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Chapters 25--Examples. Problem. The current in a wire varies with time according to the relation I=55A-(0.65 A/s 2 )*t 2 How many coulombs of charge pass through a cross-section of wire in the time interval from t=0 s to t=8 s?

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problem
Problem

The current in a wire varies with time according to the relation

I=55A-(0.65 A/s2)*t2

  • How many coulombs of charge pass through a cross-section of wire in the time interval from t=0 s to t=8 s?
  • What constant current would transport the same amount of charge?
integratin means summin
Integratin’ means summin’
  • Need to sum the charge between t=0 to t=8 i.e. integrate
problem1
Problem

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

  • The current carried by the wire?
  • The potential difference between two points 6.4 m apart?
  • The resistance of a 6.4 m length of this wire?
problem2
Problem

A 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.

  • What is the magnitude and direction of the current density J?
  • Can you calculate the total current i in the beam? If not, what else do you need to know?
part a1
Part 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 vd
part b1
Part B
  • i=J*A but what is A?
problem3
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 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?

n

p

p

part a2
Part 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 A
part b2
Part B
  • J=i/A
  • A=pr2
    • where r=0.165 mm = 0.165 x 10-3 m
  • J=(9.2 x 10-4)/(p* (0.165 x 10-3 )2
  • J=1.08 x 104 A/m2
problem4
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 cm2 and length 0.85 m?

prep stuff
Prep Stuff
  • r for Cu = 1.69 x 10-8 W*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/m3
solution
Solution
  • 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 min
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