Chapters 25--ExamplesProblemIntegratin’ means summin’Slide 4Part APart BPart CProblemSlide 9Slide 10Slide 11Slide 12Slide 13Slide 14Prep StuffSolution1Chapters 25--Examples2ProblemThe current in a wire varies with time according to the relationI=55A-(0.65 A/s2)*t2a) How many coulombs of charge pass through a cross-section of wire in the time interval from t=0 s to t=8 s?b) What constant current would transport the same amount of charge?3Integratin’ means summin’Need to sum the charge between t=0 to t=8 i.e. integrate AtqiCqttdttqtidtiqdtdqi141832932936505565055650558038022.currentconstantFor...Recall4ProblemA current-carrying gold wire has a diameter of 0.84 mm. The electric field is 0.49 V/m . What isa) The current carried by the wire?b) The potential difference between two points 6.4 m apart?c) The resistance of a 6.4 m length of this wire?5Part AAIrAAUofAEIEJorJEwhereJAI11110442284049021084010442822228...*...6Part BVVLrAAUofIALIIALRwhereIRV133284046104421114621084010442111282228...*......7Part C2 8011 11 33...AVIVRALRwhereIRV8ProblemA beam contains 2 x 108 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 1x105 m/s.a) What is the magnitude and direction of the current density J?b) Can you calculate the total current i in the beam? If not, what else do you need to know?9Part AJ=n*q*vdn= 2 x 108 ions/cc and 1 cc= 1/1003 m3 so n= 2x 1014 ions/m3q=2e = 2 * 1.602x10-19vd = 105 m/sJ=(2x1014)(2*1.602x10-19)*105J=6.4 A/m2 and J is in same direction as vd10Part Bi=J*A but what is A?11ProblemA pn junction is formed from two different semiconducting materials in the form of identical cylinders with a radius of 0.165 mm (as shown below). In one application, 3.5 x 1015 electrons/second flow from the n side to the p side while 2.25 x 1015 holes (positive charge carriers) flow from the p to the n side. What is a) the total current and b) the charge density?pnp12Part AFind net charge=(ne+nh)*e=(3.5 x 1015 +2.25 x 1015)*1.602 x 10-19Both of these were charge rates (rates/second)So i= =(3.5 x 1015 +2.25 x 1015)*1.602 x 10-19 =9.2 x 10-4 A13Part BJ=i/AA=r2 where r=0.165 mm = 0.165 x 10-3 mJ=(9.2 x 10-4)/( 0.165 x 10-3 )2J=1.08 x 104 A/m214ProblemHow long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and the electrons travel through a copper wire with cross-sectional area 0.21 cm2 and length 0.85 m?15Prep Stuff for Cu = 1.69 x 10-8 *mA=0.21 cm2 where 1 cm2 =(1/100)2 m2n=number of electrons/ m3Assume 1 conduction electron for each Cu atomMass/m3 *(mol/mass)*(atoms/mol)=atoms/m3The atomic weight is the mass/mol= 64 x 10-3 kg/m3The number of atoms per mol is Avogadro’s number (6.02 x 1023)Density of Cu = 9000 kg/m3n=9000*(1/64)*6.02 x 1023=8.47 x 1028 atoms/m3 or e/m316SolutionJ=n*q*vdvd =J/(n*q) where J=i/Avd =i/(n*q*A)d=v*t or t=d/vdt=d*(n*q*A)/(i)t=(0.85)*(8.47 x 1028*1.602x 10-19* 0.21x104)/300t=8.1 x 102 s=13