Chapter 29 - Resolução Sears - Vol 3

29-10 Chapter 29 EXECUTE: |ℰs| = |NΔΦpΔt| = |N[AB2-B1]Δt| = |NAΔBzπlΔt| ℰs = μo(I2(0.0780x10-2 m2 )(9000 m-1 )(0.350 A)) = 9.50 x 10-2 V. 0.0400 s EVALUATE: An emf is induced in the second winding even though the magnetic field of the solenoid is zero at the location of the second winding. The changing magnetic field induces an electric field outside the solenoid and that induced electric field produces the emf. 29.34 IDENTIFY: Apply Eq.(29.14). SET UP: δεdΕ.dt = 3.50x10-11 δΦm dt EXECUTE: id = δεdΕ.dt = (3.5x1011 F/m)(24.0x101 V m/s)2/s2/2y2 ; = 21x104 A gives t = 5.0 s. EVALUATE: id depends on the rate at which Φ is changing. 29.35 IDENTIFY: Apply Eq.(29.14), where ε=κε(encoded text) . SET UP: c/Φ_= _Δ/48.76x10-3 V m/s )/2 = 8.854x10-12 EMm. = 1.280x10-14 δCi 6/ c/FB (48.76 x10/ Vm/s )(26.1x10-3 y = 2.07x10-11 E. The dielectric constant is κε = 23.4. EVALUATE: The larger the dielectric constant, the larger is the displacement current for a given ΦdEh. 29.36 IDENTIFY and SET UP: Eqs.(29.13) and (29.14) show that id = ic, and also relate ic to the rate of change of the electric field flux between the plates. Use this to calculate ΔE/; and apply the generalized form of Ampere’s law (Eq.29.15) to calculate B. (a) EXECUTE: ic=£10, B=Joicμ = Jo ; ic = Jo ; A = 0.280 A = 55.7 A/m2 0.7000x000 m) (b) is incd ōE dE A ic= Jo;d = Jo;d = m = 6.29x107 V/m s; = 8.854x1012; c/Nm = j, σrs (c) SET UP: Apply Ampere’s law ∆B;dB = 45.7 A/m)(0.0200 m). An end view of the solenoid is given in Figure 29.36. Figure 29.36 EXECUTE: Thus ∮B; da = £Bda B ; fṫB = B(27r). ic = (no conduction current flows through the air space between the plates) The displacement current enclosed by the path is jσr2. Thus B(27r) = Jo(list9)+Jσr2r7 ) = Jojbσrr7* x107 T 2/m A/S}{7.35 x1072 1 T/m A){(55.7 ; G0.0200 m) = 7.00 x107mic) T B7 +J/Joic ; Now r is 2 per the value in 45.7 t/s 2iC(t encoding), so B is also; EVALUATE: The definition of displacement current allows the current to be continuous at the capacitor. The magnetic field between the plates is zero on the axis (r=0) and increases as r increases. 29.37 IDENTIFY: q = CV . For a parallel-plate capacitor, C = , where ε = KeKc ic = d5/dE dt. it is E dt /d5 SET UP: E = q/aeDST so E/mdt in jet AC). EXECUTE: (a) q =CV E d Æ = 6.70 (3,00xltir m2 )(0 20 V) = 5.99x10. CH ̅C6/ 7M m 2.50x10-7