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Week8_1 Drexel PHYS 102
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  • Title: Week8_1
  • Type: Notes
  • School: Drexel
  • Course: PHYS 102
  • Term: Spring

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Law Ampere's Ampere's Law r r states that the line integral of B ds around any closed r path ds equals oI where I is the total steady current passing through any surface bounded by the closed path r r B ds r r B ds = o I Ampere's Law, cont Ampere's Law describes the creation of magnetic fields by all continuous current configurations Most useful for this course if the current configuration has a high degree of symmetry Put the thumb of your right hand in the direction of the current through the amperian loop and your figures curl in the direction you should integrate around the loop Amperian Loops Each portion of the path satisfies one or more of the following conditions: The value of the magnetic field can be argued by symmetry to be constant over the portion of the path The dot product can be expressed as a simple algebraic product B ds The vectors are parallel Amperian Loops, cont Conditions: The dot product is zero The vectors are perpendicular The magnetic field can be argued to be zero at all points on the portion of the path Field Due to a Long Straight Wire From Ampere's Law Want to calculate the magnetic field at a distance r from the center of a wire carrying a steady current I The current is uniformly distributed through the cross section of the wire Field Due to a Long Straight Wire Results From Ampere's Law Outside of the wire, r > R r r B ds = B(2 r ) = oI oI B= 2 r Inside the wire, we need I', the current inside the amperian circle r r B ds = B(2 r ) = oI ' I B = o 2 r 2 R r2 I' = 2 I R Field Due to a Long Straight Wire Results Summary The field is proportional to r inside the wire The field varies as 1/r outside the wire Both equations are equal at r = R Magnetic Field of a Toroid Find the field at a point at distance r from the center of the toroid The toroid has N turns of wire o NI B= 2 r r r B ds = B(2 r ) = oNI Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix A reasonably uniform magnetic field can be produced in the space surrounded by the turns of the wire Each of the turns can be modeled as a circular loop The net magnetic field is the vector sum of all the fields due to all the turns Magnetic Field of a Solenoid, Description The field lines in the interior are Approximately parallel to each other Uniformly distributed Close together This indicates the field is strong and almost uniform Magnetic Field of a Tightly Wound Solenoid The field distribution is similar to that of a bar magnet As the length of the solenoid increases The interior field becomes more uniform The exterior field becomes weaker Ideal Solenoid Characteristics An ideal solenoid is approached when The turns are closely spaced The length is much greater than the radius of the turns For an ideal solenoid, the field outside of solenoid is negligible The field inside is uniform Ampere's Law Applied to a Solenoid Ampere's Law can be used to find the interior magnetic field of the solenoid Consider a rectangle with side l parallel to the interior field and side w perpendicular to the field The side of length l inside the solenoid contributes to the field This is path 1 in the diagram Ampere's Law Applied to a Solenoid, cont Applying Ampere's Law gives r r B ds = path1 r r B ds = B path1 ds = Bl The total current through the rectangular path equals the current through each turn multiplied by the number of turns r r B ds = Bl = oNI Magnetic Field of a Solenoid, final Solving Ampere's Law for the magnetic field is N B = o I = o nI l n = N / l is the number of turns per unit length This is valid only at points near the center of a very long solenoid Consider a solenoid that is very long compared with the radius. Of the following choices, the most effective way to increase the magnetic field in the interior of the solenoid is to 1. 2. 3. double its length, keeping the number of turns per unit length constant, reduce its radius by half, keeping the number of turns per unit length constant, or overwrap the entire solenoid with an additional layer of current-carrying wire. N B = o I = o nI l Chapter 23 Faraday's Law and Induction Michael Faraday 1791 1867 Great experimental physicist Contributions to early electricity include Invention of motor, generator, and transformer Electromagnetic induction Laws of electrolysis Induction An induced current is produced by a changing magnetic field There is an induced emf associated with the induced current A current can be produced without a battery present in the circuit Faraday's Law of Induction describes the induced emf EMF Produced by a Changing Magnetic Field, 1 A loop of wire is connected to a sensitive ammeter When a magnet is moved toward the loop, the ammeter deflects The deflection indicates a current induced in the wire EMF Produced by a Changing Magnetic Field, 2 When the magnet is held stationary, there is no deflection of the ammeter Therefore, there is no induced current Even though the magnet is inside the loop EMF Produced by a Changing Magnetic Field, 3 The magnet is moved away from the loop The ammeter deflects in the opposite direction EMF Produced by a Changing Magnetic Field, Summary The ammeter deflects when the magnet is moving toward or away from the loop The ammeter also deflects when the loop is moved toward or away from the magnet An electric current is set up in the coil as long as relative motion occurs between the magnet and the coil This is the induced current that is produced by an induced emf Faraday's Experiment Set Up A primary coil is connected to a switch and a battery The wire is wrapped around an iron ring A secondary coil is also wrapped around the iron ring There is no battery present in the secondary coil The secondary coil is not electrically connected to the primary coil Faraday's Experiment Findings At the instant the switch is closed, the galvanometer (ammeter) needle deflects in one direction and then returns to zero When the switch is opened, the galvanometer needle deflects in the opposite direction and then returns to zero The galvanometer reads zero when there is a steady current or when there is no current in the primary circuit Faraday's Experiment Conclusions An electric current can be produced by a timevarying magnetic field This would be the current in the secondary circuit of this experimental set-up The induced current exists only for a short time while the magnetic field is changing This is generally expressed as: an induced emf is produced in the secondary circuit by the changing magnetic field The actual existence of the magnetic field is not sufficient to produce the induced emf, the field must be changing Magnetic Flux To express Faraday's finding mathematically, the magnetic flux is used The flux depends on the magnetic field and the area: r r B = B dA Flux and Induced emf An emf is induced in a circuit when the magnetic flux through the surface bounded by the circuit changes with time This summarizes Faraday's experimental results Faraday's Law Statements Faraday's Law of Induction states that the emf induced in a circuit is equal to the time rate of change of the magnetic flux through the circuit Mathematically, d B =- dt Faraday's Law Statements, cont If the circuit consists of N identical and concentric loops, and if the field lines pass through all loops, the induced emf becomes d B = -N dt The loops are in series, so the emfs in the individual loops add to give the total emf A circular loop of wire is held in a uniform magnetic field, with the plane of the loop perpendicular to the field lines. Which of the following will not cause a current to be induced in the loop? 1. 2. 3. p 4. th e or ie p in g lo o us h th e th e tin g in g cr ep ro ta ke pu lli n g th e lo o p ou t crushing the loop rotating the loop about an axis perpendicular to the field lines keeping the orientation of the loop fixed and moving it along the field lines pulling the loop out of the field 25% 25% 25% 25% lo o ab ou nt at i.. of ... ... 10 Figure 23.7 shows a graphical representation of the field magnitude versus time for a magnetic field that passes through a fixed loop and that is oriented perpendicular to the plane of the loop. The magnitude of the magnetic field at any time is uniform over the area of the loop. Rank the magnitudes of the emf generated in the loop at the five instants indicated, from largest to smallest. 1. 2. 3. c, d = e, b, a a, d = e, b, c a, d, c, e, b 33% 33% 33% 10 a c e, e, = = d d c, a, a, d, c, e, b b, b, Faraday's Law Example Assume a loop enclosing an area A lies in a uniform magnetic field The magnetic flux through the loop is B = B A cos The induced emf is d = - ( BA cos ) dt Ways of Inducing an emf The magnitude of the field can change with time The area enclosed by the loop can change with time The angle between the field and the normal to the loop can change with time Any combination of the above can occur Applications of Faraday's Law Pickup Coil The pickup coil of an electric guitar uses Faraday's Law The coil is placed near the vibrating string and causes a portion of the string to become magnetized When the string vibrates at the some frequency, the magnetized segment produces a changing flux through the coil The induced emf is fed to an amplifier Motional emf A motional emf is one induced in a conductor moving through a magnetic field The electrons in the conductor experience a force that is directed along l r r r FB = qv B Motional emf, cont Under the influence of the force, the electrons move to the lower end of the conductor and accumulate there As a result of the charge separation, an electric field is produced inside the conductor The charges accumulate at both ends of the conductor until they are in equilibrium with regard to the electric and magnetic forces Motional emf, final For equilibrium, q E = q v B or E = v B A potential difference is maintained between the ends of the conductor as long as the conductor continues to move through the uniform magnetic field If the direction of the motion is reversed, the polarity of the potential difference is also reversed You wish to move a rectangular loop of wire into a region of uniform magnetic field at a given speed so as to induce an emf in the loop. The plane of the loop must remain perpendicular to the magnetic field lines. In which orientation should you hold the loop while you move it into the region of magnetic field so as to generate the largest emf? 1. 2. 3. with the long dimension of the loop parallel to the velocity vector with the short dimension of the loop parallel to the velocity vector either way because the emf is the same regardless of orientation di lo ng 33% 33% 33% ... .. m en s. or td im en t.. ay be 10 sh th e th e ith ith w w ei th e rw ca us e In Active Figure 23.11, a given applied force of magnitude Fapp results in a constant speed v and a power input P. Imagine that the force is increased so that the constant speed of the bar is doubled to 2v. Under these conditions, what are the new force and the new power input P? 25% 25% 25% 25% 1. 2. 3. 4. 2F and 4F and 2F and 4F and 2P 2P 4P 4P 10 4P 2P 2P d an an d 2F 4F 2F 4F an d an d 4P Sliding Conducting Bar A bar moving through a uniform field and the equivalent circuit diagram Assume the bar has zero resistance The work done by the applied force appears as internal energy in the resistor R Sliding Conducting Bar, cont The induced emf is d B dx = -Bl = -Blv =- dt dt Since the resistance in the circuit is R, the current is Blv I= = R R Sliding Conducting Bar, Energy Considerations The applied force does work on the conducting bar This moves the charges through a magnetic field The change in energy of the system during some time interval must be equal to the transfer of energy into the system by work The power input is equal to the rate at which energy is delivered to the resistor 2 = Fappv = ( I lB ) v = R AC Generators Electric generators take in energy by work and transfer it out by electrical transmission The AC generator consists of a loop of wire rotated by some external means in a magnetic field Induced emf In an AC Generator The induced emf in the loops is d B = -N dt = NAB sin t This is sinusoidal, with max = N A B Lenz' Law Faraday's Law indicates the induced emf and the change in flux have opposite algebraic signs This has a physical interpretation that has come to be known as Lenz' Law It was developed by a German physicist, Heinrich Lenz Lenz' Law, cont Lenz' Law states the polarity of the induced emf in a loop is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop The induced current tends to keep the original magnetic flux through the circuit from changing In equal-arm balances from the early 20th century (Fig. 23.18), it is sometimes observed that an aluminum sheet hangs from one of the arms and passes between the poles of a magnet, which causes the oscillations of the equal arm balance to decay rapidly. In the absence of such magnetic braking, the oscillation might continue for a very long time and the experimenter would have to wait to take a reading. Why do the oscillations decay? 1. 2. 3. The aluminum sheet is attracted to the magnet. Currents in the aluminum sheet set up a magnetic field that opposes the oscillations. Aluminum is ferromagnetic. 33% 33% 33% ... m ... ti a. .. um is fe rr om in u m um ur re n ts in th e 10 ee sh Th C A lu m in e al al u Lenz' Law Example 1 When the magnet is moved toward the stationary loop, a current is induced as shown in a This induced current produces its own magnetic field that is directed as shown in b to counteract the increasing external flux Lenz' Law Example 2 When the magnet is moved away the stationary loop, a current is induced as shown in c This induced current produces its own magnetic field that is directed as shown in d to counteract the decreasing external flux Induced emf and Electric Fields An electric field is created in the conductor as a result of the changing magnetic flux Even in the absence of a conducting loop, a changing magnetic field will generate an electric field in empty space This induced electric field has different properties than a field produced by stationary charges Induced emf and Electric Fields, cont The emf for any closed path can r r be expressed as the line integral of E ds over the path Faraday's Law can be written in a general form r r d B = E ds = - dt Induced emf and Electric Fields, final The induced electric field is a nonconservative field that is generated by a changing magnetic field The field cannot be an electrostatic field because if the field were electrostatic, and hence conservative, the line integral would be zero and it isn't In a region of space, a magnetic field is uniform over space but increases at a constant rate. This changing magnetic field induces an electric field that 1. 2. 3. 4. ns er se ct io di re re a co th e in ha is s a co ns in c is ta n tm ag increases in time, is conservative, is in the direction of the magnetic field, or has a constant magnitude. 25% 25% 25% 25% e, iv e , tim in va t s n of n. .. ... 10 Self-Induction When the switch is closed, the current does not immediately reach its maximum value Faraday's Law can be used to describe the effect Self-Induction, 2 As the current increases with time, the magnetic flux through the circuit loop due to this current also increases with time The corresponding flux due to this current also increases This increasing flux creates an induced emf in the circuit Self-Inductance, 3 The direction of the induced emf is such that it would cause an induced current in the loop which would establish a magnetic field opposing the change in the original magnetic field The direction of the induced emf is opposite the direction of the emf of the battery Sometimes called a back emf This results in a gradual increase in the current to its final equilibrium value Self-Induction, 4 This effect is called self-inductance Because the changing flux through the circuit and the resultant induced emf arise from the circuit itself The emf L is called a self-induced emf Self-Inductance, Equations An induced emf is always proportional to the time rate of change of the current dI L = -L dt L is a constant of proportionality called the inductance of the coil It depends on the geometry of the coil and other physical characteristics Inductance of a Coil A closely spaced coil of N turns carrying current I has an inductance of NB L= =- I dI dt The inductance is a measure of the opposition to a change in current Compared to resistance which was opposition to the current Inductance Units The SI unit of inductance is a Henry (H) V s 1H = 1 A Named for Joseph Henry Joseph Henry 1797 1878 Improved the design of the electromagnet Constructed one of the first motors Discovered the phenomena of selfinductance Inductance of a Solenoid Assume a uniformly wound solenoid having N turns and length l Assume l is much greater than the radius of the solenoid The interior magnetic field is N B = o nI = o I l Inductance of a Solenoid, cont The magnetic flux through each turn is Therefore, the inductance is 2 N B o N A L= I = l NA I B = BA = o l This shows that L depends on the geometry of the object RL Circuit, Introduction A circuit element that has a large selfinductance is called an inductor The circuit symbol is We assume the self-inductance of the rest of the circuit is negligible compared to the inductor However, even without a coil, a circuit will have some self-inductance RL Circuit, Analysis An RL circuit contains an inductor and a resistor When the switch is closed (at time t=0), the current begins to increase At the same time, a back emf is induced in the inductor that opposes the original increasing current RL Circuit, Analysis, cont Applying Kirchhoff's Loop Rule to the previous circuit gives dI - IR - L = 0 dt Looking at the current, we find -Rt L I= (1 - e R ) RL Circuit, Analysis, Final The inductor affects the current exponentially The current does not instantly increase to its final equilibrium value If there is no inductor, the exponential term goes to zero and the current would instantaneously reach its maximum value as expected RL Circuit, Time Constant The expression for the current can also be expressed in terms of the time constant, , of the circuit I (t ) = (1 - e ) R -t where = L / R Physically, is the time required for the current to reach 63.2% of its maximum value RL Circuit, Current-Time Graph, 1 The equilibrium value of the current is /R and is reached as t approaches infinity The current initially increases very rapidly The current then gradually approaches the equilibrium value RL Circuit, Current-Time Graph, 2 The time rate of change of the current is a maximum at t = 0 It falls off exponentially as t approaches infinity In general, dI -t = e dt L The circuit in Figure 23.28 includes a power source that provides a sinusoidal voltage. Thus, the magnetic field in the inductor is constantly changing. The inductor is a simple air-core solenoid. The switch in the circuit is closed and the lightbulb glows steadily. An iron rod is inserted into the interior of the solenoid, which increases the magnitude of the magnetic field in the solenoid. As that happens, the brightness of the lightbulb 1. 2. 3. increases, decreases, or is unaffected. 33% 33% 33% 10 , or se s s, cr ea in de is un af fe c cr ea se te d . Two circuits like the one shown in Active Figure 23.26 are identical except for the value of L. In circuit A, the inductance of the inductor is LA, and in circuit B, it is LB. The switch has been in position b for both circuits for a long time. At t = 0, the switch is thrown to a in both circuits. At t = 10 s, the switch is thrown to b in both circuits. The resulting graphical representation of the current as a function of time is shown in Figure 23.29. Assuming that the time constant of each circuit is much less than 10 s, which of the following is true? 1. 2. 3. LA > LB. LA < LB There is not enough information to determine the relative values. 33% 33% 33% LB . LB > < i.. no te no . 10 LA Th er e is ug h LA Energy in a Magnetic Field In a circuit with an inductor, the battery must supply more energy than in a circuit without an inductor Part of the energy supplied by the battery appears as internal energy in the resistor The remaining energy is stored in the magnetic field of the inductor Energy in a Magnetic Field, cont Looking at this energy (in terms of rate) dI 2 I = I R + LI dt I is the rate at which energy is being supplied by the battery I2R is the rate at which the energy is being delivered to the resistor Therefore, LI dI/dt must be the rate at which the energy is being delivered to the inductor Energy in a Magnetic Field, final Let U denote the energy stored in the inductor at any time The rate at which the energy is stored is dUB dI = LI dt dt To find the total energy, integrate and UB = L I2 Energy Density of a Magnetic Field Given U = L I2, B 1 B2 U = o n 2 Al Al = 2 2 o o n 2 Since Al is the volume of the solenoid, the magnetic energy density, uB is U B2 uB = = V 2o This applies to any region in which a magnetic field exists not just the solenoid You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Which of the following increases the energy density? (More than one choice may be correct.) 1. 2. 3. th e on ly th e g g g si n si n si n cr ea cr ea cr ea in in in in cr ea si n g th e 4. increasing the number of turns per unit length on the solenoid increasing the crosssectional area of the solenoid increasing only the length of the solenoid while keeping the number of turns per unit length fixed increasing the current in the solenoid 25% 25% 25% 25% s... m be l.. rr en cu t. . ... . 10 cr os nu th e Inductance Example Coaxial Cable Calculate L and energy for the cable The total flux is B = BdA = b a o I Il b ldr = o ln 2 r 2 a Therefore, L is L= The total energy is 1 2 o lI 2 b U = LI = ln 2 4 a B o l b = ln 2 a I

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Statics HW _13, Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, April 20th, 2008. Statics MCE 220 Assignment # 13 Solve the following problems from the textbook (Engineering Mechanics: Statics ...
Tensile Test - Al 1
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Length: Diameter: Tensile Test Without Extensometer Group2 AUS Material Sc. Lab 9/28/2003 Circular 1 aluminum 95 mm 6.1 mm Extension mm 0.000478316 0.01973053 0.03683033...
MCE 230 -Static Bending Test
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: MCE 230 L Materials Science Static Bending Test Gagan Gururaj ID: 8526 Group 2 Date of Experiment: 16/11/03 2 Table of Contents 1) Discussion.3 2) Introduction..3 3) Theoretical Background.3 4) Theory.3 5) Experimental Procedure.4 6) Experimental...
Static Bending Test - 90 Scratches
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Width: Thickness: Support span: Span ratio: Fixture type: Comment: Static Bending Test Fall 2003 Gourp 2 AUS Materials Science 11/16/2003 Rectangular 1 Wood 25 mm 25 mm ...
Static Bending Test - Without Scratches
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Width: Thickness: Support span: Fixture type: Comment: Static Bending Fall 2003 Gourp 2 AUS Materials Science 11/16/2003 Rectangular 1 Wood 25 25 360 3-point Without Scr...
Presentation ARA 101 'Abbasids [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: Caliphal Succession: The `Abbasid Caliphate The `Abbasid Caliphate The `Abbasid period in its entirety covers approximately half a millennium spanning the years 132/750 to 656/1258. Consequently, historians have divided the period into three sub-per...
MCE 230 Copied
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: SUMMARY: Fatigue is a form of failure that occurs in structures subjected to dynamic and fluctuating stresses (e.g., bridges, aircraft, and machine component); in this experiment we see that four different metals aluminum, copper, brass and steel, wh...
MCE 230 - Impact Test (Table)
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Specimen Aluminum Aluminum Brass Brass Aluminum Aluminum Aluminum Notch type V U V U V V V Temperature (degrees celsius) 25 25 25 25 100 200 300 Width (cm) 0.5 0.49 0.5 0.49 0.5 0.5 0.5 Depth (cm) 0.49 0.41 0.49 0.41 0.49 0.49 0.49 Net Impact Wo...
Compression Test - 2 (Secondary Parallel)
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Width: Thickness: Length: Compression Test with Secondary Parallel Sample Fall 2003-Group 2 AUS Material Sc. Lab 10/19/2003 Rectangular area=.000625 1 Wood 25 mm 25 mm 1...
MCE 230 - Hardness Test
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: MCE 230 L Materials Science Hardness Test Gagan Gururaj ID: 8526 Group 2 Date of Experiment: 26/10/03 2 Table of Contents 1) Abstract.3 2) Introduction..3 3) Theory.3 4) Experimental procedure.3 5) Data..4 6) Calculations.4 7) Discussion.5 8) Refe...
Statics HW _12, Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, April 13th, 2008. Statics MCE 220 Assignment # 12 Solve the following problems from the textbook (Engineering Mechanics: Statics ...
MCE 230 - Cover Page
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: MCE 230 Materials Science Gagan Gururaj ID: 8526 Email: b00008526@aus.ac.ae Instructor: Dr. Hany El-Kadi Date due: 14/12/03 Problems: Ch.4 4.4 4.11 4.18 4.23 4.34 Ch.5 5.3 5.6 5.11 5.25 ...
MCE 230 - Fatigue Test
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Abstract: Three samples of unpolished aluminum and three samples of unpolished copper were used in the fatigue test. Under an applied load of 80N, the first sample of aluminum fractured after 13892 load cycles. Similarly, the second sample of aluminu...
Compression Test (All graphs)
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Stress (Pa) Primary parallel Secondary parallel Perpendicular Strain Stress Strain stress Strain Stress 0 6.05E+01 0 492.917 0 5803.364 0.000209 8.94E+04 0.000413 91217.01 6.26E-05 95780.96 0.000417 1.78E+05 0.000832 223763 0.000108 196940.9 0.00062...
NaḳÄ'id EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: Na i Meaning \"contradicting poems, flytings\", pl. of na a (from the verb na a a \"to destroy, undo, rebut, oppose\"), synonymous with mun a t (from the verbal form III n a a): a form of poetic duelling in which tribal or personal insults are exchange...
Jahiliyya EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: hilliya A term used, in almost all its occurrences, as the opposite of the word islm , and which refers to the state of affairs in Arabia before the mission of the Prophet, to paganism (sometimes even that of non-Arab lands), the pre-Islamic perio...
ARA 101 Essay Questions [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: AMERICAN UNIVESITY OF SHARJAH COLLEGE OF ARTS AND SCIENCES Department of Arabic Studies ARA 101 Readings in Arabic Heritage Essay Questions for Fall 2007 To be submitted by 16.00 on the 5th of December 2007. The essays should be submitted in soft co...
Introduction ARA 101E (1)
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: ARA 101 E Readings in Arabic Heritage (English Sections) Dr. Gavin Picken Office: NAB 222 Office Tel: ext. 2873 Office Hours: Sunday 11-12, Monday 11-12, Tuesday 3-4 and Wednesday 3-4. E-mail: gpicken@aus.edu B. Sc. Biological Sciences University ...
al-Akhtal EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: - al b. al- alt, who died probably before 92/710. He belonged to the great tribe of the lib [q.v.] of northern Syria, which remained entirely Christian, of the Monophysite persuasion. By his mother Layla he was connected to another Christian tribe...
Presentation ARA 101 Umayyad Poetry [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: Poetry in the Umayyad Period: Al-Farazdaq, Jarir and al-Akhtal Poetry in the Umayyad Period Tammm b. Ghalib (Ab Firs) al- Farazdaq \"the lump of dough\", famous Arab satirist and panegyrist, died at Basra about 110/728 or 112/730. Born in Yamma on a ...
al-Farazdaq EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: al- Farazda \"the lump of dough\", properly Tammm b. lib (Ab Firs), famous Arab satirist and panegyrist, died at Ba ra about 110/728 or 112/730. Born in Yamma (Eastern Arabia) on a date which is uncertain (probably after 20/640), this poet i, of the Dr...
ḤassÄn b. Thabit EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: bit - ir b. arm Of the tribe of rib (later Medna), traditionally known as the \"poet laureate\" of the Prophet, is more correctly the most prominent of several poets who were associated with the rise of Islam , and one who already had an establis...
Ḳaṣīda EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: a da 1. In Arabic. a da collective a d is the name given in Arabic to some poems of a certain length. It is derived from the root a ada, \"to aim at\", for the primitive a da was intended to eulogize the tribe of the poet and denigrate the opposi...
Ḳabīla EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: abla Denotes a large agnatic group, the members of which claim to be descended from one common ancestor; this word is generally understood in the sense of tribe. It derives from the Arabic root -b-l, of which the form bala signifies to meet, to be...
Jarir EI [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: arr b. A iyya b. al- a ayfa) b. Badr Was among the most important -writers of the Umayyad period (the other two were his rivals al and al-Farazda [qq.v.], and may be considered one of the greatest Islamic-Arabic poets of all time. He belonged...
Hasan%20b.%20Thabit%20EAL%20%5b1%5d
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: ...
al-Farazdaq [1]
Path: American University of Sharjah >> ARA >> 101,102 Spring, 2008
Description: From the wafayat al-a`yan of Ibn Khallikan Abu-l ,Abbas Ahmad b. Khallikan (1211-1282 CE) was born at Arbela, Iraq. By his talents and his writings he received the title of the most learned man and the ablest historian of that city. He studied jurisp...
Statics HW _6 Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, February 24th, 2008. Statics MCE 220 Assignment # 6 Solve the following problems from the textbook (Engineering Mechanics: Static...
Statics HW _5 Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Tuesday, February 19th, 2008. Statics MCE 220 Assignment # 5 Solve the following problems from the textbook (Engineering Mechanics: Stati...
Statics HW _3 Springr 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Tuesday, February 12th, 2008. Statics MCE 220 Assignment # 3 Solve the following problems from the textbook (Engineering Mechanics: Stati...
Statics HW _1 Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, February 3rd, 2008. Solve the following problems 1. Round off the following numbers to three significant figures: (a) 4.65735 m, (b...
Statics HW _2 Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, February 10, 2008. Statics MCE 220 Assignment # 2 Solve the following problems from the textbook (Engineering Mechanics: Statics ...
Statics HW _7, Spring 2008
Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, March 2nd, 2008. Statics MCE 220 Assignment # 7 Solve the following problems from the textbook (Engineering Mechanics: Statics by...
Tensile Test - Al 2
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Length: Diameter: Tensile Test for Metallic Materials With Extensometer Group 2 AUS Material Sc. Lab 9/28/2003 Circular 1 aluminum 95 6.2 Extension mm Load N Strain Stre...
Tensile Test - steel
Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Length: Diameter: Tensile Test Group 2 AUS Material Sc. Lab 9/28/2003 Circular 1 Steel 95 mm 5.9 mm Extension mm 0 0.01147958 0.02062737 0.0307318 0.0409558 0.05207665 0...

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