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# University Physics II PHYS 260

Mason

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This 35 page Class Notes was uploaded by Sonny Breitenberg on Monday September 28, 2015. The Class Notes belongs to PHYS 260 at George Mason University taught by Neil Goldman in Fall. Since its upload, it has received 44 views. For similar materials see /class/215187/phys-260-george-mason-university in Physics 2 at George Mason University.

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Date Created: 09/28/15

Exam 2 Topics DC Circuits Current amp Ohm s Law Macro and Microscopic Power Kirchhoff s Loop Rules ChargingDischarging Capacitor RC Circuits Magnetic Fields Force due to Magnetic Field Lorentz Force Magnetic Dipoles Generating Magnetic Fields BiotSavart Law amp Ampere s Law General Exam Suggestions You should be able to complete every problem m If you are confused ask If it seems too hard you aren t thinking enough Look for hints in other problems If you are doing math you re doing too much Read directions completely before amp after Write down what you know before starting Draw pictures define label variables Make sure that unknowns drop out of solution Don t forget units Current amp Ohm s Law gt dQ 69 A I E I t gt 39 d j i A gtE h b W Epjltgtj AVIR Rz O Series vs Parallel Series Parallel Current same Currents add Voltages add Voltages same Current Voltage amp Power Battery 8 I P 1d21AV218 II b supple AV8 2 Pdissi ated 12R 9 W p R AV IR b Capacitor pabsorbed AV 2022 39 dt C 9 I I Q 2 39 I z iQdU AV QC dt2C dt Kirchhoff s Rull9 A I DisCharging A Capacitor W Q Q 11 w m I 01 Ice 3 ZAVZ g Q IR0 Q nal T CE Q RC gzo dt General Comment RC All Quantities Either Value Value0 t T Valuet ValueFmal 1 e W Valuet Valueoe t C can be obtained from differential equation prefactor on ddt eg C RC Right Hand Rules 1 Torque Thumb torque Fingers show rotation 2 Feel Thumb I Fingers B Palm F 3 Create Thumb l Fingers curl B 4 Moment Fingers curl l Thumb Moment B inside loop Magnetic Force FBqVgtltB A g d fmxrz FELL VQi FB1LgtltB Magnetic Dipole Moments EIA EIA Feel 1 Torque to align with external field 2 Forces as for bar magnets Helmholtz Coil Common Concept Question Parallel Helmholtz makes uniform field torque no force Antiparallel makes zero non uniform field force no torque Helmholtz Currents The BiotSavart Law Current element of length ds carrying current I or equivalently charge q with velocity v produces a magnetic field 4371 uo qVXI 2 BiotSavart 2 Problem Types Notice that r is the same for every point on the loop You don t really need to integrate except to find path length Ampere s Law 4 d 2 yo Long Circular Solenoid 2 Current Sheets TorusCoax SAMPLE EXAM PPPP 23 Problem 1 Wire Loop P0 In 20 A current flowing in the circuit pictured produces a magnetic field at point P pointing out of the page with magnitude B a What direction is the current flowing in the circuit b What is the magnitude of the current flow Solution 1 Wire Loop a The current is flowing counterclockwise as shown above b There are three segments of the wire the semicircle the two horizontal leads and the two vertical leads The two vertical leads do not contribute to the B field ds H r The two horizontal leads make an infinite wire a distance D from the field point Solution 1 Wire Loop For infinite wire use Ampere s Law q zyOIeijZ Dzon B 2 HO 27Z39D u Id gtltf D A For the semicircle dB 2 4 0 2 I andd J 1 7 r 2 use BiotSavart A uo Id gtltr B dB J47 r2 I I I 47 r2 4r 2D Solution 1 Wire Loop Adding together the two parts 32 010101 1 27Z39D 2D 2D 7r They gave us B and want I to make that B 2DB yo 1i 7r 1 Problem 2 RC Circuit Initially C is uncharged 1 When the switch is first closed what is the current is 2 After a very long time how much charge is stored on the capacitor 3 Obtain a differential equation for the charge on the capacitor Here only let R1R2R3R Now the switch is opened 4 Immediately after opening the switch what is i1 i2 i3 5 How long before i2 falls to 1e of this initial value Solution 2 RC Circuit Initially C is uncharged gt Looks like short 1 8 R 12 gti eq 3 3 ii Req Solution 2 RC Circuit After a long time C is full gt i2 O C8 R1 R3 Solution 2 RC Circuit R R Kirchhoff s Loop Rules m2 Left i3R8 i1R0 I18 q Rightz i3R8 i2R qc 0 39q Current i3 i1 i2 R Want to have i2 and q only L 2R O i1i2R8 i1R2i1i2R 282i2R2 3i2R 82 2 dt dt 3R 3RC Solution 2 RC Circuit Now open the switch Q R l R1 1 C ll 2 8 C 11213 R1R2 R1R3R1R2 Solution 2 RC Circuit How long to fall to 1e of initial current The time constant This is an easy circuit since itjust looks like a resistor and capacitor in senesso C1 q 721121 R2C Notice that this is different than the charging time constant because there was another resistor in the circuit during the charging Problem 3 NonUniform Slab y Consider the slab at left with it of Page nonuniform current density e x A JJ0 k d Find B everywhere 4 2d gt Solution 3 NonUniform Slab if Jog Direction Up on right down on left B o B Inside at 0ltxltd 4365 ojenc II gt 5 C Bd BEOOO 6 x J x x 35 0 enc 0 0 d I w 0 d 2 W B u 2d 1u0 d 2 p Solution 3 NonUniform Slab IC A Direction Up on right down on left y JJ0 k JL d a B Outside x gt d q B d olenc q dg36000 56 3 E E a a de i I JdAz 0 de x 35 0 enc 0 d I Lo 0 d 2 xVA BOJ0dup 2d gt Problem 4 Solenoid A current I flows up a very long solenoid and then back down a wire lying along its axis as pictured The wires are negligibly small ie their radius is O and are wrapped at n turns per meter a What is the force per unit length magnitude and direction on the straight wire due to the current in the solenoid b A positive particle mass m charge q is launched inside of the solenoid at a distance r a to the right of the center What velocity direction and nonzero magnitude must it have so that the field created by the wire along the axis never exerts a force on it Solution 4 Solenoid SUPERPOSITION You can just add the two fields from each part individually a Force on wire down axis Since the current is antiparallel to the field produced by the solenoid there is no force FO on this wire b Launching Charge q The central wire produces a field that wraps in circles around it To not feel a force due to this field the particle must always move parallel to it it must move in a circle of radius a since that is the radius it was launched from Solution 4 Solenoid R H b Launching Charge q So first we should use Ampere s law to calculate the field due to the solenoid ci d BlyONI 5 I v I B UOZ uOnI up the solenoid Now we just need to make a charge q move in a circular orbit with r a 2 2 F 7 v mv B q xB qu mg A B I v qa 2W out of the page m m Problem 5 Coaxial Cable Consider a coaxial cable of with inner conductor of radius a and outer conductor of inner radius b and outer radius 0 A current lflows into the page on the inner conductor and out of the page on the outer conductor What is the magnetic field everywhere magnitude and direction as a function of distance rfrom the center of the wire Solution 5 Coaxial Cable Everywhere the magnetic field is clockwise To figure out the magnitude use Ampere s Law Cilia zyolenc gt B27rr zyol 710 I gt B luO enc Drawn for a lt rlt b 271739 The amount of current penetrating our Amperian loop depends on the radius r 2 r rSa 16m 21 2 gt B a 27m 017 2 clockwise Solution 5 Coaxial Cable B IuOIenc 271739 clockwise I gt B u Jclookwise 2727 rz bz bSrSc End I 1 62b2 2 2 gt Bz u OIG r bjolockwise 2 2 6

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