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,

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Mabel Skinner
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1 Chapter 6 1 (a) The charge that passes through any cross section is the product of the current and time Since t = 4 min = (4 min)(6 s/min) = 4 s, q = it = (5 A)(4 s) = 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 = (1 C)/( C) = Suppose the charge on the sphere increases by Δq in time Δt Then, in that time its potential increases by Δq Δ V =, 4πε r where r is the radius of the sphere This means Δ q= 4 πε rδ V Now, Δq = (i in i out ) Δt, where i in is the current entering the sphere and i out is the current leaving Thus, t Δ = = = iin iout iin iout s = ( 1 m)( 1 V) 9 ( F/m)( 1 A 1 A) Δq 4πε rδv 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 = σvw, which leads to i σ = = vw A 3 ms 5 1 b gc mh 6 = 67 1 Cm 4 We express the magnitude of the current density vector in SI units by converting the diameter values in mils to inches (by dividing by 1) and then converting to meters (by multiplying by 54) and finally using i i 4 i J = = = A π πd 139

2 14 CHAPTE 6 For example, the gauge 14 wire with D = 64 mil = 16 m is found to have a (maximum safe) current density of J = A/m 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 5 (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 = 1 8 /cm 3 = 1 14 m 3, the charge is and the drift speed is m/s Thus, q = e = ( C) = C, c hc hc h J = 1 / m 3 1 C 1 1 m/s = 64 A/m (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 6 (a) Circular area depends, of course, on r, so the horizontal axis of the graph in Fig 6-3(b) is effectively the same as the area (enclosed at variable radius values), except for a factor of π The fact that the current increases linearly in the graph means that i/a = J = constant Thus, the answer is yes, the current density is uniform (b) We find i/(πr ) = (5 A)/(π m ) = A/m 7 The cross-sectional area of wire is given by A = πr, where r is its radius (half its thickness) The magnitude of the current density vector is so J = i A= i r / / π,

3 141 i 5 A r = = = π J π 4 ( 44 1 A/m ) m The diameter of the wire is therefore d = r = ( m) = m 8 (a) The magnitude of the current density vector is 1 ( ) 3 ( ) i i 41 1 A J = = = = A π d /4 π 5 1 m (b) The drift speed of the current-carrying electrons is A/m J vd = = ne A/m / m c hc Ch = m/s 9 We note that the radial width Δr = 1 μm is small enough (compared to r = 1 mm) that we can make the approximation Brπrdr BrπrΔr Thus, the enclosed current is πbr Δr = 181 μa Performing the integral gives the same answer 1 Assuming J is directed along the wire (with no radial flow) we integrate, starting with Eq 6-4, i= J da= ( kr )πrdr= kπ ( 656 ) 9 /1 where k = and SI units are understood Therefore, if = m, we 3 obtaini = 59 1 A 11 (a) The current resulting from this nonuniform current density is J i = J da = r rdr = J = 3 3 = 133 A (b) In this case, 3 4 cylinder a π π π(34 1 m) (55 1 A/m )

4 14 CHAPTE 6 r i J cylinder bda J 1 rdr J (34 1 m) (55 1 A/m ) π 3 π 3 π = = = = = 666 A (c) The result is different from that in part (a) because J b is higher near the center of the cylinder (where the area is smaller for the same radial interval) and lower outward, resulting in a lower average current density over the cross section and consequently a lower current than that in part (a) So, J a has its maximum value near the surface of the wire 1 (a) Since 1 cm 3 = 1 6 m 3, the magnitude of the current density vector is c hc h F J = nev = H G 87 I K J 16 1 C 47 1 m/s = A/m m (b) Although the total surface area of Earth is 4π E (that of a sphere), the area to be used in a computation of how many protons in an approximately unidirectional beam (the solar wind) will be captured by Earth is its projected area In other words, for the beam, the encounter is with a target of circular area π E The rate of charge transport implied by the influx of protons is ( ) ( ) i= AJ = π J = π m A/m = A 13 We use v d = J/ne = i/ane Thus, E ( ) ( 14 ) ( m 1 1 m / m )( C) L L LAne t = = = = vd i/ Ane i 3A = 81 1 s = 13min 14 Since the potential difference V and current i are related by V = i, where is the resistance of the electrician, the fatal voltage is V = (5 1 3 A)( Ω) = 1 V 15 The resistance of the coil is given by = ρl/a, where L is the length of the wire, ρ is the resistivity of copper, and A is the cross-sectional area of the wire Since each turn of wire has length πr, where r is the radius of the coil, then L = (5)πr = (5)(π)(1 m) = 1885 m If r w is the radius of the wire itself, then its cross-sectional area is A r w = π = π ( m) = m

5 143 8 According to Table 6-1, the resistivity of copper is ρ = Ω m Thus, 16 We use /L = ρ/a = 15 Ω/km c hb g 8 ρl Ω m m = = 6 A m = 4 Ω (a) For copper J = i/a = (6 A)(15 Ω/km)/( Ω m) = A/m (b) We denote the mass densities as ρ m For copper, (m/l) c = (ρ m A) c = (896 kg/m 3 ) ( Ω m)/(15 Ω/km) = 11 kg/m (c) For aluminum J = (6 A)(15 Ω/km)/( Ω m) = A/m (d) The mass density of aluminum is (m/l) a = (ρ m A) a = (7 kg/m 3 )( Ω m)/(15 Ω/km) = 495 kg/m 17 We find the conductivity of Nichrome (the reciprocal of its resistivity) as follows: 1 L L Li 1 m 4 A σ = = = = = ρ A V / i A VA V 1 1 m b g b gb g 6 b gc h 18 (a) i = V/ = 3 V/ Ω = A 6 = 1 / Ω m 1 (b) The cross-sectional area is A = π r = 4 π D Thus, the magnitude of the current density vector is 3 i 4i 4153 ( 1 A) 7 J = = = = A/m 3 A π D π ( 6 1 m) (c) The resistivity is 3 3 A (15 1 Ω ) π (6 1 m) ρ = = = Ω L 4(4 m) m (d) The material is platinum 19 The resistance of the wire is given by L A = ρ /, where ρ is the resistivity of the material, L is the length of the wire, and A is its cross-sectional area In this case,

6 144 CHAPTE 6 Thus, ( ) 3 7 A= πr = π 5 1 m = m ( Ω ) ( m ) A ρ = = = Ω L m 8 1 m The thickness (diameter) of the wire is denoted by D We use L/A (Eq 6-16) 1 and note that A= π D D The resistance of the second wire is given by 4 F A L D 1 1 L 1 = H G I A K J F H G I L K J F = H G I D K J F H G I L K J F = H G I K J = 1 1 bg 1 The resistance at operating temperature T is = V/i = 9 V/3 A = 967 Ω Thus, from = α (T T ), we find 1 T = T + = Ω = 3 1 C C 3 α 45 1 K 11Ω Since a change in Celsius is equivalent to a change on the Kelvin temperature scale, the value of α used in this calculation is not inconsistent with the other units involved Table 6-1 has been used Let r = mm be the radius of the kite string and t = 5 mm be the thickness of the water layer The cross-sectional area of the layer of water is A= π ( r+ t) r = [(5 1 m) ( 1 m) ] = 77 1 m π Using Eq 6-16, the resistance of the wet string is The current through the water layer is ( Ω )( ) 1 ρl 15 m 8 m = = = Ω A 77 1 m V 1 V i = = = Ω A 3 We use J = E/ρ, where E is the magnitude of the (uniform) electric field in the wire, J is the magnitude of the current density, and ρ is the resistivity of the material The electric field is given by E = V/L, where V is the potential difference along the wire and L is the length of the wire Thus J = V/Lρ and

7 145 V ρ = = LJ 115 V 4 = 8 1 Ω m 4 b1 mgc14 1 A m h 4 (a) Since the material is the same, the resistivity ρ is the same, which implies (by Eq 6-11) that the electric fields (in the various rods) are directly proportional to their current-densities Thus, J 1 : J : J 3 are in the ratio 5/4/15 (see Fig 6-4) Now the currents in the rods must be the same (they are in series ) so J 1 A 1 = J 3 A 3, J A = J 3 A 3 Since A = πr, this leads (in view of the aforementioned ratios) to 4r = 15r 3, 5r 1 = 15r 3 Thus, with r 3 = mm, the latter relation leads to r 1 = 155 mm (b) The 4r = 15r 3 relation leads to r = 1 mm 5 Since the mass density of the material does not change, the volume remains the same If L is the original length, L is the new length, A is the original cross-sectional area, and A is the new cross-sectional area, then L A = LA and A = L A /L = L A /3L = A /3 The new resistance is ρl ρ3l ρl = = = 9 = 9, A A / 3 A where is the original resistance Thus, = 9(6 Ω) = 54 Ω 6 The absolute values of the slopes (for the straight-line segments shown in the graph of Fig 6-5(b)) are equal to the respective electric field magnitudes Thus, applying Eq 6-5 and Eq 6-13 to the three sections of the resistive strip, we have J 1 = i A = σ 1 E 1 = σ 1 (5 1 3 V/m) J = i A = σ E = σ (4 1 3 V/m) J 3 = i A = σ 3 E 3 = σ 3 (1 1 3 V/m) We note that the current densities are the same since the values of i and A are the same (see the problem statement) in the three sections, so J 1 = J = J 3 (a) Thus we see that σ 1 = σ 3 = (3 1 7 (Ω m) 1 ) = (Ω m) 1

8 146 CHAPTE 6 (b) Similarly, σ = σ 3 /4 = (3 1 7 (Ω m) 1 )/4 = (Ω m) 1 7 The resistance of conductor A is given by L A = ρ pr, A where r A is the radius of the conductor If r o is the outside diameter of conductor B and r i is its inside diameter, then its cross-sectional area is π(r o r i ), and its resistance is The ratio is A B r = ri r o A B = π ρl ( ro ri ) b1 mmg b5 mmg = 5 mm b g = 3 8 The cross-sectional area is A = πr = π( m) The resistivity from Table 6-1 is ρ = Ω m Thus, with L = 3 m, Ohm s Law leads to V = i = iρl/a, or V = i ( Ω m)(3 m)/ π( m) which yields i = 97 A or roughly 3 ma 9 First we find the resistance of the copper wire to be With potential difference 8 ( Ω )( ) m m ρl = = = Ω A π ( 1 m) V = 3 nv, the current flowing through the wire is V A V i = = = 69 1 Ω Therefore, in 3 ms, the amount of charge drifting through a cross section is Δ Q= iδ t = ( A)(3 1 s) = C 3 We use L/A The diameter of a -gauge wire is 1/4 that of a 1-gauge wire Thus from = ρl/a we find the resistance of 5 ft of -gauge copper wire to be = (1 Ω)(5 ft/1 ft)(4) = 4 Ω 31 (a) The current in each strand is i = 75 A/15 = A

9 Since the current spreads uniformly over the hemisphere, the current density at any given radius r from the striking point is J = I /πr From Eq 6-1, the magnitude of the electric field at a radial distance r is ρwi E = ρwj =, π r where ρ w = 3 Ω m is the resistivity of water The potential difference between a point at radial distance D and a point at D+ Δr is ρ I ρ I ρ I r Δ V = Edr = dr = = π π π D +Δ r D +Δ r w w 1 1 w D D Δ r D+ Δ r D D( D+Δr), which implies that the current across the swimmer is Substituting the values given, we obtain Δ V ρ I w Δ i = = r π DD ( + Δ r) 4 (3 Ω m)(78 1 A) 7 m 3 (4 1 Ω ) (35 m)(35 m + 7 m) i = = π 5 1 A 37 From Eq 6-5, ρ τ 1 v eff The connection with v eff is indicated in part (b) of Sample Problem Mean free time and mean free distance, which contains useful insight regarding the problem we are working now According to Chapter, veff T Thus, we may conclude that ρ T 38 The slope of the graph is P = W Using this in the P = V / relation leads to V = 1 Vs 39 Eq 6-6 gives the rate of thermal energy production: P= iv = (1A)(1V) = 1 kw Dividing this into the 18 kj necessary to cook the three hotdogs leads to the result t = 15 s 4 The resistance is = P/i = (1 W)/(3 A) = 111 Ω 41 (a) Electrical energy is converted to heat at a rate given by P= V /, where V is the potential difference across the heater and is the resistance of the heater Thus,

10 15 CHAPTE 6 ( 1 V) 3 P = = 1 1 W= 1 kw 14 Ω (b) The cost is given by (1kW)(5h)(5cents/kW h) = US$5 4 (a) eferring to Fig 6-3, the electric field would point down (toward the bottom of the page) in the strip, which means the current density vector would point down, too (by Eq 6-11) This implies (since electrons are negatively charged) that the conduction electrons would be drifting upward in the strip (b) Equation 4-6 immediately gives 1 ev, or (using e = C) J for the work done by the field (which equals, in magnitude, the potential energy change of the electron) (c) Since the electrons don t (on average) gain kinetic energy as a result of this work done, it is generally dissipated as heat The answer is as in part (b): 1 ev or J 43 The relation P = V / implies P V Consequently, the power dissipated in the second case is 44 Since P = iv, the charge is 15 V 3 V ( 54 W) 135 W P = F H G I K J = q = it = Pt/V = (7 W) (5 h) (36 s/h)/9 V = C 45 (a) The power dissipated, the current in the heater, and the potential difference across the heater are related by P = iv Therefore, P 15 W i = = = 1 9 A V 115 V (b) Ohm s law states V = i, where is the resistance of the heater Thus, V 115 V = = = 1 6 Ω i 19 A (c) The thermal energy E generated by the heater in time t = 1 h = 36 s is 6 (15W)(36s) 45 1 J E= Pt= = 46 (a) Using Table 6-1 and Eq 6-1 (or Eq 6-11), we have

11 151 8 A E = ρ J = ( Ω m) = V/m 6 1 m (b) Using L = 4 m, the resistance is found from Eq 6-16: = ρl/a = 338 Ω The rate of thermal energy generation is found from Eq 6-7: P = i = ( A) (338 Ω) = 135 W Assuming a steady rate, the amount of thermal energy generated in 3 minutes is found to be (135 J/s)(3 6 s) = 43 1 J 47 (a) From P = V / = AV / ρl, we solve for the length: 6 AV ( 6 1 m )( 75 V) L = = = 585 m 7 ρp ( 5 1 Ω m)(5 W) F I HG K J = (b) Since L V the new length should be L = L V V F H G I = KJ 1 V ( 585 m) V m 48 The mass of the water over the length is m 3 5 = ρ AL= (1 kg/m )(15 1 m )(1 m) = 18 kg, and the energy required to vaporize the water is Q 4 = Lm= (56 kj / kg)(18 kg) = 46 1 J The thermal energy is supplied by joule heating of the resistor: Q= PΔ t = I Δ t Since the resistance over the length of water is ( Ω )( ) 5 ρ 15 m 1 m w = L = = Ω, A 15 1 m the average current required to vaporize water is

12 15 CHAPTE 6 I J 5 3 (1 1 )( 1 s) Q = = = 13 A Δ t Ω 49 (a) Assuming a 31-day month, the monthly cost is (1 W)(4 h/day)(31days/month) (6 cents/kw h) = 446 cents = US$446 (b) = V /P = (1 V) /1 W = 144 Ω (c) i = P/V = 1 W/1 V = 833 A 5 The slopes of the lines yield P 1 = 8 mw and P = 4 mw Their sum (by energy conservation) must be equal to that supplied by the battery: P batt = ( ) mw = 1 mw 51 (a) We use Eq 6-16 to compute the resistances: C L 1 m = ρc = ( 1 Ω m) = 55 Ω π C 6 rc π (5 m) The voltage follows from Ohm s law: V1 V = VC = ic = ( A)(55 Ω ) = 51V (b) Similarly, D L 1 m = ρd = (1 1 Ω m) = 59 Ω π D 6 rd π (5 m) and V V3 = VD = id = ( A)(59 Ω ) = 1V 1V (c) The power is calculated from Eq 6-7: P C = = 1W i C (d) Similarly, P D = = W i D 5 Assuming the current is along the wire (not radial) we find the current from Eq 6-4: i = J da = kr π rdr = 1 kπ4 = 35 A where k = A/m 4 and = 3 m The rate of thermal energy generation is found from Eq 6-6: P = iv = 1 W Assuming a steady rate, the thermal energy generated in 4 s is Q= PΔ t =(1 J/s)(36 s) = J 53 (a) From P = V / we find = V /P = (1 V) /5 W = 88 Ω (b) Since i = P/V, the rate of electron transport is

Electric Current. You must know the definition of current, and be able to use it in solving problems.

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