Homework 12 Solutions

Thus a dv id 0 d dt now 0ad is the capacitance c of

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: refore, we need to change the voltage difference across the capacitor at the rate of 7.5 ×105 V/s . 14. Consider an area A, normal to a uniform electric field E . The displacement current density is uniform and normal to the area. Its magnitude is given by Jd = id/A. For this situation , id = ε 0 A(dE / dt ) , so dE dE 1 Jd = ε 0 A = ε0 . A dt dt 15. The displacement current is given by id = ε 0 A(dE / dt ), where A is the area of a plate and E is the magnitude of the electric field between the plates. The field between the plates is uniform, so E = V/d, where V is the potential difference across the plates and d is the plate separation. Thus, ε A dV id = 0 . d dt Now, ε0A/d is the capacitance C of a parallel-plate capacitor (not filled with a dielectric), so dV id = C . dt 16. We use Eq. 32-14: id = ε 0 A(dE / dt ). Note that, in this situation, A is the area over which a changing electric field is present. In this case r > R, so A = πR2. Thus, i i dE 2.0 A V . = d = d 2= = 7.2 ×1012 2 m ⋅s dt ε 0 A ε 0π R π ( 8.85 × 10−12 C2 /N ⋅ m 2 ) ( 0.10 m ) 17. (a) Using Eq. 27-10, we find E = ρJ = ρi A . c162 × 10 = −8 hb Ω ⋅ m 100 A −6 5.00 × 10 m 2 g = 0.324 V m . (b) The displacement current is id = ε 0 dΦE dE d ⎛ ρi ⎞ di = ε0 A = ε 0 A ⎜ ⎟ = ε 0 ρ = 8.85 × 10−12 F/m 1.62 × 10−8 Ω ( 2000 A s ) dt dt dt ⎝ A ⎠ dt ( )( ) = 2.87 × 10−16 A. (c) The ratio of fields is B ( due to id ) μ0id 2π r id 2.87 ×10−16 A = == = 2.87 ×10−18. B ( due to i ) μ0i 2π r i 100 A 18. From Eq. 28-11, we have i = (ε / R ) e− the capacitor. Equation 32-16 gives t/ τ since we are ignoring the self-inductance of 1249 B= μ0id r . 2π R 2 Furthermore, Eq. 25-9 yields the capacitance C= ε 0π (0.05 m) 2 0.003 m = 2.318 × 10−11 F , so that the capacitive time constant is τ = (20.0 × 106 Ω)(2.318 × 10−11 F) = 4.636 × 10−4 s. At t = 250 × 10−6 s, the current is i= 12.0 V t/ e− τ = 3.50 × 10−7 A . 20.0 x 106 Ω Since i = id (see Eq. 32-15) and r = 0.0300 m, then (with plate radius R = 0.0500 m) we find μ i r (4π× 10−7 T ⋅ m/A)(3.50 × 10−7 A)(0.030 m) B= 0d2 = = 8.40 × 10−13 T . 2 2π R 2π (0.050 m) 2 19. (a) Equation 32-16 (with Eq. 26-5) gives, with A = πR , B= μ0id r μ0 J d Ar μ0 J d (π R 2 )r 1 = = = μ0 J d r 2π R 2 2π R 2 2π R 2 2 1 (4π× 10−7 T ⋅ m/A)(6.00 A/m 2 )(0.0200 m) = 75.4 nT . 2 μi μ J π R2 = 67.9 nT . (b) Similarly, Eq. 32-17 gives B = 0 d = 0 d 2π r 2π r = 20. (a) Equation 32-16 gives B = (b) Equation 32-17 gives B = μ0id r = 2.22 μ T . 2π R 2 μ0id = 2.00 μ T . 2π r 21. (a) Equation 32-11 applies (though the last term is zero) but we must be careful with id,enc . It is the enclosed portion of the displacement current, and if we related this to the displacement current density Jd , then r r ⎛1 r3 ⎞ id enc = ∫ J d 2π r dr = (4.00 A/m 2 )(2π ) ∫ (1 − r / R ) r dr = 8π ⎜ r 2 − ⎟ 0 0 3R ⎠ ⎝2...
View Full Document

Ask a homework question - tutors are online