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