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Chapters 25--Examples.

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1 Chapters 25--Examples

2 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?

3 Integratin’ means summin’
Need to sum the charge between t=0 to t=8 i.e. integrate

4 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?

5 Part A

6 Part B

7 Part C

8 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?

9 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

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

11 Problem A pn junction is formed from two different semiconducting materials in the form of identical cylinders with a radius of 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

12 Part A Find net charge=(ne+nh)*e
=(3.5 x x 1015)*1.602 x 10-19 Both of these were charge rates (rates/second) So i= =(3.5 x x 1015)*1.602 x =9.2 x 10-4 A

13 Part B J=i/A A=pr2 J=(9.2 x 10-4)/(p* (0.165 x 10-3 )2
where r=0.165 mm = x 10-3 m J=(9.2 x 10-4)/(p* (0.165 x 10-3 )2 J=1.08 x 104 A/m2

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 cm2 and length 0.85 m?

15 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

16 Solution J=n*q*vd vd =J/(n*q) where J=i/A d=v*t or t=d/vd
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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