Homework 12 Solutions

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

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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...
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## This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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