law 𝐅𝐞 = Where:

𝒌𝒒𝟏 𝒒𝟐 𝒓𝟐

k = proportionality constant, 9.00 x 109Nm2/C2 Fe = Electric Force, N q = charge, C r = distance between the charges, m

Electric Field 𝐄= Where:

𝐅𝐨𝐧 𝐪𝐨 𝐪𝐨

E = electric field, F/C F = force, N qo = charge, C

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Electric Field Due to a Point Charge q 𝐤𝐪 𝐄 = 𝐫𝟐 Where:

E = electric field, F/C k = proportionality constant, 9.00 x 109Nm2/C2 q = charge, C r = distance between the charges, m

Electric Field Between Two Closely Spaced Parallel Plates 𝟒𝛑𝐤𝐐 𝐄= 𝐀 Where:

E = electric field, F/C k = proportionality constant, 9.00 x 109Nm2/C2 Q = charge, C A = area, m2

Example 1: If you shuffle across a carpeted floor on a dry day and acquire a net charge of -2.0uC, will you have a deficiency or excess of electrons? How many missing or extra electrons will you have? Given: -6

Q = -2.0 µC = -2.0 ×10 C e = -1.60 ×10-19 C

Required: (a) whether you have lost or gained electrons (b) number of missing or excess electron

Solution: (a) Since the sign of your net charge is negative and electrons carry a negative charge, you have acquired an excess of electron. (b) n =

Q e

=

−2.0 ×10−6 C −1.60× 10−19 C/electron

= 𝟏. 𝟑 × 𝟏𝟎𝟏𝟑 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧

Example 2: Two point charges of -1.0x10-6 C and +2.0x10-6 C are separated by a distance of 0.30m. What is the electrostatic force on each particle? q1 = -1.0 µC

Given:

q2 = +2.0µC

Required: q1 = -1.0x10-6 C q2 = +2.0x10-6 C r = 0.30 m

F12 and F21 F3

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a. F12 = F21 = Fe =

kq1 q 2 = r2

(9.00 x 109 N.

m2 ) (1.0x10-6 C)(2.0x10-6 C) C2 = 𝟎. 𝟐𝟎𝐍 (0.3m)2

(b) A configuration of three charges is shown below. What is the electrostatic force on q3? (0,0.30m)

q1 = +2.5µC

q3 = +3.0 µC (0.40, 0) (0,-0.30m)

q2 = +2.5µC

F13 = F23 = Fe =

kq1 q 3 = r2

Force 0.27 N

(9.00 x 109 N.

Fx 4 (0.27 𝑁) = 0.22 N 5

Fy 3 − 5 (0.27 𝑁) = −0.16 N

(0.27 𝑁) = 0.22 N ∑Fx= 0.44 N

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4

0.27 N

m2 ) (2.50 x10-6 C)(3.0x10-6 C) C2 = 𝟎. 𝟐𝟕𝐍 (0.5m)2

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(0.27 𝑁) = +0.16 N ∑Fy= 0 N

R= √Fx2 +Fy2 =√0.442 +02 =0.44N tan θ =

x y

=

.44 .0

= undefined

F3 = 0.44 N ,due E Example 3: What is the magnitude of the repulsive electrostatic force between two protons in a nucleus? Take the distance between the centers of nuclear protons to be 3.0x10-15 m. If these protons were released from rest, how would the magnitude of their initial acceleration compare to that of the acceleration due to the gravity of the earth’s surface? Given: r= 3.0×1015 m q1= q2= +1.60×10-19C mp= 1.67× 10-27 kg

Required: (a) Fe = ? (b) a/g = ?

Solution: 4

a. Fe = b. F

kq1 q 2 = r2

(9.00 x 109 N.

-19 m2 10-19 C)(1.60 x 10 C) 2 ) (1.60 x C =25.66 N

(3.0×10-15 m)

2

25.66 N

𝑎 = me = = 1.53 × 1028 m/s2 p 1.67×10-27 kg a g

=

1.53×1028 m/s2 9.81 m/s2

= 𝟏. 𝟓𝟔 × 𝟏𝟎𝟐𝟕

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Name: Course/Year/Section: Assignment No.: 1. What is the electrostatic force on each particle?

(0,0.30m)

q1 = +2.5µC

q3 = +3.0 µC (0.40, 0) (0,-0.30m)

2.

q2 = +2.5µC

Two charges are attracted by a force of 25 N when separated by 10cm. What is the force between the charges when the distance between them is 50 cm?

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Example 4: Compare the magnitudes of the electric and gravitational forces between a proton and an electron. Express your answer as a ratio of electric force to gravitational force. Given:

Required: Fe

qe = -1.60×10-19C

Fg

(ratio of forces)=?

qp = +1.60×10-19C me = 9.11 × 10-31 kg mp = 1.67 × 10-27kg Solution: Fe =

kqe qp r2

Fg =

Gme mp r2

m2 (9.00 x 109 N. 2 ) (1.602 x 10−19 C)(1.602 x 10−19 C) kq e q p Fe C = = Fg Gme mp (6.67 x 10−11 N. m2) ( 9.11 × 10−31 kg )(1.67 × 10−27 kg) C2 𝟑𝟗 = 𝟐. 𝟐𝟕 × 𝟏𝟎 Example 5: Two point charges are placed on the x- axis as shown. Find the location on the axis where the electric field is zero. q1 = +1.5 µC q2 = +2.0µC x(m) 0 0.6 Given: Required: d = 0.60m q1 = +1.50×10-6C q2 = +2.0×10-6C

x = ? (location of zero E)

Solution:

q1 q2 = x2 (0.6-x)2 1.5 2 = x2 (0.6 − x)2 (0.6 − x)2 2 √ √ = x2 1.5

0.6 − x = 1.15 x 0.6 − x = 1.15x 0.6 = 2.15x 0.6 x= = 𝟎. 𝟐𝟖 𝐦 2.15

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Example 6: What is the electric field at the origin for a 3- charge configuration?

4.0m

Given: q1 = -1.0×10-6C q2 = +2.0×10-6C q3 = -1.5×10-6C r1= 3.50 m r2= 5.00 m r3= 4.00 m

-1.5uC

-5.0m 2uC

3.5m -1.0uC

Required: E=?

m2 9 -6 (9.0×10 N. kq1 C ) (-1.0×10 C) E1 = 2 = = 7.35 × 102 N/C 2 r1 (3.50 m) m2 9 −6 (9.0 × 10 N. kq 2 C ) (+2.0 × 10 C) 𝐸2 = 2 = = 7.20 × 102 N/C (5.0 m)2 r2 m2 9 −6 (9.0 × 10 N. kq 3 C ) (−1.5 × 10 C) 𝐸3 = 2 = = 8.44 × 102 N/C (4.0 m)2 r3 Ex = E1 + E2 = 7.35 ×

102 N C

+ 7.20 × 102 N/C = 1.46 × 103 N/C

Ey = E3 = 8.44 × 102 N/C E = √(1.46 × 103 N/C) + (8.44 × 102 N/C) = 𝟏. 𝟔𝟗 × 𝟏𝟎𝟑 𝐍/𝐂 Example 7: Show that the electric field far from a dipole, on its perpendicular bisector (the x axis) is given by kqd/x3. Given: q1 = + q q2 = - q

Required: E=?

Solution: 𝐸 =

kqd kq (x) kqx = 2∗ = 3 (x) x3 x x

x ≈d 𝐄=

𝐤𝐪𝐝 𝟏 ∝ 𝐱𝟑 𝐱𝟑

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Example 8: The electric field required to ionize air is about 1.0x104N/C when the field reaches this value, the least bound electron begin leaving their molecules, eventually creating lightning. Assume that this value of field E exist between a negatively charged lower cloud surface and the positively charged ground. If we take the storm clouds to squares 10 miles on each side, estimate the total negative charge on the lower cloud surface. Given: E = 1.0×104 N/C d = 10 mi ≈1.6× 104 m

Required: Q=?

Solution:

2 4N 4 EA (1.0×10 C ) (1.6× 10 m) Q= = = 𝟐𝟑 𝐂 m2 4πk 9 4π(9.0×10 N. 2 ) C

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Chapter 2: Electric Potential, Energy and Capacitance Important Terms: Electric Potential Energy = It is the work done on the charge. Electric Potential Difference = It is the change in potential energy per unit positive test charge Volt = The Si unit of electric potential difference is the joule per coulomb, and renamed in honor of an Italian Scientist, Alessandro Volta Voltage = Potential Difference. Electric Potential = It as the same as electric potential difference however it is not measurable because of the choice of reference point Equipotential Surfaces = A plane on which the value of the electric potential is constant. Electron Volt = The kinetic energy acquired by an electron accelerated through a potential difference of exactly 1V. Capacitor = A typical arrangement of conductors. Capacitance = It expresses the charge stored per volt. Farad = The combined unit of capacitance is the coulomb per volt which was named after an English scientist, Michael Faraday. Dielectric = A sheet of insulating material which is placed between the plates and serves a several purpose. Dielectric Constant = It is the characteristics of the capability of the chosen material. Dielectric Permittivity = It is always greater than the permittivity of the vacuum by a factor of K. Equivalent Series Capacitance = The value of a single capacitor that could replace the series combination. Equivalent Parallel Capacitance = The value of a single capacitor that could replace the parallel combination and holds the same total charge when connected to the battery.

Important Equations 10

Electric Potential Difference 𝚫𝐕 = Where:

𝚫𝐔𝐞 𝐖 = 𝐪𝐨 𝐪𝐨

𝚫V = electric potential difference, v ∆Ue = electric potential energy, J = qoEd E = electric field, N/C q = charge, C d = distance, m W = work, J qo = charge, C

Electric Potential Difference between Parallel Plates 𝚫V = Ed Where:

𝚫V = electric potential difference, v E = electric field, N/C d = distance, m

Electric Potential Due to a Point Charge 𝐕= Where:

𝐤𝐪 𝐫

V = electric potential, v k = proportionality constant, 9.00 x 109Nm2/C2 r = distance, m q = charge, C

Electric Potential Energy of a Configuration of Point Charges U = U12 + U23 + U13 + …….. Where:

U = electric potential energy, J

Relationship between Potential and Electric Field 𝜟𝑽 𝑬 = −( ) 𝜟𝒙 Where:

E = electric field, N/C 𝚫V = electric potential difference, v 𝚫x = distance, m

Capacitance Q = CV Where: Q = Charge, C C = Capacitance, f 11

V = voltage, v Capacitance of a Parallel-Plate Capacitor 𝐂= Where:

Є𝐨𝐀 𝐝

C = capacitance A = Area, m2 d = distance, m Єo = permittivity of a vacuum, 8.85 x10-12C2/Nm2

Energy in a Charged Capacitor 𝐔𝐜 = Where:

𝐐𝐕 𝐐𝟐 𝐂𝐕 𝟐 = = 𝟐 𝟐𝐂 𝟐

Uc = energy stored in a capacitor, J Q = charge, C V = voltage, v C = capacitance, f

Dielectric Effect on Capacitance C = KCo Where:

C = capacitance, f K = dielectric constant

Equivalent Capacitance for Capacitors in Series 𝟏 𝟏 𝟏 𝟏 = + + + … .. 𝑪𝒔 𝑪𝟏 𝑪𝟐 𝑪𝟑 Where:

Cs = equivalent capacitance connected in series

Equivalent Capacitance for Capacitors in Parallel Cp = C1 + C2 + C3 + … Where:

Cp = equivalent capacitance connected in parallel

Example 1: A proton is moved from negative to the positive plate of a parallel- plate arrangement. The plates are 1.50 cm apart, and the dielectric field is uniform with a magnitude of 1500 N/C. (a) What is the proton’s potential energy charge? (b)What is the potential difference between the plates? (c) Between the negative plate and a point midway between the plates? (d) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?

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Given: d= 1.5cm = 0.015m E= 1500N/C

Required: UPROTON = ? ∆V = ? ν=?

Solution: 1500N Nm J 𝑏. ∆V = Ed = ( ) (0.015m) = 22.5 = 22.5 = 22.5 V C C C a. ∆V =

U q

U = Vq = 𝑐. ∆V =

22.5J (1.602 x 10−19 C) = 3.62 x 10−18 J (Kinetic Energy) C

1500N = 11.25J/C C(0.0075m)

U = Vq = 11.25J/C(1.602 x 10−19 C) = 1.81 x 10−18 J 1 𝑑. KE = mυ2 2 3.62 x 10−18 =

1 (1.672 x 10−27 ) v 2 2

2(3.62 x 10-18) 𝑣=√ = 6.50 x104 m/s 1.672 x 10−27 Example 2: We want to accelerate an electron to one percent of the speed of light in a space of 2.50mm between a pair of large horizontal, parallel plates. If the top plate is positively charged, what voltage between the plates is required? What is the magnitude of the electric field? Given: c = 3.00 x 108m/s ν = 0.01 x 3.00 x 108m/s = 3.0 x 106m/s d = 2.5mm = 0.0025m

Required: V=? E =?

Solution:

1 2 1 106 m −31 𝑎. KE = mν = (9.109 x 10 kg) (3.0 x ) = 4.10 x10−18 J = U 2 s 2 ∆𝑉 =

∆𝑈 4.10𝑥10−18 = = 25.48𝐽/𝐶 𝑞𝑜 1.609𝑥10−19

𝑏. V = Ed; V 25.48 J/C 10236𝑣 E= = = = 10236N/C d 0.0025 m m 13

Name: Course/Year/Section: Assignment No.: What is the difference in potential between two points, one at 20 cm and the other at 40cm from a charge of 5.5 micro coulomb? Which point is at a higher potential?

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Example 3: In the Bohr model of the hydrogen atom, the electron in the orbit around the proton can exists only in certain circular orbits. The smallest has a radius of 0.0529nm and the next largest has a radius of 0.212nm, what is the potential difference between the two orbits? Which orbit is at a higher potential? Given: r1= 0.0529 nm= 5.29 x 10-11m r2= 0.212nm = 2.12 x 10-10m Required: a. V = ? b. ∆V =?

Solution: kq 9.00 x 109Nm2/C2(1.609 x 10-19C) V1 = = = 27.26 v r 5.29x10-11 kq 9.00 x 109Nm2/C2(1.609 x 10-19C) V2 = = = 6.8 v r 2.12x10-10 ∆V= V1 - V2 = 27.26 - 6.8 = 20.46v Example 4: The water molecule is the foundation of life as we know it. Many of its important properties are related to the fact that it is a polar molecule. Although the molecule’s net charge is zero, it is separated into positive and negative regions. A very simple picture of water molecule along with the charges is given in the diagram below. The distance from each hydrogen atom to the oxygen atom is 9.60x10-11m and the angle between the hydrogen-oxygen bond directions is 104o. What is the total electrostatic energy of the water molecule?

H q1= 5.20x10-20 c

9.6x10-11 m

O

q3= -10.4x10-20 c

104o

H

q2=5.20x10-20 c

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Given: θ= 104o d = 9.60x10-11m Solution: sin

Required: Utotal = ?

1 ℎ 104 2 = 2 9.6x10-11m

h =2 (9.6x10-11m)sin 52 = 1.51x10-10m

UT = U12 + U13 + U23 9.00 x 109Nm2/C2(5.2x10-20C)(-10.4x10-20C) U13 = U23 = = −5.09x10−19 J -11 9.6x10 m

U12 =

9.00 x 109Nm2/C2(5.2x10-20C)2 = 1.61x10−19 J 1.51x10-10m

Utotal = 2(−5.09x10−19 ) + 1.61x10−19 = −8.57x10−19 J Example 5: Under normal atmospheric conditions, the Earth is electrically charged and creates an approximately constant electric field of about 150 V/m pointing down near its surface. (a)How far apart are two equipotential surfaces that have a 1000-volt difference between them? Which has a higher potential – the one farther from the Earth or the one closer? (b)During an approaching lightning storm, the polarity changes and the field can rise to many times the normal value,Under these conditions, if the field is 900 V/m and points up, now how far apart are the two surfaces in (b)? Given: 150 V/m = constant electric Field V = 1000v Solution: E =

∆V ∆x

a. ∆x =

∆V 1000V = = 6.67m E 150V/m

b. ∆x =

1000V = 1.11m 900V/m

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Example 6: What would be the plate area if air-filled 1.0 farad parallel plate capacitor is if the plate separation were 1.0 mm? Would it be realistic to consider building one? Given: C = 1F d = 1mm= 0.001m Solution:

Required: A=?

ϵoA d Cd 1F (0.001m) A = = = 1.13x108 m2 ϵ0 8.85x10-12C2/Nm2 C =

Example 7: Following a heart attack, the heart sometimes beats in erratic (and a useless fashion, called fibrillation). One way to get the heart back to its normal rhythm is to shock it with electric energy supplied by an instrument called a cardiac defibrillator. Experiment show that about 300 J of energy is required to produce the desired effect. Typically, a defibrillator stores this energy in a capacitor charge by 5000 V power supply. (a)What capacitance is required? (b)What is the charge on the capacitor plate? Given: 300J = U 5000v= V

Required: C=? Q=?

Solution: C =

2U 2(300J) = = 2.4 x 10−5 F = 24μF 2 (5000𝑣)2 V

𝑄 = 𝐶𝑉 = (24x10−6 C)(5000v) = 0.12C Example 8: Consider a capacitor with dielectric underneath a computer key. It is connected to a 12volt power supply, has normal plate separation of 3.00mm and a plate area of 0.750cm2. What is the required dielectric constant if the capacitance is 1.10pF? How much charge is stored on the plates under normal conditions? How much charge flows onto the plates (that is, what is the change in charge) if they are compressed to a separation of only 2.00mm? Given: V = 12V d = 3mm = 0.003m (1m)2

A = 0.75cm2 x

(100cm)2

Required: K=? Q=? = 7.5x10-5m2

∆Q = ?

C = 1.1pF d’ = 2mm Solution: 17

∈0A 8.85x10-12C2/Nm2(7.5x10-5m2) Co = = = 2.21x10−13 F d 0.003m K =

C 1.1x10-12 F = = 4.98 Co 2.21x10-13F

Q = CV = 1.1x10-12F(12V) = 1.32x10-11C K∈0A 4.98(8.85x10-12C2/Nm2)(7.5x10-5m2) C = = = 1.65x10−12 F d' 0.002m ′

Q' = C'V = 1.65x10-12F(12V) = 1.98x10-11C ∆Q = Q'-Q = 1.98x10-11C - 1.32x10-11C = 6.60x10-12C Example 9: You have two capacitors, one 2.50 F and the other one is 5.00 F. What are the charges on each and the total charge stored if they are connected across a 12.0- volt battery (a) in series and (b) in parallel. Given: C1= 2.50 F C2 = 5.00 F V= 12.0V

Required: Q1 =? Q2 =? QT =?

Solution: a.

1 1 1 1 1 2+1 = + + = CS C1 C2 2.50 F 5.00 F 5.00 F

CS =

5 F 3

Q = CV =

5 F(12V) = 2.00x10-5C = Q1 = Q 2 3

Q 2.00x10-5C = -6 = 8V C1 2.50 x10-5 F Q 2.00x10 C V2 = = = 4V C2 5.00 x10-6F V1 =

b. CP = C1 + C2 = 2.50 F + 5.00 F = 7.5F QT = CV = 7.5 x10-6F(12V) = 9.0x10-5C VT = V1 = V2 = 12V Q1 = C1V 1 = 2.50 F (12V) = 3.0x10-5 C Q2 = C2 V2 = 5.00 F (12V) = 6x10-5 C 18

Example 10: Three Capacitors are connected in a circuit as shown below. What is the voltage across the capacitor?

C2= 0.60 uF

12V C1= 0.20 uF C3= 0.10 uF

Given: C1 = 0.20F C2 = 0.60F C3 = 0.10F

Required: V1 = ? & Q1 = ? V2 = ? & Q2 = ? V3 = ? & Q3 = ?

Solution: CP = C1 + C3 = 0.20F +0.10F = 0.3F 1 1 1 1 1 1+2 = + = + = CT C2 CP 0.60F 0.3F 0.6F 0.6F CT = = 0.2F 3

C2= 0.60 uF

12v CP= 0.30 uF

QT = CT VT = 0.2 x10-6F (12V)= 2.4x10-6C = Q2 = QP V2 =

Q2 2.4x10-6C = = 4v C2 0.60x10-6F

QP 2.4x10-6C Vp = = = 8v = V1 = V3 CP 0.30x10-6F Q1 = C1 V1 = 0.20x10-6F (8v) = 1.6x10-6C Q3 = C3 V3 = 0.10x10-6F (8v) = 8x10-7C

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Name: Course/Year/Section: Assignment No.: What is the equivalent capacitance of two capacitors, 0.40 micro farad and 0.60 micro farad when they are connected in series and in parallel?

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Chapter 3: Electric Current and Resistance Important Terms: Battery = It is a device that converts chemical energy into electrical energy. Electrolyte = A solution that conducts electricity. Circuit- Any complete loop consisting of wires and electrical devices. Cathode = The electrode that has the larger number of excess electrons. It designated the negative terminal of the battery. Anode = It has still excess negative charge, has a smaller excess than the cathode. It is the positive (+) terminal of the battery. Electromotive force (emf) = The potential difference across the terminals of a battery when it is not connected to a circuit. Terminal voltage = The operating voltage of a battery. Direct current = The flow of electrons is from negative (-) terminal to the positive(+) terminal. Complete Circuit = It forms from a battery or some other voltage source connected to a continuous conducing path. Conventional Current = The direction opposite to the electron flow. Electric Current = The rate of flow of net charge. Ampere = The unit of current is coulombs per second in honor of the French scientist, Andre Ampere. Drift Velocity = The net electron flow is characterized by an average velocity which is smaller than the random velocities of the electron themselves. Resistance = The ratio of the voltage to the resulting current. Ohm’s Law = A resistor that has constant resistance. Ohm = The combined unit of volts per ampere which was named in honor of a German physicist, George Ohm. Resistivity = It is the intrinsic atomic properties which acts as a constant of proportionality for the relationship of resistance to length and area. Conductivity = The reciprocal of resistivity. 21

Temperature Coefficient of Resistivity = It is a constant over a small temperature range. Superconductivity = The electrical resistance of metals and alloys generally decreases with decreasing temperature. Electric Power = The rate at which energy is delivered to the external circuit by the battery. Joule Heat (I2R) Loss = The thermal energy expended in a current – carrying resistor. Kilowatt – Hour (KWH) = The thing that is being paid for electricity measurement.

Important Equations Electric Current

Where:

I = current, A q = charge, C t = time, s

Electric Resistance

Where:

R = resistance, Ω V = voltage, v I = current, A

Resistivity

Where:

ρ = resistivity, Ωm R = resistance, Ω A = area, m2 L = length, m

Conductivity

Where:

σ = conductivity, per Ωm ρ = resistivity, Ωm

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Temperature Dependence of Resistance

Where:

Rf = final resistance, Ω Ro = initial resistance, Ω T = temperature, oC α = temperature coefficient of resistivity, per oC

Electric Power

Where:

P = power, watts R = resistance, Ω V = voltage, v I = current, A

Example 1: There is a steady current of 0.50A in a flashlight bulb for 2.0 min. How much charge passes through the bulb during this time? How many electrons does this represent? Given: I = 0.5 A t = 2mins = 120s

Required: Q=? n=?

Solution: q t q = It = 0.5A (120s) = 60C I =

n =

Q 60C = = 3.75x1020 e e 1.602x10-19C

Example 2: When a 12-volt automobile battery is connected across a resistor of unknown resistance, a current of 2.5 mA results. What is the value of the resistor? Given: V = 12V I = 2.5 mA = 0.0025A

Required: R=?

Solution: 𝑅=

V 12V = = 4800Ω I 0.0025A 23

Example 3: A 1.5 meter insulated extension cord, which consists of two wires to make a complete circuit, is used to connect a lamp to an outlet. If the copper wire of the cord has a diameter of 2.3mm.what is the resistance of the wires at room temperature? Given: L = 1.5m LT = 3m Ø = 2.3mm = 0.0023m ρCu = 1.70x10-8Ωm

Required: R=?

Solution: ρL 1.70x10-8Ωm x 3m 𝑅= = = 𝟏. 𝟐𝟑𝐱𝟏𝟎−𝟐 𝛀 π (0.0023)2 2 A 𝑚 4 Example 4: A platinum wire has a resistance of 0.50 Ω at 0 C. It is placed in a water bath where its resistance rises to a final value of 0.60Ω. What is the temperature of the bath? Given: RO = 0.50Ω Rf = 0.60Ω T1 = 0ºC α = 3.9x10-3/C

Required: Tf = ?

Solution: 𝑅𝐹 = 𝑅𝑂 (1 + 𝛼∆𝑇) TF =

RF −1 RO

α

=

0.6Ω 0.50Ω

-1

3.9x10-3/C

= 𝟓𝟏. 𝟐𝟖𝐨 𝐂\

Example 5: A desktop computer system includes a color monitor and the central processing unit with keyboard, each operating at an average voltage of 120V. a typical 17’’ color monitor has a power requirement of 130 W, and a CPU’s requirement has a power of 960W.(a). Calculate the average current that each component carries.(b). What is the resistance of each component under this operating condition? Given: V = 120V PMONITOR = 130W PCPU = 960W

Required: IMONITOR = ? ICPU = ? R MONITOR = ? R CPU = ?

Solution: P = IV IMONITOR =

PMONITOR 130W = = 1.08 A V 120V 24

ICPU =

PCPU 960W = = 8A V 120V

R MONITOR = R CPU =

V ICPU

V IMONITOR =

=

120V = 111.11Ω 1.08 A

120V = 15Ω 8A

Name: Course/Year/Section: 25

Assignment No.: 1. How long would it take for a net charge of 2.5C to pass a location in a wire so as to produce a steady current of 5.0milliampere?

2. What is the percentage variation of the resistivity copper over the range from room temperature (20 degrees Celsius) to 100 degrees Celsius?

Example 6: A hair dryer is rated at 1200W for 115V. The uniform wire filament breaks near one end, and the owner repairs it by removing one section near the break and simply reconnecting it. 26

The filament is then 10.0% shorter than its original length. What will be the change in the heater’s power output? Given: P = 1200 W V = 115V

Required: PF = ?

Solution: LF = 90% Lo PO 1200W IO = = = 10.43A V 115V Ro =

V 115V = = 11.03Ω I 10.43A

Rf =

ρLF ρ( 0.90LO) = = 0.90R o A A

Vo = Vf = 115V Io R o = If R f Io R o = If (0.90R o ) Io = If (0.90) IF =

10.43A = 11.59A 0.90

Pf = IF Vf = 11.59A (115v) = 1,332.85W Example 7: If the motor of a frost- free refrigerator runs 15% of the time, how much does it cost to operate per month if the power company changes 11с per kWh? Given: PREFRIGERATOR = 500W 30 days 24 hours t = 0.15 ( )( ) = 108hrs/mo 1 month 1 day

Required: Cost of refrigerator per month

Solution: 108hrs 500W ( month )

= 54 1000 54KWH 11¢ 594¢ $1 𝑃𝑟𝑖𝑐𝑒 = ( month ) (KWH) = (month) (100¢) = $5.94/mo KWH =

Example 8: Most modern power plants produce electricity at a rate of about 1.0GW. Estimate to two significant figures how many fewer power plants the state of California would need if all its 27

households switched from the 500watt refrigerators in the previous example to more efficient 400watt refrigerators. Assume exactly 6 million homes with an average of 1.2 refrigerators per home. Given: PP = 1.0GW = 1.0x109W = 1.0x106KW 500

24hrs

Energy Req’t of 500W refrigerator = 1000 (15%) ( day ) = 1.8 KWH/day # of homes = 6.0x106 # of refrigerators/home = 1.2 ref/home Required: # of power plants Solution: Total 1.8KWH 1.2ref KWH day (6,000,000homes) ( ) = 12,960,000 refrigerator home day Switch to 400W 0.80 (12,960,000 12,960,000 1.0x106

KWH KWH ) = 10,368,000 day day

KWH KWH KWH - 10,368,000 = 2, 592, 000 day day day

KW 24hrs KWH/plant ( ) = 2.4x107 plant day day

KWH day # of power plants = = 0.108 plants KWH/plant 7 2.4x10 day 2, 592, 000

Chapter 4: Basic Electric Circuits 28

Important Terms: Equivalent Series Resistances = The sum of the individual resistances. Equivalent Parallel Resistances = The value of a single resistor that could replace any number of resistors and maintain the same current through the battery. Kirchhoff’s Rules = A general method for analyzing multi-loop circuits which embody the conservation of charge and the conservation of energy. Junction (Node) = A point at which three or more wires are joined. Branch = A path connecting two junctions. Kirchhoff’s Junction Theorem = The algebraic sum of the currents at any junction is zero. Kirchhoff’s Loop Theorem = The algebraic sum of the potential differences across all of the elements of any closed loop is zero. RC Circuit = A circuit that is made of a resistor and a capacitor connected in series. Time constant = It is a special value to express the charging time. Ammeter = It measures the current through circuit elements. Voltmeter = It measures the voltage across circuit elements Galvanometer = A basic component of both ammeter and voltmeter. Fuse = Common to older homes which limits the current in a circuit. Circuit Breaker = They are used extensively in wiring new homes which uses bimetallic strip which also limits the current in a circuit. Grounded Plug = It is the large round prong connects with the dedicated grounding wire. Polarized Plug = A two prong plug that fits in the socket only one way because one prong is wider than the other and one of the slits of the receptacle is also larger.

Important Equations 29

Equivalent Series Resistance 𝐑 𝐒 = 𝐑 𝟏 + 𝐑 𝟐 + 𝐑 𝟑 + … .. Where:

R = resistance, Ω

Equivalent Parallel Resistance 1 1 1 1 = + + + … .. RP R1 R 2 R 2 Where:

R = resistance, Ω

Kirchhoff’s Rules ∑ 𝐈𝐢 = 𝟎 𝐢

∑ 𝐕𝐢 = 𝟎 𝐢

Where:

I = Current, A V = Voltage, v

Charging Voltage across a Capacitor in an RC Circuit 𝐕𝐂 = 𝐕𝐎 (𝟏 − 𝐞−𝐭/𝐑𝐂 ) Where:

Vc = voltage of a charging capacitor, v Vo = Voltage of the battery, v t = time, s R = resistance, Ω C = capacitance, F

Time Constant for an RC Circuit τ = RC Where:

τ = time constant R = resistance, Ω C = Capacitance, F

Discharging Voltage across a Capacitor in an RC Circuit 𝐕𝐂 = 𝐕𝐎 𝐞−𝐭/𝐑𝐂 = 𝐕𝐎 𝐞−𝐭/𝛕 Where:

Vc = voltage of a charging capacitor, v Vo = Voltage of the battery, v t = time, s R = resistance, Ω C = capacitance, F τ = time constant

30

Galvanometer Current used as Ammeter Ig = Where:

IR s r + Rs

Ig = galvanometer currnt, A I = current, A Rs = shunt resistance, Ω r = resistance of the galvanometer coil

Galvanometer Current used as Voltmeter Ig = Where:

V r + Rm

Ig = galvanometer currnt, A V= voltage, v Rm = resistance, Ω r = resistance of the galvanometer coil

Example 1: What is the equivalent resistance of three resistors (1.0Ω, 2.0Ω, and 3.0Ω) when they are connected (a) in series and (b) in parallel? (c) What current will be delivered from a 12 – volt battery for each of these arrangements? Given: R1 = 1.0Ω , R2 = 2.0Ω , R3 = 3.0Ω ; V = 12v;

Required: I=?

Solution: Series 𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 = 1 + 2 + 3 = 𝟔Ω VT 12 IT = = = 𝟐𝐀 = I1 = I2 = I3 RT 6 V1 = I1R1 = 2(1Ω) = 2v V2 = I2R2 = 2(2) = 4v V3 = I3R3 = 2(3) = 6v VT = 12v PT = IT VT = 2x12 = 𝟐𝟒𝐖 P1 = I1 V1 = 2x2 = 𝟒𝐖 P2 = I2 V2 = 2x4 = 𝟖𝐖 P3 = I3 V3 = 2x6 = 𝟏𝟐𝐖 PT = 𝟐𝟒𝐖 Parallel

31

1 1 1 1 1 1 1 6 + 3 + 2 11 = + + = + + = = 𝑅𝑇 𝑅1 𝑅2 𝑅3 1 2 3 6 6 𝟔 𝑅𝑇 = Ω 𝟏𝟏 IT =

VT 12 = = 𝟐𝟐𝐀 6⁄ RT 11

VT = V1 = V2 = V3 = 𝟏𝟐𝐕 V1 = R1 V2 I2 = = R2 V3 I3 = = R3 I1 =

12V = 𝟏𝟐𝐀 1Ω 12V = 𝟔𝐀 2Ω 12V = 𝟒𝐀 3Ω IT = 𝟐𝟐𝐀

PT = IT VT = 22A x 12V = 𝟐𝟔𝟒𝐖 P1 = I1 V1 = 12A x 12V = 𝟏𝟒𝟒𝐖 P2 = I2 V2 = 6A x 12V = 𝟕𝟐𝐖 P3 = I3 V3 = 4A x 12V = 𝟒𝟖𝐖 PT = 𝟐𝟔𝟒𝐖

Name: 32

Course/Year/Section: Assignment No.: 1. Three resistors with values 1.0 ohm, 2.0 ohm, and 4.0 ohm are connected in parallel in a circuit with a 6.0v battery. What are the (a) the total equivalent resistance, (b) the voltage across each resistors, and (c) the power dissipated in the resistors?

33

2.. Three resistors with values 5.0 ohm, 2.5 ohm, and 4.0 ohm are connected in series in a circuit with a 36.0v battery. What are the (a) the total equivalent resistance, (b) the voltage across each resistors, and (c) the power dissipated in the resistors?

34

Example 2: What are the voltages across and the current in each of the resistors (R1 to R5) in the figure below? How much power is dissipated in each of the resistors? R1= 6.00 Ώ R5= 2.50 Ώ R2= 4.00 Ώ R4= 2.00 Ώ

R3= 6.00 Ώ

Given:

Required:

R1= 6.00 Ώ R2= 4.00 Ώ R3= 6.00 Ώ R4= 2.00 Ώ R5= 2.50 Ώ

Currents & Voltages

Solution: I I I 1 1 1+3 4 = + = + = = R34 R3 R4 6 2 6 6 R34 = 1.5Ω

I2 = I345

R345 = 2.5 + 1.5 = 4Ω

V2 6𝑣 = = 1.5𝐴 R2 4 V345 6𝑣 = = = 1.5𝐴 = I5 = I34 𝐴 R 345 4

V5 = I5R5 = 1.5(2.5) = 3.75v V34 = I34R34 = 1.5(1.5) = 2.25v = V3 = V4 𝑉3 2.25 = = 𝟎. 𝟑𝟕𝟓𝐀 𝑅3 6 𝑉4 2.25 I4 = = = 𝟏. 𝟏𝟐𝟓𝐀 𝑅4 2 I3 =

1 R 2345 R2345

1 1 1 1 2 + = + = R 2 R 345 4 4 4 = 2Ω

=

PT = VTIT = 24(3) = 72w

RT =R1+ R2345 = 6 + 2 = 8Ω 𝑉𝑇 24 = = 3A = I1 = I2345 𝑅𝑇 8 V1 = I1R1 = 3(6) = 18v V2345 = I2345R2345 = 3(2) = 6v =V2 = V345 IT =

P1 = I1V1 = 3(8) = 254w P2 = I2V2 = 1.5(6) = 9w P3 = I3V3 = 0.375 (2.25) = 0.84375w P4 = I4V4 = 1.25(2.25) = 2.53125w P5 = I5 V5 = 1.5(3.75) = 5.625w PT =72.00000w

35

Example 3: Three resistors are connected in parallel across a 36.0-volt battery. Use Kirchhoff’s rule to calculate (a) the current through each resistor and the battery and (b) the equivalent resistance of the pair.

36v

6.0

3.0

@PT.1 A–B–C=0

2.50

@ Loop 3 3D – 2.5E = 0

@PT.2 C–D–E=0

3(12) – 2.5E = 0 36 E = = 𝟏𝟒. 𝟒𝐀 2.5

@ Loop 1 -6B + 36 = 0

C=D+E = 12 + 14.4 = 26.4A

B = 6A @ Loop 2 6B – 3D = 0 6(6) – 3D = 0

A=B+C = 6 + 26.4 A = 32.4A

36

D = 3 = 𝟏𝟐𝐀

Example 4: What are the voltages across the current in each of the resistors (R1 to R4) in the figure below? (Use the Kirchhoff’s law). 9Ω R3=2.00Ω

2Ω

Given: R3=2.00Ω V2=12.00V Required: Voltages

V2=12.00V 6Ω

6V

36

@PT.1 A–B–C=0

eq. 1

@ Loop 1 -9B – 2A + 12 = 0 2A + 9B = 12 eq. 2 @ Loop 2 6 + 6C – 12 + 2A = 0 2A + 6C=6 2 A + 3C = 3 eq. 3 *1&2 9 (A – B – C = 0) 2A + 9B = 12 9A – 9B – 9C =0 2A + 9B = 12 11A – 9C =12 eq.4

*3&4 3(A + 3C = 3) 11A – 9C = 12 3A + 9C = 9 11A – 9C = 12 14A = 21 A = 3/2 or 1.5A 3

2 (2) +9B = 12 3 + 9B = 12 9B = 12 – 3 B = 1A 3 + 3C = 3 2 3 3 3C = 3 – = 2 2 3 1 C = = A 2(3) 2

37

Name: Course/Year/Section: Assignment No.: Use the Kirchhoff’s law to determine the power dissipated in the resistors.

7.0

10.0

9.0 36 v

6.0

8.0

5.0

38

Example 5: The capacitance resistance in the RC circuit is 6.0 µF and 0.25MΩ, respectively, and the battery has a 12-volt terminal voltage. (a) The capacitor was initially uncharged. What is the voltage across it one time constant after the switch is closed? (b) What are the voltage across the capacitor and the capacitor’s charge at t = 5.0 s? a) Vc = 0.63Vº = 0.63(12) = 7.56v T = Rc = (6 x 10 -6F) (0.25 x 106Ω) T = 1.5s

b) Vc = Vº (1 – e -t/T) = 12(1 - e -5/1.5) = 11.57V Q = CVc = 6 x 10-6 F(11.57) Q = 6.94 x 10 -5C

Example 6: A galvanometer that can safely carry a maximum coil current of 200 µA (called the full-scale sensitivity) has a coil resistance of 50 Ω. It is to be used in an ammeter designed to read currents up to 3.0 A (at full scale). What is the required shunt resistance? Given: Ig = 200µA = 200 x 10 -6A R = 50Ω 1 = 3A

Required: Rs = ?

Solution: Ig =

IRs ; 200 x 10 R+Rs

−6 =

3Rs 50 + Rs

200 x 10 -6 (50) + 200 x 10-6 Rs = 3Rs 0.01

= 3,0003 Rs 0.01 Rs= = 𝟑. 𝟑𝟑 𝐱 𝟏𝟎 − 𝟑Ω 3,0003

39

Example 7: Suppose that the galvanometer in latter problem is to be used instead in a voltmeter with a full-scale reading of 3.0 V. What is the required multiplier resistance? Given: V = 3V

Required: Rm =

Solution: Ig =

V R + Rm 200 x 10-6 = 3 50 + Rm Rm = 3 - 50 200 x 10-6 Rm = 1.49 x 104Ω

40

Name: Course/Year/Section: Assignment No.:

In a flashing neon light circuit, at a time constant of 2.0 s is desired. If you have a 1.0 micro farad capacitor, what resistance should you use in the circuits?

Ig =

V R + Rm

V - Igr Ig 1 gr + 1gRm = V 1gRm = Rm = Rm =

V - Igr Ig 6 – 200 x 10 -6(50) 200 x 10-6

Rm = 3.0 x 104Ω

A galvanometer has a coil resistance of 20 ohm. A current of 200 microamperes deflects the needle through 10 divisions of full scale. What resistance is needed to convert the galvanometer to a full scale 10-volt meter?

41

A voltmeter has a resistance of 300k-ohms. What is the current through the meter when it is connected across a 10 ohm resistor that is wired to a 6.0 volt source?

42

Chapter 5: Magnetism Important Terms: Pole – Force Law or Law of Poles = Like magnetic poles repel each other and unlike magnetic poles attract each other. Magnetic Field = It surrounds every magnet. Electromagnetism = It is the study of interactions between electrically charged particles and magnetic field. Tesla = The combined unit, newton per ampere meter, of magnetic field is named in honor of Nikola Tesla. Right Hand Force Rule = When the fingers of the right hand are pointed in the direction of v and then curled toward the vector B, the extended thumb points in the direction of F for a positive charge. Right Hand Source Rule = If a current carrying wire is grasped with the right hand with the extended thumb pointing in the direction of the current I, the curled fingers indicate the circular sense of the magnetic field direction. Ferromagnetic Materials = Strong magnetic materials where the electron spins do not pair or cancel completely. Magnetic Domains = A strong coupling or interaction between neighboring atom. Magnetic Permeability = It is the core material of the magnet. Hard Iron = Irons that retains some magnetism after being in an external magnetic field. Permanent Magnet = It magnetism can be lost. Curie Temperature = The critical temperature where the domain coupling is destroyed by the increased in thermal oscillations and a ferromagnetic loses its ferromagnetism. Right Hand Force Rule for a Current Carrying Wire = When the fingers of the right hand are pointed in the direction of the conventional current I and then curled toward the vector B, the extended thumb points in the direction of the magnetic force on the wire. Magnetic Moment = It is the quantity IA. DC Motor = It has a continuous rotation for continuous energy output. Cathode Ray Tube = A vacuum tube that is used in an oscilloscope. 43

Mass Spectrometer = It is one form of electric and magnetic field. Magnet = composed of two “centers” of force called “poles”, one or near each end of the magnet these poles are distinguished as positive and negative, these are called the north (N) and south poles (S). This terminology comes from the early used of the magnetic compass. Magnetic Poles = With two bar magnets we notice a pattern of attraction and repulsion between their various ends.

Important Equations: Magnitude of the Magnetic Force on a Moving Charged Particle F = qvBSinθ Where:

F = force, N q = charge, C v = velocity, m/s B = magnetic field, T

Magnitude of the Magnetic Field near a Long, Straight, Current-Carrying Wire μ𝐨 𝐈 𝐁= 𝟐𝛑𝐝 Where:

B = magnetic field, T I = current ,A µo = magnetic permeability of free space = 4π x 10-7Tm/A d = distance, m

Magnitude of the Magnetic Field at the Center of a Circular Loop of CurrentCarrying Wire μ𝐨 𝐈 𝐁= 𝟐𝐫 Where:

B = magnetic field, T I = current, A µo = magnetic permeability of free space = 4π x 10-7Tm/A r = radius, m

Magnitude of the Magnetic Field at the Center of a Solenoid μ𝐨 𝐍𝐈 𝐁= 𝐋

44

B = µonI Where:

B = magnetic field, T I = current ,A µo = magnetic permeability of free space = 4π x 10-7Tm/A L = length, m N = turn or loop n = linear turn density, turn/m = N/L

Magnitude of Force on a Straight, Current-Carrying Wire F = ILBSinθ Where:

F = force, N I = current, A L = length, m B = magnetic field, T

Magnitude of Torque on a Single Current- Carrying Loop τ = IABSinθ Where:

τ = torque, I = current, A A = area, m2 B = magnetic field, T

Magnitude of Torque on a Single Current- Carrying Coil τ = NIABSinθ Where:

τ = torque, I = current, A A = area, m2 B = magnetic field, T N = loops, turns

45

Example 1: A particle with a charge of -5.0 x 10-4C move at a speed of 1.0 x 103 m/s in the + x direction toward a uniform magnetic field of 0.20T in the +y direction. What is the force on the particle just as it enters the magnetic field? Given: Q= -5x10-4C v= 1.0 x 10 -5 m/s β = 0.2 T

Required: F=?

Solution: F= QvβSinθ= (-5x10-4C)( 1.0 x 10 -5 m/s)(0.2 T)Sin 90 = 0.1 N Example 2: The maximum household current in a wire is about 15 A. for simplicity; assume that it is a steady (dc) current (even though it is not). Assume also that this current is a west-to-east direction. What are the magnitudes and the directions of magnetic field it produces 1.0 cm directly below the wire? Given: I= 15A d= 1cm = 0.01 m

Required: β= ?

Solution: B=

μo I 4πx10−7 Tm/AI 2x10−7 TmI 2x10−7 Tm/A(15A) = = = = 𝟑 𝐱 𝟏𝟎−𝟒 𝐓 2πr 2πd d 0.01m

Example 3: A solenoid is 0.30 m long with 103 turns per meter and carries a current of 5.0 A. What is the magnitude of the magnetic field at the center of this solenoid? Compare your result with the field near the single wire in the previous example (notice that this wire carries three times the current of the solenoid in this Example!), and comment. Given: L=0.3 m n= 1000 turns/m I= 5A

Required: B=?

Solution: B=μ𝐨 nI = 4π x 10-7 Tm/A (1000m) (5A) = 0.00628T

46

Name: Course/Year/Section: Assignment No.: 1. A positive charge of 0.25 C moves horizontally to the right at a speed of 2.0x102 m/s and enters a magnetic field directed vertically downward. If it experiences a force of 20N, what is the magnetic field strength?

2. Two long parallel wires separated by 50 cm carry currents of 4.0 amperes each in a horizontal direction. Find the midway between the wires if the currents are (a) in the same direction and (b) in opposite directions.

3. What is the magnitude of the magnetic field of the circular orbit of the electron in a hydrogen atom, which has an orbital radius of 0.0529nm?

47

Example 4: Two long parallel wires separated by a distance d and carry currents I1 and I2 in the same direction.(a) Derive an expression for the magnetic force per unit length on one of the wires due to the other and show that, in keeping with Newton’s third law, the forces on the two wires are equal in magnitude. (b) Show that for these current directions, the forces are mutually attractive, in keeping with Newton’s third law that the forces of an action-at reaction pair must be opposite in direction. Given: I1 I2 d Solution: 𝐹 = 𝑄𝑣𝛽𝑆𝑖𝑛𝜃 𝑄 𝐼 = ; 𝑄 = 𝐼𝑡 𝑡 𝑑 𝐿 𝑣= = 𝑡 𝑡 𝜃 = 90 𝐿 𝐹 = (𝐼𝑡) ( ) 𝛽𝑆𝑖𝑛90 𝑡 𝐹 = (𝐼)(𝐿)𝛽 F

= Iβ L F1 = I1β1 L μo I2 β = 2πd F1 L F1 L

= =

I1 µo I2 2πd

μo I1 I2 2πd

=

F2 L

48

Name: Course/Year/Section: Assignment No.: 1. Two long straight parallel wires 10cm apart carry equal currents pf 3.0 ampere in opposite directions. What is the force per unit length of the wires?

49

Example 5: An orbiting electron possesses an orbital magnetic moment, and the motion can produce a magnetic field. In the Bohr model of the hydrogen atom, the single electron travels in a circle with a radius of 0.0529 nm and has a period of 1.5 x 10-16s. Derive a general expression for the orbital magnetic moment in terms of the orbital radius r, the charge on the electron e, and the orbital period T. What is the magnitude of the orbital magnetic moment of the electron in the hydrogen atom? Solution: a) Τ = IA A = πr2 I= Τ= b) Τ=

q t

=

eπr2 t

e t

=

πer2 t

π(1.609 x 10-19 )(0.0529x10-9 )

2

1.5x10-6 s T= 9.43 x 10−24 Am2

Example 6: One electron is removed from a methane molecule prior to entering the mass spec arrangement. After passing through the velocity selector, the charged molecule has a speed of 1.0 x 103m/s. It then enters the final magnetic field region, in which the field is 6.7 x 10-3T and follows a circular path. The molecule lands on the detector 5.0cm from the entrance to the field. Determine the mass of this molecule. Neglect the mass of the removed electron. Given: V= 1.0 x 103m/s B= 6.7 x 10-3T d= 5cm = 0.05m q= 1.602 x 10-19 C θ = 90 Sol’n: qBr 1.602 x 10−19 (6.7x10−3 )(0.025) m= = = 2.68x10−26 𝑘𝑔 3 V 1.0x10

50

Chapter 6: Electromagnetic Induction Important Terms: Electromagnetic Induction = It is the current induced in a loop caused by an induced electromotive force (emf) due to this process. Magnetic Flux = It is a measure of the number of field lines passing through an area. Faraday’s Law of Induction = It is the time rate of change of the magnetic flux through all the loops. Lenz’s Law = An induced emf in a metal loop or coil gives rise to a current whose magnetic field opposes the change in magnetic flux that produces it. Alternating Current = The polarity of the voltage and the direction of the current periodically change. AC Generators = The electricity used in homes and industry is primarily ac, this is sometimes called as an alternator. Back EMF = It is the induced emf in motors and generators. Transformer = It consists of two coils of insulated wire would on the same (closed) iron core. Primary Coil = When AC voltage is applied to the input coil, the alternating current gives rise to an alternating magnetic flux that is concentrated in the iron core. Secondary Coil = The changing flux also passes through the output coil, inducing an alternating voltage and current in it. Step-up Transformer = If the secondary coil has more windings than the primary coil does (Ns > Np, or Ns/Np > 1), the voltage is stepped up or (Vs > Vp). Step-down Transformer = The secondary coil has fewer turns than the primary and the voltage is stepped down, or reduced and the current is increased. Eddy Currents = The fourth cause of energy loss in the transformer core. Electromagnetic Waves = It is considered as a means of heat transfer. Maxwell’s Equation = A set of equation which combines the electric field and the magnetic field into a single electromagnetic field. Radiation Pressure = The radiation force per area. 51

52

Important Equations: Magnetic Flux Φ = BACosθ Where:

φ = magnetic flux, Tm2 B = magnetic field strength, T A = area, m2

Faraday’s Law of Induction 𝛏 = −𝐍 Where:

𝚫𝚽 𝚫𝐭

ξ = induced emf, v N = loops, turns φ = magnetic flux, Tm2 t = time, s

Generator emf 𝛏 = 𝛏𝐨 𝐒𝐢𝐧𝛚𝐭 = 𝛏𝐨 𝐒𝐢𝐧𝟐𝛑𝐟𝐭 Where:

ξ = induced emf, v ξo = emf, v = NBAω ω = angular velocity, rad/s t = time, s f = frequency, hz N = loops, turns B = magnetic field strength, T A = area, m2

Back emf of a Motor ξb = V - IR Where:

ξb = back emf, v V = voltage, v I = current, A R = Resistance, Ω

Currents, Voltages and Turn Ratio for a Transform 𝐈𝐏 𝐕𝐒 𝐍𝐒 = = 𝐈𝐒 𝐕𝐏 𝐍𝐏 Where:

IP = primary current, A IS = secondary current, A VP = primary voltage, v 53

VS = secondary voltage, v NP = primary turns, turns NS = secondary turns, turns Example 1: Consider a laptop computer’s speaker that is near a single wire carrying a household current. At a distance of 30 cm, maximum current of 10 A would produce a maximum magnetic field of about 7.0x10-6 Tesla. Both the current and the field would reverse direction every half cycle, or Δt= 1/120s. The speaker’s coil consists of 1000 circular wire loops and has a total resistance of 1.0 ohm and a radius of 3.0 cm. Since the coil is small, assume for simplicity that the magnetic field strength at any instant is constant over coils’ area, which is perpendicular to the to the field direction. According to the manufacturer, the coil can carry only 25mA before it is damaged. What is the magnitude of the average induced emf in the coil during one half cycle and the magnitude of the average induced current in the coil? Will the speaker’s coil survive? Given: d = 30cm = 0.3m Imax = 10 A (household) β = 7.0 x 10-6 T ∆t = 1⁄120 S N = 1000 loops R┌ = 1 Ω ┌ = 3cm = 0.03m Imax = 25m A = 0.025A (Manufacturer Limit)

Required: Emf = v = ? I (induced) =?

Solution. V = emf =

N |∆ø| ∆t

∆ø = øf -øi øi = β o A Cosθ = 7.0 x 10-6 (π) (0.03)2 cos 0 øi = 1.97 x 10-8 Tm2 / loop øf = - 1.97 x 10-8 Tm2 / loop ∆ø = øf -øi = - 1.97 x 10-8 - 1.97 x 10-8 = -3.94x10-8 Tm2 / loop V = emf =

N ⁄ ∆ø ⁄ 1000(−3.94𝑥10−8 ) = = 4.728𝑥10−3 𝑣 1⁄ ∆t 120 𝑠

𝑉 4.728𝑥10−3 𝐼= = = 4.728𝑥10−3 𝐴 = 4.728𝑚𝐴 < 25𝑚𝐴 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑏𝑟𝑒𝑎𝑘 𝑅 1

54

55

Example 2: A farmer decides to use a waterfalls on his property to create a small hydroelectric power plant. He builds a coil consisting of 1500 closely packed circular loops of wire with a radius of 20 cm and constructs his generator to spin at 60Hz in a magnetic field. To generate a maximum emf of 120 V, what must be the magnitude of the magnetic field? Given: N = 1500 loops T = 20cm = 0.2m F = 60hz Emf v = 120V

Required : β=?

Solution : V = ε = π (0.2)2 (1500) (60) β = 120v β = 1.69 x 10-3T Example 3: A dc motor with a resistance of 8.0 ohm in its windings operates on 120 V. With a normal load, there is a back emf of 100 V when the motor reaches full speed . What are (a) the starting current drawn by the motor and (b) the armature current at operating speed under a normal load? Given: R = 8Ω V = 120V εb = 100V

Required Ia =?

Solution : V = IR I20 = Ia (8) 120

Ia = 8 = 15A

ε

= V – IR 100 = 120 – Ia (8) 20 = 8𝐼𝑎 Ia = 2.5 A Example 4: A transformer has 50 turns on its primary coil and 100 turns on its secondary coil. (a) if the primary is connected to 120- volt source, what is the voltage output of the secondary? (b) If the transformer is operated in reverse and the 120-volt input is applied to the 100-turn coil, what would be the voltage output? Given: Np = 50 turns Ns = 100 turns Vp = 120 volt

Required: Vs = ?

56

Solution: 100

Vs

100

120 Vs 2 = 120 Vs = 240v

50

=

50

=

2=

120

Vs 120 Vs

Vs = 60v

Example 5: A small hydroelectric power plant produces energy in the form of electric current at 10 A and a voltage of 440 V. The voltage is stepped up to 4400 V (by an ideal transformer) for the transmission over 40 km of power line, which has a resistance of 0.50Ώ/km. (A power transmission line has two wires for a complete circuit. (a) What percentage of the original energy would have been lost in transmission if the voltage had not been stepped up? (b) What percentage of the original energy is lost with the voltage stepped up? Given:

I = 10 A V = 440v v = 4400v L = 40km

Required:

ρ=?

LT = 80km. R/L = 0.5Ω⁄km RT = 40Ω

Solution: PL = I2 R = 102(40) = 4000 watt P = IV = 10(440) = 4400 watt %PL = 10𝐴 𝐼𝑠

=

4400 440(10) 440

;

4400

4000 x 100 = 𝟗𝟎. 𝟗𝟏% 4400

= 𝐼𝑠 = 𝟏𝐀

PL = I2 R = 12 (40) = 40w %PL =

40 x100 = 𝟎. 𝟗𝟏% 4400

Example 6: Viking space probes landed on Mars in 1976 and sent radio and TV signals back to Earth. How much longer did it take for a signal to reach us when Mars was farthest from Earth than when it was closest to us? The average distance of Mars and Earth from the Sun are 229 million km and 150 million km respectively. Assume circular orbits with these average distances as radii. Given: d=229x109 m, 150x109 m

Required:

𝑡 =? 57

dM = 229x109 m dE = 150x109 m Solution: ∆d = (dM + dE ) − (dM − dE ) = 2dE = 𝟑𝟎𝟎𝐱𝟏𝟎𝟗 𝐦

c = speed of light = 3.0x108

c=

d t

∆d 300x109 1min. t= = = 1000x = 𝟏𝟔. 𝟔𝟕𝐦𝐢𝐧 c 3.0x108 60sec.

58

Chapter 7: AC Circuits Important Terms: Peak Voltage = The voltage oscillates between values of positive Vo and negative Vo. Peak Current = The current oscillates between values of positive Io and negative Io rms Current = effective current rms Voltage = effective voltage Capacitive Reactance = The quantitative measure of the opposition of a capacitor to current. Inductance = It is the measure of the opposition a circuit element presents to a time varying current. Inductor = It is a coil that exhibits a reverse voltage, or back emf, in opposition to the changing current. Henry = The combined unit, volt-second per ampere, for inductance named after Joseph Henry. Inductive Reactance = The opposition that an inductor presents to the current in an AC circuit depends on the value of the inductance and on the frequency of the voltage. Phase Diagram = The resistance and reactance of the circuit are given in a vectorlike properties. Phasors = The magnitude of the resistance and reactance represented by an arrow. Impedance = The phasor sum is the effective, or net, opposition to the current. Phase Angle = The angle between the voltage from the source and the current in the circuit. Inductive Circuit = If the inductive reactance is greater than the capacitive reactance, the phase angle is positive. Capacitive Circuit = If the capacitive reactance is greater than the inductive reactance, the phase angle is negative. Power Factor = It is the ratio of the resistance to the impedance. Resonance Frequency = The frequency for the condition of minimum impedance. 59

Important Equations: Instantaneous Voltage in an AC Circuit V = VOSinωt = VOSin2πft Where:

V = instantaneous voltage, v VO = peak voltage, v ω = angular velocity, rad/s t = time, s f = frequency, hz Instantaneous Current in an AC Circuit I = IOSin2πft

Where:

I = instantaneous current, A IO = peak current, A t = time, s f = frequency, hz Effective Current and Power IO = √𝟐𝐈rms VO = √𝟐𝐕rms

Where: VO = peak voltage, v Vrms = effective voltage, v Irms = effective current, A IO = peak current, A Effective Power in an AC Circuit P = I2rmsR Relationship between Voltage and Current in an AC Circuit Vrms = IrmsR Where: Vrms = effective voltage, v Irms = effective current, A R = resistance, Ω Capacitive Reactance 𝟏 𝟏 𝐗𝐂 = = 𝟐𝛑𝐟𝐂 𝛚𝐂 Where:

XC = capacitive reactance, Ω C = capacitance, F f = frequency, hz 60

ω = angular velocity, rad/s Relationship between Voltage and Current in an AC Circuit V = IXC Where: V = voltage, v I = current, A XC = capacitive reactance, Ω Inductive Reactance XL = 2πfL = ωL Where:

XL = inductive reactance, Ω L = inductance, H f = frequency, hz ω = angular velocity, rad/s Relationship between Voltage and Current in an AC Circuit V = IXL

Where: V = voltage, v I = current, A XL = inductive reactance, Ω Ohm’s Law Generalized to AC Circuits V = IZ Where: V = voltage, v I = current, A Z = impedance, Ω Impedance for a Series RLC Circuit 𝐙 = √𝐑𝟐 + (𝐗 𝐋 − 𝐗 𝐂 )𝟐 Z = impedance, Ω R = resistance, Ω XL = inductive reactance, Ω XC = capacitive reactance, Ω Where:

Phase Angle Between Voltage and Current in a Series RLC Circuit 𝐗𝐋 − 𝐗𝐂 𝐓𝐚𝐧𝚽 = 𝐑 φ = phase angle, o R = resistance, Ω XL = inductive reactance, Ω Where:

61

XC = capacitive reactance, Ω Power Factor of a Series RLC Circuit 𝐑 𝐂𝐨𝐬𝚽 = 𝐙 Z = impedance, Ω R = resistance, Ω Φ = phase angle, o

Where:

Power in Terms of Power Factor 𝐏 = 𝐈𝐕𝐂𝐨𝐬𝚽 = 𝐈𝟐 𝐙𝐂𝐨𝐬𝚽 Where:

P = power, watts I = current, A V = voltage, v Cosφ = power factor Z = impedance, Ω Resonance Frequency of a Series RLC Circuit 𝟏 𝒇𝒐 = 𝟐𝝅√𝑳𝑪

Where:

fo = resonance frequency, hz L = inductance, H C = capacitance, F

Example 1: A lamp with a 60watt bulb is plugged into a 120 V outlet. (a) What are the Irms and peak currents through the lamp? (b) What is the resistance of the bulb? (Neglect the resistance of the wiring.) Given:

Required: Irms Ipeak

P = 60watts V= 120v Solution: A) Irms =

P V

=

60 120

= 0.5𝐴

B)

R =

V 120 = = 240Ω I 0.5

Ipeak = √2 Irms = √2 (0.5A) = 0.71A Example 2: A 15.0 micro farad capacitor is connected to a 120V, 60Herts source. What are (a). The capacitive reactance and (b). The current (rms and peak) in the circuit? Given: C = 15μF = 15x10−6 F V = 120v

Required: Irms Ipeak 62

F = 60hz Solution: a. xc =

1 1 = = 2πfc 2𝜋(60)(15𝑥10−6 𝐹)

𝑏. Irms =

V Xc

=

176.84Ω

120 = 0.68A 176.84

Ipeak = √2 Irms = √2(0.68) = 0.96A Example 3: A 125-mH inductor is connected to a 120-volt, 60-Hertz source. What are (a) the inductive reactance? And (b) rms current in the circuit? Given: L = 125mH = 125x10−3H V = 120v f = 60Hz

Required: XL Irms

Solution: 𝑎. XL = 2πfL = 2π (60) (125x10−3 ) = 47.12Ω b. V = XL Irms 120 = 47.12 Irms Irms = 2.55A Example 4: A series RC circuit has a resistance of 100 ohm and a capacitance of 15.0 micro farad. (a). what is the (rms current) in the circuit when driven by a 120 volt 60-H source? (b) Compute the (rms) voltage across each circuits element and the two element combined. Compare it to that of the voltage source. Is Kirchoff’s loop theorem satisfied? Given: R = 100Ω C = 15 Uf = 15x10-6 F V = 120 V f = 60 Hz

Required: Irms

Solution: Xc =

1 1 = = 176.84Ω 2πfc 2π(60)(15X10−6 )

Z = √1002 + 176.842 = 203.16 Ω 63

Irms =

V 120 = = 0.59 A Z 203.16

VR = IR = 0.59 (100) = 59V VC = IXC = 0.59 (176.84) = 104.34 163.34 ≠ 120 = √592 + 104.342 = 119.87 Example 5: A series RLC circuit has a resistance of 25.0 ohm, capacitance of 50 microfarad, in an inductance of 0.300 H. if the circuit is driven by a 120-volt, 60-hz source, what are (a) the impedance of the circuit (b) the current in the circuit, and (c) the phase angle between the current and the voltage. (d) How much power is dissipated in the circuit? Given: R = 25 Ω C = 50 uF = 50×10-6 F L = 0.3 H V = 120 V f = 60 hz

Required: Z I Ø P

Solution: Xc =

1 = 53.05 Ω 2π(60)(50X10−6 )

XL = 2π( 60 )(0.3) = 113.1 Ω Z = √252 + (113.1 − 53.05)2 = 65.05 Ω Irms =

120 = 1.84 A 65.05

TanØ =

XL − 𝑋𝐶 113.1-53.05 = 𝑅 25

Ø = 67.40°

P = IV cos Ø = 1.84 (120)cos 67.40 = 84.85 W

64

Example 6: A series RCL circuit has a resistance of 50.0 ohms, capacitance of 6 nanofarad, and an inductance of 28 mH. The circuit is corrupted to a wide range adjustable frequency – voltage source with an output of 25 volts. (a) What is the resonance frequency of the circuit? (b) How much current is in the circuit when it is resonance? (c) What is the voltage across each circuit element for this condition? Given: Required: R = 50 Ω f C = 6nF = 6×10-9 F Z L = 28m H VR V = 120 V Solution: f =

1 2π√LC

=

1 2π√28 × 10−3 (6.0 × 10−9 )

= 12,279.07 hz

Z = √R2 + (×L −×C )2 ×L = ×C Z= R = 50 Ω V

𝐼=Z

= 25 = 50

0.5 A

VR = IR = 0.5 (50) = 25 V VC = VL XL = 2π(12,279.07)(28 × 10−3 H ) = 2160.25 Ω VC =VL = XL(I) = 2160.25 x 0.5 = 1080.12v

65

Chapter 8: Geometrical Optics: Reflection and Refraction of Light Important Terms: Wave Front = It is a line or surface defined by adjacent portions of a wave that are in phase. Plane wave front = The curvature of a short segment of a circular or spherical wave front is small. Such a segment might be approximated as a linear wave front in two dimensions or a plane wave front. Ray = A line drawn perpendicular to a series of wave fronts and pointing in the direction of propagation. Geometrical optics = The use of the geometrical representations of wave fronts and rays to explain phenomena. Reflection = It involves the absorption and re-mission of light by means of complex electromagnetic vibrations in the atoms of the reflecting medium. Angle of incidence = It is the angle between a ray incident on a surface and the line perpendicular to the reflecting surface. Angle of reflection = It is the angle measured from the reflected ray to the surface normal. Law of Reflection = It defines the relationship of angle of reflection and incidence. Regular (Specular) Reflection = When the reflecting surface is smooth, the reflected rays of a beam of light are parallel. Irregular (Diffuse) Reflection = If the reflecting surface is rough, the light rays are reflected in non parallel direction because of the irregular nature of the surface. Refraction = It is the change in direction of a wave at a boundary where the wave passes from one medium into another. Huygens Principle = Every point on an advancing wave front can be considered to be a source of secondary waves, or wavelets, and the line or surface tangent to all these wavelets defines a new position of the wavefront. Angle of Refraction = The change in the direction of wave propagation. Index of Refraction = It is the relationship between the speed of light in a vacuum and the speed of light in a medium. 66

Critical Angle = It is the limit for a certain angle of incidence. Total Internal Reflection = It is an optical phenomenon that occurs when a ray of light strikes a medium boundary at an angle larger than the critical angle with respect to the normal to the surface. Fiber Optics = It is a fascinating field centered on the use of transparent fibers to transmit light. Dispersion = It is a transparent material with different indices of refraction for different wavelengths of light, in refracting there is a separation of the wavelengths. Indices of Refraction of some substances Substance Air Water Ethyl Alcohol Fused Quartz Glycerine Polystyrene Oil (typical value) Glass (by type) Crown Flint Zircon Diamond

n 1.00029 1.33 1.36 1.46 1.47 1.49 1.50 1.45-1.70 1.52 1.66 1.92 2.42

Important Equations: Law of Reflection θi = θr Where:

θi = angle of incidence Θr = angle of reflection Snell’s Law 𝐒𝐢𝐧 𝛉𝟏 𝐯𝟏 = 𝐒𝐢𝐧 𝛉𝟐 𝐯𝟐 𝐧𝟏 𝐒𝐢𝐧 𝛉𝟏 = 𝐧𝟐 𝐒𝐢𝐧 𝛉𝟐

Where:

v = velocity, m/s n = index of refraction

67

Index of Refraction 𝐜 𝛌 𝐧= = 𝐯 𝛌𝐦 Where:

n = index of refraction c = speed of light, 3.0 x 108m/s v = velocity, m/s Critical Angle at Boundary between Two Materials (where n1 > n2) 𝐧𝟐 𝐒𝐢𝐧𝛉𝐂 = 𝐧𝟏

Where:

n = index of refraction Critical Angle at Medium – Air Boundary 𝟏 𝐒𝐢𝐧𝛉𝐂 = 𝐧

Where:

n = index of refraction

Example 1: What is the speed of light in water? Given: n = 1.33 c = 3.0x108m/s

Required: v =?

Solution: 3.0x108 𝑣= = 2.26x108 𝑚/𝑠 1.33 Example 2: Light in air is incident on a piece of crown glass at angle of 37.00 (relative to the normal). What is the angle of reflection of the glass? Given: 𝑛𝑎𝑖𝑟 = 1.00 𝑛𝑔𝑙𝑎𝑠𝑠 = 1.52 𝜃𝑎𝑖𝑟 = 37.00° Required” 𝜃𝑔𝑙𝑎𝑠𝑠 =? Solution: 𝑛1 𝑆𝑖𝑛𝜃1 = 𝑛2 𝑆𝑖𝑛𝜃2 1.00𝑆𝑖𝑛37 = 1.52𝑆𝑖𝑛𝜃2 𝜃2 = 23.33° 68

Example 3: A beam of light traveling in air strikes the plate-glass top of a coffee table at an angle of incidence of 45.00 the glass has an n= 1.5 a.) What is the angle of refraction for the light transmitted into the glass? b.) if the glass plate is 2.0 cm thick, what is the lateral distance between the point where they enters the glass---x in the figure? c.) prove that the emergent beam is parallel to the incident beam, that is θ4= θ2 Given: 𝑛𝑎𝑖𝑟 = 1.00 𝑛𝑔𝑙𝑎𝑠𝑠 = 1.5 𝜃𝑎𝑖𝑟 = 45.00° thickness = 2cm Required” 𝜃𝑔𝑙𝑎𝑠𝑠 =? X=? 𝜃1 = 𝜃4 𝑝𝑟𝑜𝑣𝑒 Solution: 𝑎. 𝑛1 𝑆𝑖𝑛𝜃1 = 𝑛2 𝑆𝑖𝑛𝜃2 𝑛1 𝑆𝑖𝑛𝜃1 1𝑆𝑖𝑛45 𝜃2 = 𝑆𝑖𝑛−1 [ ] = 𝑆𝑖𝑛−1 [ ] = 28.13° 𝑛2 1.5 𝑏. 𝑇𝑎𝑛𝜃2 =

𝑦 𝑥

2 𝑇𝑎𝑛28.13° = 𝑥 2 𝑥= = 3.74𝑐𝑚 𝑇𝑎𝑛28.13° 𝑐. 𝑛1 𝑆𝑖𝑛𝜃1 = 𝑛2 𝑆𝑖𝑛𝜃2 = 𝑛4 𝑆𝑖𝑛𝜃4 1.0𝑆𝑖𝑛45 = 1.5𝑆𝑖𝑛28.13° = 1𝑆𝑖𝑛𝜃4 𝜃4 = 45° Example 4: What is the critical angle for light traveling in water and incident on a waterair boundary? (b)if a diver submerged in a poll looked up at the water surface at an angle of θ< θc what would see?(Neglect ant thermal or motional in the water). Given: 𝑛𝑎𝑖𝑟 = 1.00 𝑛𝑤𝑎𝑡𝑒𝑟 = 1.33

Required: 𝜃𝑐 =?

Solution: 𝑆𝑖𝑛𝜃𝑐 =

𝑛1 𝑛2

𝜃𝑐 = 𝑆𝑖𝑛−1 [

𝑛1 1 ] = 𝑆𝑖𝑛−1 [ ] = 48.75° 𝑛2 1.33 69

Example 5: The index of refraction of a particular transparent material is 1.4053 for the red end (λ= 700nm) of the visible spectrum and 1.4698 for the blue end ( λ= 400nm). If white light is incident on a prism of this material at an angle of 45.00, what is the angular dispersion of the visible spectrum inside the prism? Given: Red End 𝑛1 = 1.00 𝜃1 = 45° 𝑛2 = 1.4053

Blue End 𝑛1 = 1.00 𝜃1 = 45° 𝑛2 = 1.4698

Required: 𝜃2 =? Solution: 𝑎. 𝑛1 𝑆𝑖𝑛𝜃1 = 𝑛2 𝑆𝑖𝑛𝜃2 1𝑆𝑖𝑛45 = 1.4053𝑆𝑖𝑛𝜃2 1𝑆𝑖𝑛45 𝜃2 = 𝑆𝑖𝑛−1 [ ] = 29.18° 1.4053 𝑏. 𝑛1 𝑆𝑖𝑛𝜃1 = 𝑛2 𝑆𝑖𝑛𝜃2 1𝑆𝑖𝑛45 = 1.4698𝑆𝑖𝑛𝜃2 1𝑆𝑖𝑛45 𝜃2 = 𝑆𝑖𝑛−1 [ ] = 28.76° 1.4698

70

Chapter 9: Mirrors and Lenses Important Terms: Mirrors = Are smooth reflecting surfaces: usually made of polished metal or glass that has been coated with some metallic substances. Image = It is the visual counterpart of an object produced by reflection (mirrors), refraction (lenses), or the passage of rays through a small hole. Plane mirror = A mirror with a flat surface. Virtual image = The image appears to be behind the mirror. Real image = Image through which light actually passes. Lateral Magnification = It refers to the comparison between the size of the image to its object. Spherical mirrors = It is a reflecting surface with spherical geometry. Concave mirror = It is the inside reflections. Convex mirror = It is the outside reflections. Optic axis = It is the radial line through the center of the spherical mirror and it intersects the mirror surface at the vertex of the spherical section. Center of curvature (c) = It is the point on the optic axis that corresponds to the center of the sphere of which the mirror forms a section. Radius of curvature (r) = It is the distance between the vertex and the center of curvature is equal to the radius of the sphere. Converging mirror = It is a concave mirror which is a result when rays parallel to the optic axis are incident on a concave mirror, the reflected rays intersect, or converge, at a common point called, the focal point. Diverging Mirrors = It is the incident parallel rays emerge from the lens as though they emanated from a focal point on the incident side of the lens. Focal Point = When rays parallel to the optic axis are incident on a concave mirror, the reflected rays intersect, or converge at a common point. Focal length = It is the distance from the vertex to the focal point of the spherical mirror.

71

Parallel ray = It is a ray that is incident along a path parallel to the optic axis and is reflected through the focal point. Chief ray or radial ray = It is a ray that is incident through the center of curvature (c) of the mirror. Since the chief ray is incident normal to the mirrors surface, this ray is reflected back along its path, through point c. Mirror Focal ray = It is a ray that passes through the focal point and is reflected parallel to the optic axis. Spherical mirror equation = It is the distance and focal length can be shown to be related. Magnification factor =It is magnification equation for a spherical mirror Spherical aberration = It is incident parallel rays from the optic do not converge at the focal point. The farther the incident ray is from the axis, the more the more Lens = It comes from the Latin word for lentil, a seed whose shape is similar to that of a common lens. Optical lens = It is made from some transparent material. One of both surfaces usually has a spherical contour. Biconvex lens = It is a converging lens where incident light rays parallel to the lens axis converge at a focal point on the opposite side of the lens. Biconcave lens = It can be approximated by two prisms placed point to point. Diverging lens = It is the incident rays emerge from the lens as though they emanated from a focal point on the incident side of the lens. Meniscus lenses = Most commonly used for eyeglasses. Converging lenses = It is thicker at its center than periphery. Diverging lenses = It is thinner at its center than periphery. Parallel ray = It is a ray that is parallel to the lens axis on incidence and that after refraction either (a) passes through the focal point on the image side of a converging lens or (b) appears to diverge from the focal point of an object side of a diverging lens. Chief ray or Central ray(for lenses) = It is a ray that passes through the center of the lens and is undeviated.

72

Lens Focal Ray = It is a ray that passes through the focal point on the object side of a converging lens or appears to pass through the focal point on the image side of a diverging lens and after refraction is parallel to the lens axis. Lens Spherical Aberration = It is an effect that occurs when parallel rays passing through different region of a lens do not come together on a common focal plane. Circle of least confusion = It is the place where the transmitted light beam has the smallest cross section. Chromatic aberration = It is an effect that occurs because the index of refraction of the material making up a lens is not the same for all wavelengths. Astigmatism = It is the rays centering along the major and minor axis of the ellipse than focus at different points after passing through the lens. Lens Maker’s Equation = Properties which are related to the focal length of a thin lens which gives the focal length of a thin lens in air. Diopters = The unit used by Magnification Factor 𝐡𝐢 𝐌= 𝐡𝐨 Where:

M = magnification factor hi = height of image ho = height of object Focal Length of Spherical Mirror 𝐑 𝐟= 𝟐

Where:

f = focal length, m R = radius of curvature, m Spherical Mirror Equation 𝟏 𝟏 𝟏 𝟐 + = = 𝐝𝐨 𝐝𝐢 𝐟 𝐑 𝒅𝒊 =

Where:

𝒅𝒐 𝒇 𝒅𝒐 − 𝒇

do = object distance di = image distance f = focal length 73

r = radius of curvature Thin Lens Equation (where f ≠ R/2) 𝟏 𝟏 𝟏 + = 𝐝𝐨 𝐝𝐢 𝐟 𝒅𝒊 = Where:

𝒅𝒐 𝒇 𝒅𝒐 − 𝒇

do = object distance di = image distance f = focal length Magnification Factor (spherical Mirror and thin lens) 𝐝𝐢 𝐌= 𝐝𝐨

Where:

M = magnification factor hi = height of image ho = height of object Total Magnification with a Two-Lens System Mtotal = M1M2

Where:

Mtotal = total magnification M = magnification factor Lens Maker’s Equation (thin lens in air only) 𝟏 𝟏 𝟏 = (𝐧 − 𝟏) ( − ) 𝐟 𝐑𝟏 𝐑𝟐

Where:

f = focal length n = index of refraction R = radius of curvature Lens Power in Diopters (where f is in meters) 𝟏 𝑷= 𝒇

Where:

P = lens power f = focal length

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Sign convention for spherical mirrors Focal length Concave (converging) mirror : +f (or +R) Convex (diverging ) mirror : -f ( or –R ) Object distance do ( same for both convex and concave mirror ) +do when the object is in front of the object ( real object ) -di when the object is behind of the object ( virtual image) Image distance (do) +di when the image is formed in front of the mirror -di when the image is formed behind the mirror image orientation M ( same for both mirrors ) +M when the image is upright with respect to the object -M when the image is inverted with respect to the object

Sign Convention for Thin Lenses Focal length Converging lens (sometimes called a positive lens): +f Diverging lens (sometimes called a negative lens); -f Object distance (do) +do when the object is in front of the object ( real object ) -do when the object is behind of the object ( virtual image ) Image distance (do) +di when the image is formed on the opposite (image) side of the lens from the object (real image) -di when the image is formed on the same (object) side of the lens as the object (virtual image) Image orientation M +M when the image is upright with respect to the object -M when the image is inverted with respect to the object

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Example 1: What is the minimum vertical length of a plane mirror needed for a person to be able to see a complete (head to toe) image? 𝐿=

ℎ1 ℎ2 ℎ1 + ℎ2 ℎ + = = 2 2 2 2

Example 2: A concave mirror has a radius of curvature of 30 cm. If an object is placed (a) 45 cm, (b) 20 cm, and (c) 10 cm from the mirror, where is the image formed and what are its characteristics? Given: R= 30 𝑎. 𝑑𝑜 = 45 𝑏. 𝑑𝑜 = 20 𝑐. 𝑑𝑜 = 10 Required: 𝑑𝑖 =? & 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐𝑠 Solution: 𝑅 30 𝑓= = = 15 2 2 1 1 1 = + 𝑓 𝑑𝑜 𝑑𝑖 1 1 1 = + 15 45 𝑑𝑖 1 1 1 = − 𝑑𝑖 15 45 𝑑𝑖 = 22.5 (𝑟𝑒𝑎𝑙) 𝑎.

𝑀=

−𝑑𝑖 −22.5 = = −0.5(𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑑, 𝑠ℎ𝑜𝑟𝑡𝑒𝑟) 𝑑𝑜 45

1 1 1 = + 𝑓 𝑑𝑜 𝑑𝑖 1 1 1 = + 15 20 𝑑𝑖 1 1 1 = − 𝑑𝑖 15 20 𝑑𝑖 = 60 (𝑟𝑒𝑎𝑙) 𝑏.

𝑀=

−𝑑𝑖 −60 = = −3(𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑑, 𝑙𝑜𝑛𝑔𝑒𝑟) 𝑑𝑜 20

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1 1 1 = + 𝑓 𝑑𝑜 𝑑𝑖 1 1 1 = + 15 10 𝑑𝑖 1 1 1 = − 𝑑𝑖 15 10 𝑑𝑖 = −30 (𝑣𝑖𝑟𝑡𝑢𝑎𝑙) 𝑐.

𝑀=

−𝑑𝑖 −(−30) = = 3(𝑢𝑝𝑟𝑖𝑔ℎ𝑡, 𝑙𝑜𝑛𝑔𝑒𝑟) 𝑑𝑜 10

Example 3: An object is 30 cm in front of a diverging mirror that has a focal length of 10 cm. Where the image, and what is are its characteristics? Given: 𝑑𝑜 = 30 𝑓 = 10 Required: 𝑑𝑖 =? & 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐𝑠 Solution: 1 1 1 = + 𝑓 𝑑𝑜 𝑑𝑖 −1 1 1 = + 30 10 𝑑𝑖 1 −1 1 = − 𝑑𝑖 30 10 𝑑𝑖 = −7.5 (𝑣𝑖𝑟𝑡𝑢𝑎𝑙) 𝑀=

−𝑑𝑖 −(−7.5) = = 0.25(𝑢𝑝𝑟𝑖𝑔ℎ𝑡, 𝑠ℎ𝑜𝑟𝑡𝑒𝑟) 𝑑𝑜 30

Example 4: Where is the image formed by a concave mirror if the object is at infinity? 1 1 1 = + 𝑓 𝑑𝑜 𝑑𝑖 1 1 1 = + 𝑓 ∞ 𝑑𝑖 1 1 = 𝑓 𝑑𝑖 𝑑𝑖 = 𝑓

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Example 5: As the object distance of a biconvex lens is varied, at what point does the real image go from being reduced to being magnified. 1 1 1 = + 𝑓 𝑑𝑜 𝑑𝑖 1 1 1 = + 𝑓 2𝑓 𝑑𝑖 1 1 1 1 = − = 𝑑𝑖 𝑓 2𝑓 2𝑓 𝑑𝑖 = 2𝑓 = 𝑑𝑜 𝑀=

−𝑑𝑖 −2𝑓 = = −1(𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑑, 𝑠𝑎𝑚𝑒 𝑠𝑖𝑧𝑒) 𝑑𝑜 2𝑓

Example 6: Suppose the object is 20 cm in front of the lens L1, which has a focal length of 15 cm, is 26 cm from L2. What is the location of the final image, and what are its characteristics? Given: 𝑑𝑜 = 20𝑐𝑚 𝑓𝑟𝑜 𝐿1 𝑓 = 15 𝑑𝑜 = 26

Required: 𝑑𝑖 =? & 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐𝑠

Solution 1 1 1 = + 𝑓1 𝑑𝑜1 𝑑𝑖1 1 1 1 = + 15 20 𝑑𝑖1 1 1 1 = − 𝑑𝑖1 15 20 𝑑𝑖1 = 60𝑐𝑚 (𝑟𝑒𝑎𝑙) −𝑑𝑖1 −60 𝑀= = = −3( 𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑑, 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑒𝑑) 𝑑𝑜1 20 1 1 1 = + 𝑓2 𝑑𝑜2 𝑑𝑖2 1 1 1 = + 12 −34 𝑑𝑖2 1 1 1 = + 𝑑𝑖2 12 34 𝑑𝑖1 = 8.87𝑐𝑚 (𝑟𝑒𝑎𝑙) −𝑑𝑖2 −8.87 𝑀= = = 0.26( 𝑢𝑝𝑟𝑖𝑔ℎ𝑡, 𝑠ℎ𝑜𝑟𝑡𝑒𝑟) 𝑑𝑜2 −34 𝑀𝑡𝑜𝑡𝑎𝑙 = 𝑀1 𝑀2 = −3(0.26) = −0.78 78

Name: Course/Year/Section: Assignment No.:

1. A biconvex lens has a focal length of 12cm. Where the image formed, and what are its characteristics for an object (a) 18cm and (b) 4cm from the lens?

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Chapter 10: Physical Optics: The Wave Nature of Light Important Terms: Physical Optics = Physical optics or wave optics, takes into account those wave properties that geometrical optics ignores. The wave theory of light leads to satisfactory explanations of those phenomena that cannot be analyzed with rays. Young’s Double Slit Experiment = Young’s-the first method that uses interference in demonstrating the wave nature of light. Thin-Film Thickness = It is a result of the interference of light reflected from opposite surfaces of the film and may be readily understood in terms of wave interference. Optical Flats = It is used to check the smoothness of a reflecting surface. The flat is placed so that there is an air wedge between it and the surface. Newton’s Rings = It is the regular interference pattern in an optical flat that produces a set of concentric bright and dark circular fringes. Diffraction = It is the deviation of light. It generally occurs when waves pass through small openings or around sharp edges or corners. Diffraction Grating = It is the arrangements of large numbers of parallel, closely spaced slits. Bragg’s Law = It defines the relationship of 2dsinø= nλ for n=1,2,3… Polarization = Refers to the orientation of the transverse wave oscillations. Polarizing (Brewster) Angle = It is the incident angle at which this polarization occurs. Birefringence = It is the ability of a material to exhibit doubly refracting. Dichroism = It is an interesting property where in a birefringent crystal absorbs one of the polarized components more than the other. Transmission Axis = It is the direction perpendicular to the orientation of the molecular chains. Optical Activity = The ability of a material to rotate the direction of a plane polarized light. Scattering = The process when light is incident on a suspension of particle, such as the molecules of the air; some of the light may be absorbed and radiated.

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Rayleigh Scattering = It is a relationship between the scattering being inversely proportional to the wavelength to the 4th power (that is, S=1/λ4). LCD (Liquid Crystal Display) = It is a twisted-nematic display that is an application involving the optical activity of a liquid crustal and crossed polarizers. When the crystalline order is disoriented from an electric field applied voltage, the liquid crystal is optically active in that region, and the light is not transmitted and reflected.

Important Equations: Bright Fringe Condition (double slit interference) dSinθ = nλ Where:

d = distance between the slits n = order number λ = wavelength Wavelength Measurement ( double slit interference for small θ only) 𝐲𝐧 𝐝 𝛌 ≈ 𝐧𝐋

Where:

λ = wavelength yn = distance of the nth bright fringe d = distance between the slits n = order number L = distance from the slit to the point Nonreflecting Film Thickness (minimum) 𝛌 𝐭= 𝐟𝐨𝐫 𝐧𝟐 > 𝑛1 𝟒𝐧

Where:

t = thickness n = order number λ = wavelength Dark Fringe Condition (single slit diffraction) wSinθ = mλ

Where:

w = slit width λ = wavelength Lateral Displacement of Dark Fringes (single slit diffraction) 𝐦𝐋𝛌 𝐲𝐦 = 𝐟𝐨𝐫 𝐦 = 𝟏, 𝟐, 𝟑 𝐰

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Where:

ym = maximum distance L = distance from the slit to the point λ = wavelength w = slit width Interference Maxima for a Diffraction Grating dSinθ = nλ for n= 0, 1, 2, 3

Where:

d = 1/N, distance N = lines per unit length n = order number λ = wavelength Limit of the Order Number 𝒅 𝒏 ≤ 𝝀

Where:

Where:

n =order number d =distance λ = wavelength Brewster (polarizing) Angle Tanθp = n n = order number

Example 1: In a lab experiment, monochromatic light passes trough two narrow slits that are 0.050mm apart. The interference pattern is observed on a white wall 1.0 m from the slits. And the second-order fringe is 2.4 cm from the center of the central maximum (a) what is the wavelength of the light? (b) what is the distance between the second-order and the third order bright fringe? Giiven: 𝑑 = 0.05𝑚𝑚 = 0.00005𝑚 𝐿 = 1𝑚 𝑦2 = 2.4𝑐𝑚 = 0.024 𝑛=2

Required: 𝜆 =? 𝑦3 − 𝑦2 =?

Solution: 𝑦𝑛 𝑑 0.024(0.00005) 𝑎. 𝜆 = = = 6.0𝑥10−7 𝑚 = 600𝑛𝑚 𝑛𝐿 2(1) 𝑏. 𝑦3 − 𝑦2 =

𝜆𝑛 𝐿 𝜆𝑛 𝐿 𝜆𝐿 600𝑛𝑚(1000𝑚𝑚) (𝑛 + 1 − 𝑛) = − = = 1.2𝑐𝑚 𝑑 𝑑 𝑑 0.05𝑚𝑚

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Example 2:A glass lens (n=1.60) is coated with a thin, transparent film of a magnesium fluoride (Mgf2, n = 1.38) to make the lens non-reflecting. What is the minimum film thickness for the lens to be non-reflecting for incident light of wavelength 550 nm? Given: 𝑛𝑔𝑙𝑎𝑠𝑠 = 1.60 𝑛𝑀𝑔𝐹2 = 1.38 𝜆 = 550𝑛𝑚 Required: 𝑡 =? Solution: 𝜆 550𝑛𝑚 𝑡= = = 99.64𝑛𝑚 4𝑛 4(1.38) Example 3: Monochromatic blue light (λ = 425 nm) passes through a slit whose width is 0.50 mm. What is the width of the central maximum on a screen located 1.0 m from the slit? Given: 𝜆 = 425𝑛𝑚 𝑤 = 0.50𝑚𝑚 𝐿 = 1.0𝑚 Required: 2𝑦 =? Solution: 𝑚𝐿𝜆 2(1.0𝑚)(425𝑛𝑚) 2𝑦 = = = 1.7𝑚𝑚 𝑤 0.5𝑥10−3 𝑚 Example 4: A particular diffraction grating produces an n = 2 spectral order at a deviation angle of 30º for a light with wavelength of 700 nm. (a) How many lines per centimeter does the grating have? (b) If the grating were illuminated with white light, how many orders of the complete visible spectrum would be produced? Given: 𝑛 = 2 𝜃 = 30° 𝜆 = 700𝑛𝑚 Required: 𝑁 =? 𝜆= 83

Solution: 𝑎. 𝑑 = 𝑁 =

𝑛𝜆 2(500𝑛𝑚) = = 2000𝑛𝑚 = 2𝑥10−6 𝑚 = 2𝑥10−4 𝑐𝑚 𝑆𝑖𝑛𝜃 𝑆𝑖𝑛30

1 1 = = 5000 𝑙𝑖𝑛𝑒𝑠⁄𝑐𝑚 𝑑 2𝑥10−4 𝑐𝑚

𝑑 2.00𝑥10−6 𝑚 𝑏. 𝑛 ≤ = = 2.86 𝜆 700𝑥10−9 𝑚 𝜆=

2.00𝑥10−6 𝑚 = 6.6710−7 𝑚 𝑜𝑟 667𝑛𝑚 3

Example 5: Sunlight is reflected from the smooth surface of a pond. What is the Sun’s altitude when the polarization of the reflected light is greatest? Given: 𝑛 = 1.33 Required: 𝜃 =? Solution: 𝜃 = 90 − 𝜃𝑝 𝜃𝑝 = 𝑇𝑎𝑛−1 𝑛 = 𝑇𝑎𝑛−1 (1.33) = 53.06° 𝜃 = 90 − 53.06 = 36.94° Example 6: How much more is light at the blue end of the visible spectrum scattered by air molecules than is light at the red end? Given: 𝜆𝑟𝑒𝑑 = 700𝑛𝑚 𝜆𝑏𝑙𝑢𝑒 = 400𝑛𝑚 Required: 𝑆𝑏𝑙𝑢𝑒 =? 𝑆𝑟𝑒𝑑 Solution: 𝑆𝑏𝑙𝑢𝑒 𝜆𝑟𝑒𝑑 4 700𝑛𝑚 4 =( ) =( ) = 9.38 𝑆𝑟𝑒𝑑 𝜆𝑏𝑙𝑢𝑒 400𝑛𝑚

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Chapter 11: Optical Instruments Important Terms: Vitreous Humor = A jellylike substance that fill the eyeball. Sclera = The white covering of the eyeball. Cornea = It is where the light enter the eye through a curved, transparent tissue. Aqueous Humor = It is the clear fluid in the eye. Iris = It is located behind the cornea, is a circular diaphragm. Pupil = Central hole. Retina = It is located on the back interior wall of the eye ball is a light sensitive surface. Rods = Are more sensitive to light than the cones. Cones = Can distinguish frequency changes of sufficiently intense light witch the brain interprets the color. Nearsightedness = It is the ability to see distance object clearly but not distant object. Farsightedness = It is the ability to see distant objects clearly but not nearby object. Astigmatism = It is another common defect of vision in which the refractive surface, usually the cornea or crystalline lens, being out of round. Magnifying Glass = It is a instrument witch single convex lens forms a clear image of an object that is closer than the near point. Angular Magnification = It is symbolized m and it is a nation of the angular size of the object viewed though the magnifying glass. Compound Microscope = Provides greater magnification than is attained with the single lens or simple microscope. Objective = It is the converging lens with a relatively short focal length (fo < 1cm). Eyepiece = It has a longer focal length and is positioned so that the image formed by that the image formed by the objective falls just inside its focal length. Refracting Telescope = Principle of which is similar to that of a compound microscope.

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Astronomical Telescope = Form a distant object from a intermediate image at the focal point of objective (fo). Terrestrial Telescope = Utilizing a large, concave, front surface parabolic mirror. Reflecting Telescope = A type of telescope that utilizes a large, concave, front surfaced parabolic mirror. Resolution = Of two diffracted images that are to ability to distinguish both image as separate. Rayleigh Criterion = Can be express in terms of the angular separation (θ) of the sources the first maximum (m=1) for a single slit diffraction patterns satisfies this relationship. Resolving Power = This minimum distance between two points whose image can be just resolved. Additive Primary Colors = The red, blue and green from which we interpret a full spectrum of colors. Additive Method of Color of Production = When light beams of the additive primaries are projected on a white screen so they overlap, other color will be produce. Complementary Colors = Color of such pair that appear white to an eye when combined. Subtractive Method of Color Production = Mixing the pigments results in the subtraction of colors. Subtractive Primary Pigments = Cyan, magenta, and yellow which produce the three additive primary colors.

Important Equations: Angular Magnification 𝛉 𝐦= 𝛉𝐨 Where:

m = angular magnification Magnification of a Magnifying Glass with Image at Near Point (25cm) 𝟐𝟓𝐜𝐦 𝐦=𝟏+ 𝐟

Where:

m = angular magnification 86

f = focal length Magnification of a Magnifying Glass with Image at Infinity 𝟐𝟓𝐜𝐦 𝐦= 𝐟 Where:

m = angular magnification f = focal length

Angular Magnification of a Compound Microscope (with L, fo and fe in cm) (𝟐𝟓𝐜𝐦)𝐋 𝐦𝐭𝐨𝐭𝐚𝐥 = 𝐌𝐨 𝐦𝐞 = 𝐟𝐨 𝐟𝐞 Where:

mtotal = angular magnification of a compound lens Mo = angular magnification of the objective me = angular magnification of the eyepiece fo = focal length of the objective fe = focal length of the eyepiece Magnification of a Refracting Telescope 𝐟𝐨 𝐦= − 𝐟𝐞

Where:

m = angular magnification fo = focal length of the objective fe = focal length of the eyepiece Minimum Angle of Resolution for a Slit 𝛌 𝛉𝐦𝐢𝐧 = 𝐝

Where:

λ = wavelength d = distance Minimum Angle of Resolution for a Circular Aperture (diameter) 𝟏. 𝟐𝟐𝛌 𝛉𝐦𝐢𝐧 = 𝐃

Where:

λ = wavelength D = diameter Resolving Power 𝟏. 𝟐𝟐𝛌𝐟 𝐬 = 𝐟𝛉𝐦𝐢𝐧 = 𝐃

Where:

s = resolving power 87

F = focal length D = diameter λ = wavelength Example 1: A certain near sighted person cannot see objects clearly when when they are more than 78 cm from either eye. What power must corrective lens have if this person is to see distant objects clearly? Assume that the lenses are in eyeglasses and are 3.0 cm in front of the eye. Given: 𝑑𝑖 = 78𝑐𝑚 𝑑 = 3𝑐𝑚 Required: 𝑃 =? Solution: 1 1 1 𝑎. = + 𝑓 𝑑𝑖 𝑑𝑜 1 1 1 = + 𝑓 𝑑𝑖 ∞ 1 −1 = 𝑓 75 𝑓 = −75𝑐𝑚 𝑃 =

1 −1 = = −1.33𝐷𝑖𝑜𝑝𝑡𝑒𝑟𝑠(𝐷) 𝑓 0.75

Example 2: A farsighted person has a near point of 75cm for one eye and a near point of 100cm for the other. What powers should contact lenses have to allow the person to see an object clearly at a distance of 25cm? Given: 𝑑𝑖1 = −75𝑐𝑚 = −0.75𝑚 𝑑𝑖2 = −100𝑐𝑚 = −1𝑚 𝑑𝑜 = 25𝑐𝑚 = 0.25𝑚

Required: 𝑃 =?

Solution: 1 1 1 −1 1 𝑃1 = = + = + = 2.67𝐷 𝑓1 𝑑𝑖1 𝑑𝑜 0.75 0.25 𝑃2 =

1 1 1 −1 1 = + = + = 3𝐷 𝑓2 𝑑𝑖2 𝑑𝑜 1 0.25

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Example 3: Sherlock Holmes uses a converging lens with a focal length of 12cm to examine the fine detail of some cloth fibers found at the scene of crime.(a) what is the maximum magnification given by the lens?(b) what is the magnification by relaxed eye viewing? Given: 𝑓 = 12 Required: 𝑎. 𝑀𝑙𝑒𝑛𝑠 =? 𝑏. 𝑀𝑟𝑒𝑙𝑎𝑥𝑒𝑑 𝑒𝑦𝑒𝑠 =? Solution: 𝑎. 𝑀𝑙𝑒𝑛𝑠 = 1 +

25 25 = 1+ = 3.08 𝑓 12

𝑏. 𝑀𝑟𝑒𝑙𝑎𝑥𝑒𝑑 𝑒𝑦𝑒𝑠 =

25 25 = = 2.08 𝑓 12

Example 4: A microscope has an object w/ focal length of10mm and an eyepiece w/ a focal length of 4.0cm the lenses are positioned 20cm apart in the barrel. Determine the approximate total magnification of microscope. Given: 𝑓𝑜 = 10𝑚𝑚 = 1𝑐𝑚 𝑓 = 4𝑐𝑚 𝐿 = 20𝑐𝑚 Required: 𝑀𝑡𝑜𝑡 = ? Solution: (25𝑐𝑚)𝐿 (25𝑐𝑚)(20𝑐𝑚) 𝑀𝑡𝑜𝑡 = = = 125𝑥 (1𝑐𝑚)(4𝑐𝑚) 𝑓𝑜 𝑓 Example 5: An astronomical telescope has an objective lens with focal length of 30cmand eyepiece with a focal length 9.0cm a. what is the magnification of the telescope? b. if an erecting lens have a focal length of 7.5cm is use to convert the telescope to a terrestrial type, what is the overall length of the telescope tube? Given: 𝑓𝑜 = 30𝑐𝑚 𝑓𝑒 = 9𝑐𝑚 𝑓𝑖 = 7.5𝑐𝑚

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Required: 𝑎. 𝑀 = ? 𝑏. 𝐿 = ? Solution 𝑎. 𝑀 =

−𝑓𝑜 30 =− = −3.33𝑋 𝑓𝑒 9

𝑏. 𝐿1 = 𝑓𝑜 + 𝑓𝑒 = 30 + 9 = 39𝑐𝑚 𝐿 = 𝐿1 + 𝐿2 = 𝐿1 + 4𝑓𝑖 = 39 + 4(7.5) = 69𝑐𝑚 Example 6: Determine the maximum angle of resolution by Rayleigh Criterion (a) the pupil of eye (day time diameter of about 4.0mm) (b) the Yerkes observatory refracting telescope (diameter of 102cm) both a. and b. for visible light with a wavelength of 660nm and (c) a radio telescope 25m for radiation with a wave length of 21cm. Given: 𝐷𝑒𝑦𝑒 = 4𝑚𝑚 = 4𝑥10−3 𝑚 𝜆 = 660𝑥10−9 𝑚 𝐷𝑌𝑒𝑟𝑘𝑒𝑠 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑜𝑟𝑦 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑡𝑒𝑙𝑒𝑠𝑐𝑜𝑝𝑒 = 102𝑐𝑚 = 1.02𝑚 𝐷𝑟𝑎𝑑𝑖𝑜 𝑡𝑒𝑙𝑒𝑠𝑐𝑜𝑝𝑒 = 25𝑚 𝐿 = 21𝑐𝑚 = 0.21𝑚 Required: 𝜃𝑚𝑖𝑛 =? Solution: 𝑎. 𝜃𝑚𝑖𝑛

1.22𝜆 1.22(660𝑥10−9 𝑚) = = = 2.01𝑥10−4 𝑟𝑎𝑑 𝐷 4𝑥10−3 𝑚

𝑏. 𝜃𝑚𝑖𝑛 =

1.22𝜆 1.22(660𝑥10−9 𝑚) = = 7.89𝑥10−7 𝑟𝑎𝑑 𝐷 1.02𝑚

𝑐. 𝜃𝑚𝑖𝑛 =

1.22𝜆 1.22(0.21𝑚) = = 0.0102𝑟𝑎𝑑 𝐷 25𝑚

90

Chapter 12: Temperature Important Terms: Temperature = A relative measure or indication of hotness or coldness Heat =The net energy transferred from one object to another because of a temperature difference Internal Energy = The energy becomes part of the total energy of the molecules of the object or system Thermometer = A device constructed to make evident some property of a substance that changes with temperature. Thermal Expansion =A change in the dimension or volume of a substance that occurs when the temperature changes. Fahrenheit Temperature Scale = The boiling point is 212o while the freezing point is 32o Celsius Temperature Scale = The boiling point is 100o while the freezing point is 0o Mole = the quantity that contains Avogadro’s number of a molecule Absolute Zero = -273.15oC Kelvin Temperature Scale = absolute zero is the foundation. Triple Point of Water = Represents a unique set of conditions where water co – exist simultaneously in equilibrium as a solid, a liquid and a gas Kinetic Theory of Gases = the molecules undergo perfectly elastic collisions with each other and with the walls of the container. Diffusion = The process of random molecular mixing in which particular molecules moves from a region where they are present in higher concentration to one where they are in lower concentration. Osmosis = The diffusion of liquid water across a permeable membrane down a concentration gradient. Degree of Freedom = Independent way a molecule has for possessing energy. Equipartition Theorem = On average, the total internal energy of an ideal gas is divided equally among each degree of freedom its molecules possess. 91

Important Equations: Celsius – Fahrenheit Conversion

Where:

℉ =

𝟗 ℃ + 𝟑𝟐 𝟓

℃ =

𝟓 (℉ – 𝟑𝟐) 𝟗

F = Fahrenheit C = Celsius

Ideal (or Perfect) Gas Law (always absolute temperature) 𝐏𝐕 = 𝐍𝐤 𝐁 𝐓 𝐏𝟏 𝐕𝟏 𝐏𝟐 𝐕𝟐 = 𝐓𝟏 𝐓𝟐 𝐏𝐕 = 𝐧𝐑𝐓 Where:

kB = Boltzmann constant = 1.38 x 10-23J/K R = 8.31J/(mol·K) NA = Avogadro’s number = 6.02 x 1023 molecules/mol P = Pressure V = Volume T = Temperature N = Number of molecules n = Number of moles

Kelvin – Celsius Conversion 𝐊 = ℃ + 𝟐𝟕𝟑. 𝟏𝟓 Where: K = Kelvin Thermal Expansion of Solids Linear: 𝐋𝐟 = 𝐋𝐨 (𝟏 + 𝛂𝚫𝐓) Area: 𝐀 𝐟 = 𝐀 𝐨 (𝟏 + 𝟐𝛂𝚫𝐓) Volume: 𝐕𝐟 = 𝐕𝐨 (𝟏 + 𝟑𝛂𝚫𝐓) Where:

LF = Final length LO = Initial length AF = Final area AO = Initial area VF = Final volume VO = Initial volume T = Temperature α = Coefficient of thermal expansion for solids 92

Thermal Volume Expansion of Fluids 𝚫𝐕 = 𝛃𝚫𝐓 𝐕𝐨 β = Coefficient of volume expansion for fluids

Where:

Results of Kinetic Theory of Gases 𝐩𝐕 = 𝟏 𝟐

𝐦𝐯𝐫𝐦𝐬 𝟐 =

𝐔 = 𝐔 =

𝟑 𝟐 𝟓 𝟐

𝟑 𝟐

𝐤 𝐁 𝐓 (all gases)

𝐍𝐤 𝐁 𝐓 = 𝐍𝐤 𝐁 𝐓 =

Where:

𝟏 𝐍𝐦𝐯𝐫𝐦𝐬 𝟐 𝟑

𝟑 𝟐 𝟓 𝟐

𝐧𝐑𝐓 (ideal monatomic gases only) 𝐧𝐑𝐓 (diatomic gases)

v = Velocity U = Total Energy

Example 1: Convert 20oC to Fahrenheit scale and the normal body temperature and 98.6oF to the Celsius scale. Given: 𝑇1 = 20°C 𝑇2 = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑏𝑜𝑑𝑦 𝑡𝑒𝑚𝑝 = 37°𝐶 𝑇3 = 98.6°𝐹 Required: Convert to other units Solution: 9 a. ℉ = ℃ + 32 5

9 ℉ = (20) + 32 5 ℉ = 𝟔𝟖 5 b. ℃ = (℉ − 32) 9 5 ℃ = (98.6 − 32) 9 ℃ = 𝟑𝟕

Example 2: What is the absolute zero on the Fahrenheit scale? Given: 𝑇1 = 0°K Required: Absolute 0 in oF Solution: K = ℃ + 273.15 = ℃ + 273.15 ℃ = −273.15

9 9 ℉ = ℃ + 32 = (−273.15) + 32 5 5 ℉ = −𝟒𝟓𝟗. 𝟔𝟕

93

Example 3: A quantity of low density gas in a rigid container is initially at room temperature and a particular pressure. If the gas is heated to a temperature of 60oC by what factor does the pressure change? Given: Solution: T1 T2 P2 T2 T1 = 20℃ = ; = T2 = 60℃ P1 P2 P1 T1 P2 60 + 273.15 = Required: P1 20 + 273.15 P2 P2 333.15 =? = P1 P1 293.15 P2 = 1.136 P1 Example 4: A steel beam is 5.0m long at a temperature of 20oC. On a hot day, the temperature rises to 40oC. What is the change in the beam’s length due to thermal expansion? Suppose that the ends of the beam are initially in contact with rigid vertical supports, how much force will the expanded beam exert on the supports if the beam has a cross-sectional area of 60cm2? Given: L= 5m Ti = 20℃ Tf = 40℃ A = 60cm2

Required: a. ΔL =? b. F =?

Solution a. LF = Lo [1 + α(ΔT)] 12 × 10−6 LF = 5.00m [1 + ( ) (40℃ − 20℃)] ℃ LF = 5.0012 m ΔL = LF − Lo ΔL = 5.0012 m − 5.00 m ΔL = 0.0012 m

94

Example 5: A surveyor uses a steel measuring tape that is exactly 50.00 m long at a temperature of 20oC. What is its length on a hot summer day when the temperature is 35oC? Given: Lo = 50.00m Ti = 20℃ Tf = 35℃

Required: LF =?

Solution: LF = Lo [1 + α(ΔT)] 12 × 10−6 LF = 50.00m [1 + ( ) (35℃ − 20℃)] = 𝟓𝟎. 𝟎𝟎𝟗 𝐦 ℃ Example 6: In the previous example, the surveyor measures a distance when the temperature is 35oC and obtains the result 35.794m. What is the actual distance? Given: LF = 35.794 m Tf = 35℃

Required: Lo =?

Solution: LF = Lo [1 + α(ΔT)] 12 × 10−6 35.794 m = Lo [1 + ( ) (35℃ − 20℃)] ℃ 35.794m Lo = = 𝟑𝟓. 𝟖𝟎𝟎𝟒 𝐦 12 × 10−6 [1 + ( ) (35℃ − 20℃)] ℃ Example 7: What is the average (rms) speed of a helium atom (He) in a helium balloon at room temperature? Take the mass of the helium atom to be 6.65 x 10-27kg. Given: mHe = 6.65 × 10−27 kg T = 20℃ + 273.15 = 293.15 K J k B = 1.38 × 10−23 K

Solution

Required: V =?

vrms = √

vrms = √

𝐯𝐫𝐦𝐬

3k B T m J 3 (1.38 × 10−23 K) (293.15 K)

6.65 × 10−27 kg 𝐦 = 𝟏𝟑𝟓𝟎. 𝟗𝟑 𝐬

95

Example 8: More than 99% of the air we breathe consists of diatomic gases, mainly nitrogen (N2, 78%) and oxygen (O2, 21%). There are traces of other gases, one of which is radon (Rn), a monatomic gas arising from radioactive decay of uranium in the ground. Calculate the total internal energy of 1.00mole samples each of oxygen and radon at room temperature. For each sample, calculate the amount of internal energy associated with molecular translational kinetic energy. Given: n = 1 mole

T = 20℃ + 273.15 = 293.15 K J k B = 8.31 mol ∙ K Required: a. URa =? UO2 =? b. UTRANS of Ra =? UTRANS of O2 = ? Solution: URa = URa = 𝐔𝐑𝐚

2 3

nkB T

(1mol) (8.31

2 = 𝟑𝟔𝟓𝟒. 𝟏𝟏 𝐉

UO2 = UO2 = 𝐔𝐎𝟐

3

5 2 5

J mol ∙ K

) (293.15 K)

nkB T

(1mol) (8.31

2 = 𝟔𝟎𝟗𝟎. 𝟏𝟗 𝐉

J mol ∙ K

) (293.15 K)

𝐔𝐓𝐑𝐀𝐍𝐒 𝐨𝐟 𝐑𝐚 = 𝟑𝟔𝟓𝟒. 𝟏𝟏 𝐉 𝐔𝐓𝐑𝐀𝐍𝐒 𝐨𝐟 𝐎𝟐 =

3 2

nkB T = 𝟑𝟔𝟓𝟒. 𝟏𝟏 𝐉

Name: 96

Section: 1. A building with a steel framework is 150m tall when the temperature is 0oC. How much taller is the building on a hot summer day when the temperature is 36oC? Given: Lo = 150.00m Ti = 0℃ Tf = 36℃ Required: ΔL =? Solution: LF = Lo [1 + α(ΔT)] LF = Lo [1 + α(Tf − Ti )] 12 × 10−6 LF = 150.00m [1 + ( ) (36℃ − 0℃)] ℃ LF = 150.0648 ΔL = 150.0648 m − 150.00 m 𝚫𝐋 = 𝟎. 𝟎𝟔𝟒𝟖 𝐦 2. A steel bridge is built in the summer when the temperature is 35oC. At the time of construction its length is 80m. What is the length of the bridge on a cold winter day when the temperature is -12oC? Given: Lo = 80m Ti = 35℃ Tf = −12℃ Required LF =? Solution: LF = Lo [1 + α(Tf − Ti )] 12 × 10−6 LF = 80m [1 + ( ) (−12℃ − 35℃)] ℃ 𝐋𝐅 = 𝟕𝟗. 𝟗𝟓𝟒𝟖𝟖 𝐦

Chapter 13: Heat 97

Important Terms: Heat = Describes a type of net energy transferred. Kilocalorie (kcal) = the amount of heat needed to raise the temperature of a kilogram mass of water of a certain temperature in Celsius. Calorie (cal) = the amount of heat needed to raise the temperature of a pound of water of a certain temperature in Celsius. British thermal unit (Btu) = the amount of heat needed to raise the temperature of a kilogram mass of water of a certain temperature in Fahrenheit. Mechanical Equivalent of Heat = For every 4186J of work done, the temperature of the water rose 1oC per kilogram, or 4186J is equivalent to 1kcal. Specific Heat = the amount of heat energy required to raise the temperature of a 1kg of a substance by 1oC. Calorimetry = the quantitative measure of heat exchange, which allows us to determine the specific heats of substances. Solid Phase = the molecules are held together by attractive forces or bonds. Melting Point = the temperature at which solid becomes liquid Freezing Point = the temperature at which liquid becomes solid Liquid Phase = molecules of a substance are relatively free to move and a liquid assumes the shape of its container Gaseous Phase = molecules interact weakly and are separated by relatively large distances; thus has no definite shape or volume. Boiling Point = the temperature at which liquid becomes vapor/gas Condensation Point = the temperature at which gas condenses and becomes a liquid. Sublimation = the change of solids to gaseous phase directly. Deposition = A phase change from gas to a solid. Latent Heat = the heat energy involved in a phase change Latent Heat of Fusion = the latent heat for a solid – liquid phase change Latent Heat of Vaporization = the latent heat for a liquid – solid phase change 98

Latent Heat of Sublimation = the latent heat for the less common solid-gas phase change Evaporation = A cooling process for the object from which the molecules escape Conduction = energy is conductively transferred from a higher temperature region to a lower temperature region – transfer as a result of a temperature difference. Thermal Conductor = conductors of heat Thermal Insulator = poor heat conductor Thermal Conductivity = characterizes the heat-conducting ability of a material. Convection = heat transfer as a result of mass transfer, which can be natural or forced Radiation = heat transferred which does not require a medium but by electromagnetic waves Stefan’s Law = the rate at which an object radiates energy has been found to be proportional to the fourth power of the absolute temperature Emissivity = a unitless number between number 0 and 1 that is characteristics of the material. Blackbody = a body that is a good absorber of a certain wavelength of radiation is a good emitter of those same wavelengths.

Important Equations Specific Heat 𝐐 = 𝐦𝐜𝚫𝐓 Where:

Q = heat c = specific heat m = mass T = temperature

Latent Heat 𝐐 = 𝐦𝐋 Where:

Q = heat m = mass L = latent heat

Thermal Conduction 99

𝚫𝐐 𝐤𝐀𝚫𝐓 = 𝚫𝐭 𝐝 Where:

Q = heat t = time k = thermal conductivity T = temperature d = thickness

Stefan’s Law 𝐏 = Where:

𝚫𝐐 = 𝛔𝐀𝐞𝐓 𝟒 𝚫𝐭

P = power Q = heat t = time A = area σ = Stefan-Boltzmann constant e = emissivity T = temperature

Net Radiant Power 𝐏𝐧𝐞𝐭 = 𝛔𝐀𝐞(𝐓𝐬 𝟒 – 𝐓 𝟒 ) Where:

Ts = surrounding temperature

Example 1: How much mechanical work in joules would have to be done in Joule’s apparatus to raise the temperature of a liter of water from room temperature to 25oC? Given: TI = 20℃ TF = 25℃ Vol = 1L ∴ mH2 O = 1 kg Required: Wk =? Solution: Q = mc∆T kJ ) (25℃ − 20℃) kg ∙ ℃ 𝐐 = 𝟐𝟎. 𝟗𝟑 𝐤𝐉 = 𝐖 Q = (1kg) (4.186

Example 2: How much heat is required to raise the temperature of 0.20kg of water from 15oC to 45oC? 100

Given: TI = 15℃ TF = 45℃ mH2 O = 0.20 kg

Solution: Q = mc∆T J ) (45℃ kg ∙ ℃ − 15℃) 𝐐 = 𝟐𝟓𝟏𝟏𝟔. 𝟎 𝐉 Q = (0.20kg) (4186

Required: Q=?

Example 3: A half-liter of water at 30oC is cooled, with the removal of 63KJ of heat. What is the final temperature of the water? Q = −63kJ = −63000J

Given: TI = 30℃ mH2 O = 0.50 kg J cH2 O (4186 ) kg ∙ ℃

Required: TF =?

Solution: Q = mc∆T −63000J = (0.50kg) (4186 TF = −

J ) (TF − 30℃) kg ∙ ℃

63000J

J (0.50kg) (4186 ) kg ∙ ℃ 𝐓𝐅 = −𝟎. 𝟏𝟎𝟎𝟑℃ ≈ 𝟎℃

+ 30

Example 4: Equal masses of aluminum and copper are at the same temperature. Which will require the greater heat to raise its temperature by a given amount, and how many times greater is this than the heat that would have to be added to the other metal? Given: J kg ∙ ℃ J cAl = 920 kg ∙ ℃ mCu = mAl ∆TCu = ∆TAl cCu = 390

Required: QAl =? QCu

QAl mAl cAl ∆TAl = QCu mCu cCu ∆TCu Since mCu = mAl & ∆TCu = ∆TAl J 920 QAl cAl kg ∙ ℃ ∴ = = J QCu cCu 390 kg ∙ ℃ QAl = 2.36: 𝐐𝐀𝐥 = 𝟐. 𝟑𝟔𝐐𝐂𝐮 QCu 𝐐𝐀𝐥 > 𝐐𝐂𝐮

Solution:

101

Example 5: Students in a physics lab are to determine the specific heat of copper experimentally. They heat 0.150kg of copper shot to 100oC and then carefully pour the hot shot into a calorimeter cup containing 0.200kg of water at 20oC. The final temperature of the mixture in the cup is measured to be 25oC. If the aluminum cup has a mass of 0.045kg, what is the specific heat of copper? (Assume that there is no heat loss to the surroundings.) Given: TI for Cu = 100℃ TI for Al & H2 O = 20℃ TF = 25℃ J cAl = 920 kg ∙ ℃ J cH2 O = 4186 kg ∙ ℃

mCu = 0.15 kg mAl = 0.045 kg mH2 O = 0.20 kg Required: cCu =?

Solution: ΣQ = 0 QAl + QCu + Q H2 O = 0 mAl cAl ∆TAl + mCu cCu ∆TCu + mH2 O cH2 O ∆TH2O = 0 [0.045(920)(25 − 20)] + [0.15(cCu )(25 − 100)] + [0.20(25 − 20)(4186)] = 0 207 − 11.25cCu + 4186 = 0 −207 − 4186 𝐉 cCu = = 𝟑𝟗𝟎. 𝟒𝟗 −11.25 𝐤𝐠 ∙ ℃ Example 6: Heat is added to 0.500kg of water at room temperature. How much heat in joules is required to change the water to steam at 110oC? Given: m = 0.50 kg TI = 20℃ TF1 = 100℃ TF2 = 110℃ cH2 O = 4186

J for steam kg ∙ ℃ J Lv = 22.6 × 105 kg cH2 O = 2010

J @ room T kg ∙ ℃

Required: ΣQH2 O =?

Solution: J ) (100℃ − 20℃) = 167440J kg ∙ ℃ J QLv = mLv = (0.500 kg) (22.6 × 105 ) = 1130000 J kg J Q3 = mc∆T = (0.500 kg) (2010 ) (110℃ − 100℃) = 10050J kg ∙ ℃ ΣQH2 O = Q1 + QLv + Q3 = 167440 + 1130000 + 10050 𝚺𝐐𝐇𝟐 𝐎 = 𝟏𝟑𝟎𝟕𝟒𝟗𝟎 𝐉 = 𝟏. 𝟑𝟏 × 𝟏𝟎 𝟔 𝐉 Q1 = mc∆T = (0.500 kg) (4186

102

Example 7: A 0.30kg piece of ice at 0oC is placed in a liter of water at room temperature in an insulated container. Assume that no heat is lost to the container, what is the final temperature of the water? Given: mice = 0.30 kg vol = 1L: ∴ mH2 O = 1 kg J LF = 3.33 × 105 kg J cH2 O = 4186 kg ∙ ℃ Required: TF =? Solution: Qice = mLF = (0.30kg) (3.33 × 105 Qice = Q H2 O Q H2 O = mc∆T −99900J = (1kg) (4186 −99900J J (1kg) (4186 ) kg ∙ ℃

J ) = 99900J kg

J ) (TF − 20℃) kg ∙ ℃

+ 20℃ = TF

𝐓𝐅 = −𝟑. 𝟖𝟕 ℃ Q H2 O = (1kg) (4186 𝐦𝐢𝐜𝐞 =

J ) (0℃ − 20℃) = −83720 J kg ∙ ℃

Q H2 O −83720 J = = 𝟎. 𝟐𝟓 𝐤𝐠 J Lv 3.33 × 105 kg

103

Example 8: A room with a pine ceiling that measures 3.0m x 5.0m x 2.0cm thick has a layer of glass wool insulation above it that is 6.0cm thick. On a cold day, the temperature inside the room is 20oC and the temperature in the attic above the room is 8oC. Assuming that the temperatures remain constant with a steady heat flow, how much energy does the layer of insulation save in 1.0hr? Assume losses are due to the conduction only. Given: A = 3m × 5m = 15 m2 d1 = 2cm = 0.02m d2 = 6cm = 0.06m T1 = 20℃ T2 = 8℃ ∆t = 1hr = 3600s J k glass wool = 0.042 ms ∙ ℃ J k wood pine = 0.12 ms ∙ ℃ Solution: ∆T = T1 − T2 = 20 − 8 = 12℃ J 0.12 ms ∙ ℃ (15 m2 )(12℃)(3600s) k wood pine A∆T∆t ∆Q = = = 𝟑𝟖𝟖𝟖𝟎𝟎𝟎𝐉 d1 0.02 m ∆Q1 ∆Q2 = ∆t ∆t k glass wool A∆T k wood pine A∆T = d1 d2 k glass wool (T − T1 ) k wood pine (T2 − T) = d1 d2 k glass wool d2 T1 + k wood pine d1 T2 k glass wool d2 + k wood pine d1 ∆Q1 A(T2 − T1 ) = k glass wool k wood pine ∆t + d1 d2 T=

∆𝐐𝟐 =

15m2 (12℃)(3600s) = 𝟒𝟎𝟔𝟐𝟎𝟖. 𝟗𝟔 𝐉 0.06 0.02 + J J 0.042 ms ∙ ℃ 0.12 ms ∙ ℃

∆𝑄1 = ∆Q − ∆Q2 ∆𝑄1 = 3888000J − 406208.96 J = 𝟑𝟒𝟖𝟏𝟕𝟗𝟒𝟏. 𝟎𝟒 𝐉 104

%=

∆𝑄1 34817941.04J × 100 = × 100 = 𝟖𝟗. 𝟓𝟓% ∆Q 3888000J

Example 9: Suppose that your skin has an emissivity of 0.70 and that its exposed area is 0.27m2. How much net energy will be radiated per second from this area if the air temperature is 20oC? Assume your skin temperature to be the same as normal body temperature, 37oC. Given: Ts = 20℃ + 273.15 = 293.15 K Tbody = 37℃ + 273.15 = 310.15 K A = 0.27 m2 e = 0.7 𝑤 δ = 5.67 × 10−8 2 4 𝑚 𝐾 Required: E =? Solution: 𝑃𝑛𝑒𝑡 = 𝛿𝐴𝑒(𝑇𝑠 4 − 𝑇𝑏𝑜𝑑𝑦 4 ) 𝑤 𝑃𝑛𝑒𝑡 = (5.67 × 10−8 2 4 ) (0.27 𝑚2 )(0.7)(293.15 𝐾 4 − 310.15 𝐾 4 ) 𝑚 𝐾 𝑃𝑛𝑒𝑡 = −20.02 𝑤

105

Chapter 14: Thermodynamics Important Terms: Thermodynamics = transfer or the actions of heat System = a definite quantity of matter enclosed by boundaries or surfaces, either real or imaginary Thermally Isolated System = If the heat transfer into or out of the system is impossible Heat Reservoir = A system assumed to have an unlimited heat capacity Equation of State = describe the conditions of thermodynamic systems Process = change in the state or the thermodynamic coordinates of a system Irreversible Process = A process for which the intermediate steps are non-equilibrium states Reversible Process = the process path between the initial and final states would be known First Law of thermodynamics = a statement of the conservation of energy as applied to thermodynamic system Isobaric Process = a constant – pressure process Isometric Process = a constant – volume process Isothermal Process = a constant – temperature process Adiabatic Process = no heat is transferred into or out of the system Second Law of Thermodynamics = heat will not flow spontaneously from a colder body to a warmer body Entrophy = A property that indicates the natural direction of a process Isentropic Process = a constant entropy process Heat Engine = device that converts heat energy to work Thermal Cycle = a series of processes which brings the engine or system back to its original condition.

106

Thermal Pump = a generic term for any device, including refrigerators, air conditioners and heat pump, that transfers heat energy from a low-temperature reservoir to a high temperature reservoir. Coefficient of Performance (COP) = measure of a refrigerator or air conditioner’s performance Heat Pump = common commercial devices used to cool homes and offices in the summer and to heat them in the winter Carnot Cycle = consists of two isotherms and two adiabats and is conveniently represented on a T-S diagram, where it forms a rectangle Relative Efficiency = ratio of two efficiencies Third Law of Thermodynamics = it is impossible to reach a temperature of absolute zero

Important Equations First Law of Thermodynamics 𝑸 = 𝜟𝑼 + 𝑾 Where:

Q = heat U= W = work

Work Done by an Expanding Gas 𝑾 = 𝒑𝜟𝑽 = 𝒑(𝑽𝟐 – 𝑽𝟏 ) Where:

V = volume p = pressure

Change in Entropy 𝜟𝑺 = Where:

𝑸 𝑻

S = system’s entropy Q = heat T = temperature

Thermal Efficiency of a Heat Engine 𝑾𝒏𝒆𝒕 𝑸𝒉𝒐𝒕 – 𝑸𝒄𝒐𝒍𝒅 𝑸𝒄𝒐𝒍𝒅 Є𝒕𝒉 = = = 𝟏– 𝑸𝒉𝒐𝒕 𝑸𝒉𝒐𝒕 𝑸𝒉𝒐𝒕 107

Where:

Є = thermal efficiency W = work Q = heat

Coefficient of Performance 𝑸𝒄𝒐𝒍𝒅 𝑸𝒄𝒐𝒍𝒅 = 𝑾𝒊𝒏 𝑸𝒉𝒐𝒕 – 𝑸𝒄𝒐𝒍𝒅

𝑪𝑶𝑷𝒓𝒆𝒇 = For refrigerator or air conditioner

𝑸𝒉𝒐𝒕 𝑸𝒉𝒐𝒕 = 𝑾𝒊𝒏 𝑸𝒉𝒐𝒕 – 𝑸𝒄𝒐𝒍𝒅

𝑪𝑶𝑷𝒉𝒑 = For heat pump in heating mode Where:

COP = coefficient of performance Q = heat W = work

Carnot Efficiency of an Ideal Heat Engine: 𝑻𝒄𝒐𝒍𝒅 𝑻𝒉𝒐𝒕 – 𝑻𝒄𝒐𝒍𝒅 Є𝒄 = 𝟏 – = 𝑻𝒉𝒐𝒕 𝑻𝒉𝒐𝒕 Where:

Єc = carnot efficiency

Relative Efficiency Є𝒓𝒆𝒍 = Where:

Є𝒕𝒉 Є𝒄

Єrel = relative efficiency Єth = thermal efficiency Єc = carnot efficiency

Example 1: An ideal gas occupies a volume of 22.4L at STP (standard temperature and pressure). While absorbing 2.53kJ of heat from the surroundings, the gas expands isobarically to 32.4L. What is the change in the internal energy of the gas? Given: Vol1= 22.4 L at STP Q= 2.53 KJ Vol2 = 32.4L P1= P2 = 1 atm = 1.01x105 Pa

T1= 0 °C = 273 K Required: ∆𝑈 = ?

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Solution: (1𝑚)3 100𝑐𝑚3 22.4𝐿 𝑥 𝑥 = 0.0224𝑚3 (1000𝑐𝑚)3 1𝐿 (1𝑚)3 100𝑐𝑚3 32.4𝐿 𝑥 𝑥 = 0.0324𝑚3 (1000𝑐𝑚)3 1𝐿 W= PΔV = 1.01x105 N (0.0324 m3- 0.0224 m3) = 1010 J ΔU= Q – W = 2530 J - 1010 J = 1520 J Example 2: What is the change in entropy of ethyl alcohol when 0.25kg of it vaporizes at its boiling point of 78oC (latent heat of vaporization Lv = 1.0 x 105J/kg)? Given: m= 0.25 Kg T= 78 + 273.15 = 351 K LV= 1.0x105 J/Kg

Required: ∆𝑆 = ?

Solution: 𝑄 = 𝑚𝐿𝑣 = 0.25𝑘𝑔 (

1.0𝑥105 𝐽 ) 𝑘𝑔

𝑄 = 25000𝐽 ∆𝑆 =

𝑄 25000 = = 71.19𝐽/𝐾 𝑇 351.15

Example 3: A metal spoon at 24oC is placed in 1.00kg of water at 18oC. The thermally isolated system comes to equilibrium at a temperature of 20oC. Find the approximately change in the entropy of the system. Given: TMS= 24°C TW= 18°C mW= 1 Kg TF= 20°C

Required: ∆𝑆 = ?

Solution: 𝑄𝑊 = 𝑚𝑐∆𝑇 = 1𝑘𝑔 (4189 𝑇𝑊 = 𝑇𝑀𝑆 =

𝐽 ) (20°𝐶 − 18°𝐶) = 8372𝐽 𝑘𝑔°𝐶

𝑇𝐹 + 𝑇𝑊 20 + 18 = = 19 + 273.15 = 292.15°𝐾 2 2 𝑇𝐹 + 𝑇𝑀𝑆 20 + 24 = = 22 + 273.15 = 295.15°𝐾 2 2

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𝑄𝑊 8372𝐽 = = 28.66𝐽/𝐾 𝑇𝑊 292.15 𝑄𝑀𝑆 −8372𝐽 = = = −28.37𝐽/𝐾 𝑇𝑀𝑆 295.15

∆𝑆𝑊 = ∆𝑆𝑀𝑆

∆𝑆 = ∆𝑆𝑊 + ∆𝑆𝑀𝑆 = 28.66 + (−28.37) = 0.29𝐽/𝐾 Example 4: The small gasoline powered engine of a leaf blower removes 800J of heat energy from a high temperature reservoir and exhausts 700J to a low temperature reservoir. What is the engine’s thermal efficiency? Given: QHOT= 800 J QCOLD= 700 J Required: Єth =? Solution: WNET = QHOT − QCOLD = 800 − 700 = 100J Єth =

WNET 100J = = 0.125 x 100% = 12.5% QHOT 800J

Example 5: An air conditioner operating in summer extracts 100J of heat from the interior of the house for every 40J of electric energy required to operate it. Determine its COP and its COP if it is reversed in the winter, running as a heat pump to move the same amount of heat for the same amount of kinetic energy. Given: QCOLD = 100J WNET = 40J Required: COPREF = ? COPHP = ? Solution: QHOT = QCOLD + WINPUT = 100 + 40 QHOT = 140J QCOLD 100 = = 2.5 WINPUT 40 QHOT 140 = = = 3.5 WINPUT 40

COPREF = COPHP

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Example 6: An engineer is designing a cyclic heat engine to operate between temperatures of 150oC and 27oC. What is the maximum theoretical efficiency that can be achieved? Suppose the engine, when built, does 100J of work per cycle for every 500J of input heat per cycle. What is its relative efficiency? Given: THOT= 150 °C + 273.15 = 423.15 K TCOLD= 27 °C + 273.15 = 300.15 K WNET= 100 J QHOT= 500J Required: ∈𝑅𝐸𝐿 = ? Solution: 𝑇𝐶𝑂𝐿𝐷 300.15 =1− = 0.29 𝑋 100% = 29% 𝑇𝐻𝑂𝑇 423.15 𝑊𝑁𝐸𝑇 100 ∈𝑇𝐻 = = = 0.20𝑋 100% = 20% 𝑄𝐻𝑂𝑇 500 ∈ 𝑇𝐻 0.20 ∈𝑅𝐸𝐿 = = 𝑋 100% = 69% ∈𝐶 0.29 ∈𝐶 = 1 −

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