Homework 12 Solutions

10 a here the enclosed electric flux is found by

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Unformatted text preview: = ε 0 μ0π r 2 ( 0.00450 V/m ⋅ s ) . For r = 0.0200 m, this gives 1 B = ε 0 μ0 r (0.00450 V/m ⋅ s) 2 1 = (8.85 × 10−12 C2 /N ⋅ m 2 )(4π ×10−7 T ⋅ m/A)(0.0200 m)(0.00450 V/m ⋅ s) 2 = 5.01×10−22 T . (b) With r > R, the expression above must replaced by B (2π r ) = ε 0 μ0π R 2 ( 0.00450 V/m ⋅ s ) . Substituting r = 0.050 m and R = 0.030 m, we obtain B = 4.51 × 10−22 T. 10. (a) Here, the enclosed electric flux is found by integrating r r⎛ ⎛1 r⎞ r3 ⎞ Φ E = ∫ E 2π rdr = t (0.500 V/m ⋅ s)(2π ) ∫ ⎜1 − ⎟ rdr = tπ ⎜ r 2 − ⎟ 0 0 3R ⎠ ⎝ R⎠ ⎝2 with SI units understood. Then (after taking the derivative with respect to time) Eq. 32-3 leads to ⎛1 r3 ⎞ B (2π r ) = ε 0 μ0π ⎜ r 2 − ⎟. 3R ⎠ ⎝2 1247 ⎛ μ ε R 2 dE ⎞ ⎛ μ ε R 2 dV ⎞ ⎛ μ ε R2 ⎞ Bmax = ⎜ 0 0 =⎜ 0 0 = ⎜ 0 0 Vmaxω cos (ωt ) ⎟ ⎟ ⎟ ⎝ 2r dt ⎠max ⎝ 2rd dt ⎠ max ⎝ 2rd ⎠ max = μ0ε 0 R 2Vmaxω 2rd ( for r ≥ R ) (note the B ∝ r–1 dependence — see also Eqs. 32-16 and 32-17). The plot (with SI units understood) is shown below. 12. From Sample Problem — “Magnetic field induced by changing electric field,” we know that B ∝ r for r ≤ R and B ∝ r–1 for r ≥ R. So the maximum value of B occurs at r = R, and there are two possible values of r at which the magnetic field is 75% of Bmax. We denote these two values as r1 and r2, where r1 < R and r2 > R. (a) Inside the capacitor, 0.75 Bmax/Bmax = r1/R, or r1 = 0.75 R = 0.75 (40 mm) =30 mm. (b) Outside the capacitor, 0.75 Bmax/Bmax = (r2/R)–1, or r2 = R/0.75 = 4R/3 = (4/3)(40 mm) = 53 mm. (c) From Eqs. 32-15 and 32-17, μi μ i ( 4π ×10 T ⋅ m A ) ( 6.0 A ) Bmax = 0 d = 0 = = 3.0 ×10−5 T. 2π R 2π R 2π ( 0.040 m ) −7 13. Let the area plate be A and the plate separation be d. We use Eq. 32-10: id = ε 0 FG IJ HK FG IJ HK dV dΦ E d ε A dV AE = ε 0 A = ε0 =0 , dt d d dt dt dt bg or dV id d id 15 A . = == = 7.5 × 105 V s . dt ε 0 A C 2.0 × 10−6 F CHAPTER 32 1248 The...
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