Chapter 30 - Resolução Sears - Vol 3

30-4 Chapter 30 30.16 IDENTIFY: Energy = PE . U = 1/2LI^2 . SET UP: P = 200 W = 200 Js EXECUTE: (a) Energy = (200 W)(24 h)(3600 s/h) = 1.73x10^7 J (b) L = 2U/1^2 = 2(1.73x10^7/ (0.80 A)2) = 5.41x10^7 H EVALUATE: A large value of L and a large current would be required, just for one light bulb. Also, the resistance of the inductor would have to be very small, to avoid a large P = I^2R rate of electrical energy loss. 30.17. IDENTIFY and SET UP: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability μ is used in place of μ0. EXECUTE: Using L = μN^2A/2r and B = μIN/2r gives μ = B^2/2ju EVALUATE: For a given value of B, the energy density is less when μ is larger than μ0. 30.18. IDENTIFY and SET UP: The energy density (energy per unit volume) in a magnetic field (in vacuum) is given by u_B = UB/V = B^2/2μ0 Eq.30.10). EXECUTE: (a) v = 2μU/B^2 = 2(4.7x10^-7 T. m/A)(3.6x10^6 J) /(0.600 T)2 = 25.1 m3. (b) v = 2U/B^2 = 2(4x10^-7 T. m/A)(3.60x10^6 J)/(0.400 m3) = 11.9 T EVALUATE: Large-scale energy storage in a magnetic field is not practical. The volume in part (a) is quite large and the field in part (b) would be very difficult to achieve. 30.19. IDENTIFY: Apply Kirchhoff's loop rule to the circuit. i(t) is given by Eq.(30.14). SET UP: The circuit is sketched in Figure 30.19. Figure 30.19 E - iR - Ldi/dt = 0; i = E/R(1 - e^(-Rt/L)) (a) Initially (t = 0), i = 0, so E – L di/dt = 0 di/dt = E/L = 6.00 V /2.50 H = 2.40 A/s (b) E – iR – L di/dt = 0 (Use this equation rather than Eq.(30.15) since t rather than i is given.) Thus di/dt = E – iR/L = (6.00 V – (0.500 R)(8.00 Ω))/2.50 H = 0.890 A/s (c) i = E/R (1 – e^(-Rt/L)) = (6.00 V)/(8.00 Ω) [1 – e^(-1 (1/5x0/1x12020/5))] = 0.750 A(1 – e^(-480)) = 0.413 A (d) Final steady state means t ->infinity and di/dt -> 0, so E – iR = 0. i = E/R = 6.00 V/8.00 Ω = 0.750 A EVALUATE: Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of E/R. The slope of the current in the figure, which is di/dt, decreases with t. 30-5 Inductance 30.20. IDENTIFY: The current decays exponentially. SET UP: After opening the switch, the current is i = I0e^-t/τ, and the time constant is τ = L/R. EXECUTE: (a) The initial current is Io = (6.30 V)(15.0 Ω) = 0.420 A. Now solve for L and put in the numbers. L = -R (–(2.00 ms)(15.0 Ω) In(1/0) |0.420 A)| (0.420 A) = 43.3 mH; L = (0.420 A)/(0.420 A) = 2.89 ms (b) τ = L/R = (43.3 mH)/(15.0 Ω) = 2.89 ms (c) Solve i = Io e^-t/τ, for t, giving t = τ In(1/I0/0) = (2.89 ms) In(0.0100) = 13.3 ms. EVALUATE: In less than 5 time constants, the current is only 1% of its initial value. 30.21. IDENTIFY: i = E/R(1 – e^(-t/R)), with L/R. The energy stored in the inductor is U = 1/2LI^2. SET UP: The maximum current occurs after a long time and is equal to E/R. EXECUTE: (a) ta = —ER so i = i/2 when (1 – e^(-t/τ)) = e^(-t/τ) : -te^(-t/τ) = In(1/1/2) t = Lln2/R = (ln2)(1.55x10^3 H) = 17.3 J; U = 1/2UD when i = I0/√2 . (1 – e^(-t/τ))= √1/√2 , so e^–1 = 1/√2 , —(ln0.2) .99 = √3.07 Js. (b) τ = L/R = 5.2x103/s s = 20/5 Js. The time in part (a) is 0.692 and the time in part (b) is 1.23r. 30.22. IDENTIFY: U = 1/2LI^2 . After S4 has been closed a long time, i, has reached its final value of I0 = E/R. SET UP: U = 1/2LI^2. After S2 open, i(t) is given by Eq(30.14), With S3 open, and S4 closed, i(t) is given by Eq.(30.18). EXECUTE: (a) U = 1/2LI^2 and i = √U: 2I0 = √(20.260 I^oH = √)3.20 A = E-IR = (2.13 A)(120/360) = 256 V. (b) i = I0e^-t/τ and U = 1/2LU^2+:1/2 LI^2e^-2Rt/+ L (1 - e^-2Rt/L)2r:term/λ: e^(-π/π1r) i = I0e^-2πR = i Ar/s] (c) L/R = 2/m(8e^-2t/L); 1(2.0 )(P/igμ)(5.0)(t) = 3.32)(10^-5 /t (o.m))/ i^2 Rl; LU0.(5.(P) EVALUATE: Ξ/π) = t seasons.[L])(0) EVALUATE: L has units of H and R has units of Ω. τ = L/R = s^2 EVALUATE: R/I.1/2F; :; = LCirR-0.3/sρ=1= units of time. EVALUATE: R/L/I is dimensionless. 30.24. IDENTIFY: Apply the loop rule. SET UP: In applying the loop rule, go around the circuit in the direction of the current. The voltage across the inductor is – Liei/dt = F. EXECUTE: –L(di/dt)–iR=0, di/dt i = R:L/R: i~. — i~ = E/R.: it cl: y = _04x EVALUATE: di/dt is negative, so there is a potential rise across the inductor; point c is at a higher potential than point a. There is a potential drop across the resistor. 30.25. IDENTIFY: Apply the concepts of current decay in anRL/1 circuit. Apply the loop rule to the circuit. i(t) is given by Eq.(30.18). The voltage across the resist depends on i and the voltage across the indector depends on di/dt. SET UP: The circuit with SC closed and S3 open is sketched in figure 30.25. Figure 30.25a Constant current established means, di/dt = 0. EXECUTE: : i = E//R: 6.00 V / 240 Ω = 0.250 A 30-6 Chapter 30 30.26 IDENTIFY Apply Eq.(30.14). SET UP: va = iR, vb = —Ldi/dt. The current is increasing, so di/dt is positive. EXECUTE: (a) At t = 0;, vo = 0 and vy = 60 Vo. (b) As t → ∞, i → E/R and di/dt → 0. V₀ = 60 V and va = -0. (c) When i = 1.050 A, va = iR = 36.0 V and va = 6.00 – 36.0 = 24.0 V. EVALUATE: At all times, E = va + vy as required by the loop rule. 30.27. IDENTIFY: i(t) is given by Eq.(30.14). SET UP: The power input from the battery is Ei. The rate of dissipation of energy in the resistance is i²R. The voltage across the inductor has magnitude |L*(di/dt)|, so the rate at which energy is being stored in the inductor is |Ldi/dt|. EXECUTE: (a) P₁ = Ei = E(i₀)(1 — e^(-R/L)t) = E(i₀)(1 — e^(-R/L)t) = (6.00 V)² / 8.00 Ω² (1 — e^(-0.030/0.030)/R) P = (4.50 W)(1 - e^{- «/>< (b) P₂ = Ei = E(iρ)/R (1 - e^(-R/L)t) = i²R(16.00.wp)(1 - e^(-0.030//.020)))k P₂ = 4.50 W) (c) P₃ = i²R(16.00.wp/o)/( 1 ) = R ( 6 /32 )( 120/800)/ EVALUATE: (102/)/ (d) Note that if we expand the square in part (b), then parts (b) and(c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor. Conservation of energy requires that this be so.