{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Chapter-26 - Chapter 26 1(a The charge that passes through...

This preview shows pages 1–4. Sign up to view the full content.

1047 Chapter 26 1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 × 10 3 C. (b) The number of electrons N is given by q = Ne , where e is the magnitude of the charge on an electron. Thus, N = q / e = (1200 C)/(1.60 × 10 –19 C) = 7.5 × 10 21 . 2. Suppose the charge on the sphere increases by q in time t . Then, in that time its potential increases by V q r = 4 0 box4 ε , where r is the radius of the sphere. This means q r V = 4 0 box4 ε . Now, q = ( i in i out ) t , where i in is the current entering the sphere and i out is the current leaving. Thus, ( ( ( 29 ( 29 0 9 in out in out 3 0.10 m 1000 V 4 8.99 10 F/m 1.0000020 A 1.0000000 A 5.6 10 s. r V q t i i i i πε - ∆ = = = - - × - = × 3. We adapt the discussion in the text to a moving two-dimensional collection of charges. Using σ for the charge per unit area and w for the belt width, we can see that the transport of charge is expressed in the relationship i = σ vw , which leads to σ = = × × = × - - - i vw 100 10 30 50 10 6 7 10 6 2 6 A m s m C m 2 b gc h . . 4. (a) The magnitude of the current density vector is J i A i d = = = × × = × - - - box4 box4 2 10 3 2 5 4 4 12 10 2 5 10 2 4 10 / . . . . A m A / m 2 c h c h (b) The drift speed of the current-carrying electrons is

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 26 1048 v J ne d = = × × × = × - - - 2 4 10 8 47 10 160 10 18 10 5 28 19 15 . . / . . A / m m C m / s. 2 3 c hc h 5. The cross-sectional area of wire is given by A = π r 2 , where r is its radius (half its thickness). The magnitude of the current density vector is J i A i r = = / / π 2 , so r i J = = × = × - box4 box4 050 440 10 19 10 4 4 . . A A / m m. 2 c h The diameter of the wire is therefore d = 2 r = 2(1.9 × 10 –4 m) = 3.8 × 10 –4 m. 6. We express the magnitude of the current density vector in SI units by converting the diameter values in mils to inches (by dividing by 1000) and then converting to meters (by multiplying by 0.0254) and finally using J i A i R i D = = = box4 box4 2 2 4 . For example, the gauge 14 wire with D = 64 mil = 0.0016 m is found to have a (maximum safe) current density of J = 7.2 × 10 6 A/m 2 . In fact, this is the wire with the largest value of J allowed by the given data. The values of J in SI units are plotted below as a function of their diameters in mils. 7. (a) The magnitude of the current density is given by J = nqv d , where n is the number of particles per unit volume, q is the charge on each particle, and v d is the drift speed of the particles. The particle concentration is n = 2.0 × 10 8 /cm 3 = 2.0 × 10 14 m –3 , the charge is q = 2 e = 2(1.60 × 10 –19 C) = 3.20 × 10 –19 C, and the drift speed is 1.0 × 10 5 m/s. Thus,
1049 J = × × × = - 2 10 32 10 10 10 6 4 14 19 5 / . . . . m C m / s A / m 2 c hc hc h (b) Since the particles are positively charged the current density is in the same direction as their motion, to the north. (c) The current cannot be calculated unless the cross-sectional area of the beam is known. Then i = JA can be used.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}