Download presentation

Presentation is loading. Please wait.

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

Similar presentations

Presentation is loading. Please wait....

OK

Electricity and Magnetism

Electricity and Magnetism

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on motion force and pressure for class 9 Ppt on new technology free download Ppt on cloud based mobile social tv Ppt on assyrian and persian empires Ppt on digital media broadcasting company Ppt on role of foreign institutional investment in india Ppt on condition monitoring techniques Ppt on rivers of india in hindi Download ppt on endangered and extinct animals English 8 unit 13 read ppt on iphone