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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/s 2 )*t 2 a) How many coulombs of charge.

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Presentation on theme: "1 Chapters 25--Examples. 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."— Presentation transcript:

1 1 Chapters 25--Examples

2 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?

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

4 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?

5 5 Part A

6 6 Part B

7 7 Part C

8 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?

9 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 ions/m 3 q=2e = 2 * 1.602x v d = 10 5 m/s J=(2x10 14 )(2*1.602x )*10 5 J=6.4 A/m 2 and J is in same direction as v d

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

11 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 electrons/second flow from the n side to the p side while 2.25 x 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

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

13 13 Part B J=i/A A= r 2 where r=0.165 mm = x m J=(9.2 x )/( x ) 2 J=1.08 x 10 4 A/m 2

14 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?

15 15 Prep Stuff for Cu = 1.69 x *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 kg/m 3 The number of atoms per mol is Avogadros number (6.02 x ) Density of Cu = 9000 kg/m 3 n=9000*(1/64)*6.02 x =8.47 x atoms/m 3 or e/m 3

16 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 *1.602x * 0.21x10 4 )/300 t=8.1 x 10 2 s=13 min


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