ssm_ch32 - Chapter 32 3 H H ~ ~ ~ ~ (a) Use Gauss law for...

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Chapter 32 3 (a) Use Gauss’ law for magnetism: H ~ B ¢ d ~ A =0 . W r i te H ~ B ¢ d ~ A = © 1 + © 2 + © C ,whe re © 1 is the magnetic flux through the first end mentioned, © 2 is the magnetic flux through the second end mentioned, and © C is the magnetic flux through the curved surface. Over the first end, the magnetic field is inward, so the flux is © 1 = ¡ 25 : 0 ¹ Wb. Over the second end, the magnetic field is uniform, normal to the surface, and outward, so the flux is © 2 = AB = ¼r 2 B ,whe re A is the area of the end and r is the radius of the cylinder. Its value is © 2 = ¼ (0 : 120 m) 2 (1 : 60 £ 10 ¡ 3 T) = +7 : 24 £ 10 ¡ 5 Wb = +72 : 4 ¹ Wb : Since the three fluxes must sum to zero, © C = ¡ © 1 ¡ © 2 =25 : 0 ¹ Wb ¡ 72 : 4 ¹ Wb = ¡ 47 : 4 ¹ Wb : (b) The minus sign indicates that the flux is inward through the curved surface. 5 Consider a circle of radius r (= 6 : 0 mm), between the plates and with its center on the axis of the capacitor. The current through this circle is zero, so the Ampere-Maxwell law becomes I ~ B ¢ d~ s = ¹ 0 ² 0 d © E dt ; where ~ B is the magnetic field at points on the circle and © E is the electric flux through the circle. The magnetic field is tangent to the circle at all points on it, so H ~ B ¢ d~s =2 ¼rB .T h e electric flux through the circle is © E = ¼R 2 E , where R (= 3 : 0 mm) is the radius of a capacitor plate. When these substitutions are made, the Ampere-Maxwell law becomes 2 ¼rB = ¹ 0 ² 0 ¼R 2 dE dt : Thus dE dt = 2 rB ¹ 0 ² 0 R 2 = 2(6 : 0 £ 10 ¡ 3 m)(2 : 0 £ 10 ¡ 7 T) (4 ¼ £ 10 ¡ 7 H = m)(8 : 85 £ 10 ¡ 12 Fm)(3 : 0 £ 10 ¡ 3 m) 2 =2 : 4 £ 10 13 V = m ¢ s : 13 The displacement current is given by i d =
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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 ,whe re V is the potential difference across the plates and d is the plate separation. Thus i d = ² 0 A d dV dt : Now ² 0 A=d is the capacitance C of a parallel-plate capacitor without a dielectric, so i d = C dV dt : 21 (a) For a parallel-plate capacitor, the charge q on the positive plate is given by q =( ² 0 A=d ) V , where A is the plate area, d is the plate separation, and V is the potential difference between the plates. In terms of the electric field E between the plates, V = Ed ,so q = ² 0 AE = ² 0 © E ,whe re ©
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ssm_ch32 - Chapter 32 3 H H ~ ~ ~ ~ (a) Use Gauss law for...

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