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1 Chapters 25--Examples

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2 Problem The current in a wire varies with time according to the relation I=55A-(0.65 A/s 2 )*t 2 a) 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?

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3 Integratin means summin Need to sum the charge between t=0 to t=8 i.e. integrate

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4 Problem A current-carrying gold wire has a diameter of 0.84 mm. The electric field is 0.49 V/m. What is a) 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?

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5 Part A

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6 Part B

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7 Part C

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8 Problem A beam contains 2 x 10 8 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 1x10 5 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?

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9 Part A J=n*q*v d n= 2 x 10 8 ions/cc and 1 cc= 1/100 3 m 3 so n= 2x 10 14 ions/m 3 q=2e = 2 * 1.602x10 -19 v d = 10 5 m/s J=(2x10 14 )(2*1.602x10 -19 )*10 5 J=6.4 A/m 2 and J is in same direction as v d

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10 Part B i=J*A but what is A?

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11 Problem 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 10 15 electrons/second flow from the n side to the p side while 2.25 x 10 15 holes (positive charge carriers) flow from the p to the n side. What is a) the total current and b) the charge density? p n p

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12 Part A Find net charge=(n e +n h )*e =(3.5 x 10 15 +2.25 x 10 15 )*1.602 x 10 -19 Both of these were charge rates (rates/second) So i= =(3.5 x 10 15 +2.25 x 10 15 )*1.602 x 10 -19 =9.2 x 10 -4 A

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13 Part B J=i/A A= r 2 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 10 4 A/m 2

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14 Problem How 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 cm 2 and length 0.85 m?

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15 Prep Stuff for Cu = 1.69 x 10 -8 *m A=0.21 cm 2 where 1 cm 2 =(1/100) 2 m 2 n=number of electrons/ m 3 Assume 1 conduction electron for each Cu atom Mass/m 3 *(mol/mass)*(atoms/mol)=atoms/m 3 The atomic weight is the mass/mol= 64 x 10 -3 kg/m 3 The number of atoms per mol is Avogadros number (6.02 x 10 23 ) Density of Cu = 9000 kg/m 3 n=9000*(1/64)*6.02 x 10 23 =8.47 x 10 28 atoms/m 3 or e/m 3

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16 Solution J=n*q*v d v d =J/(n*q) where J=i/A v d =i/(n*q*A) d=v*t or t=d/v d t=d*(n*q*A)/(i) t=(0.85)*(8.47 x 10 28 *1.602x 10 -19 * 0.21x10 4 )/300 t=8.1 x 10 2 s=13 min

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