chapt26 - Chapter 26 Current and Resistance Problem 1...

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Unformatted text preview: Chapter 26: Current and Resistance Problem 1 During the 4.0 minutes a 5.0-A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section across the wire’s width? Answer (a) 1200 C. (b) 7.5 × 10 21 . Solution (a) Since the current is constant q = i*t = (5.0 A)*(4.0 min) = 1200 C. (b) Since the charge on one electron is e = 1.6 × 10-19 C, the number of electrons is n = q e = 1200 C 1.6 × 10-19 C = 7.5 × 10 2 1 . Problem 2 An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a current of 1.000 000 0 A out of it. How long would it take for the sphere to increase in potential by 1000 V? Answer 5.6 ms Solution The potential at the surface of a conducting sphere is V = 1 4 πε Q R If V is 1000 volt and R is 10 cm then Q is 11.1 nC. If the net rate of change in the charge on the sphere is 1.000 002 0 A - 1.000 000 0 A = 0.000 002 0 A then the time that it will take to accumulate 11.1 nC is t = Q I = 11.1 nC 0.0000020 A = 5.6 ms . Problem 7 A beam contains 2.0 × 10 8 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 1.0 × 10 5 m/s. (a) What are the magnitude and (b) direction of the current density J ? (c) What additional quantity do you need to calculate the total current i in this ion beam? 26-2 Chapter 26: Current and Resistance Answer (a) 6.4 A/m 2 , (b) north, (c) cross-sectional area of the beam. Solution (a) The magnitude of the uniform current density is J = I A = q t A . Multiplying numerator and denominator by L, the distance the charges travel in time t yields J = q L t ( ) LA = qv d V where v d is the drift speed of the ions and V is the volume of the charge that passes a plane in time t. If ρ is the charge per unit volume then J = ρ v d In this case ρ = 2(2.0 × 10 8 cm-3 )*(1.6 × 10-19 C) = 6.4 × 10-5 C/m 3 since the ions are doubly charged. Therefore J = ( 6.4 × 10-5 C/m 3 )*(1.0 × 10 5 m/s) = 6.4 A/m 2 . (b) The direction of the current density is the direction in which positive charge carriers move - north. (c) We need the cross-sectional area of the beam in order to determine the beam current. Problem 10 Near Earth, the density of protons in the solar wind(a stream of particles from the Sun) is 8.7 cm-3 and their speed is 470 km/s. (a) Find the current density of these protons. (b) If the earth's magnetic field did not deflect them, the protons would strike Earth. What total current would Earth receive? Answer (a) 0.654 μ A/m 2 . (b) 83.4 MA. Solution (a) From Equation 26-7 J = (ne)v d = (8.7 cm-3 )*(1.6 × 10-19 C)*(470 km/s) = 0.654 μ A/m 2 . (b) For a uniform charge density I = JA = J π R e 2 , where R e is the radius of the earth. Therefore I = (0.654 μ A/m 2 ) π (6.37 × 10 6 m) 2 = 83.4 MA. Chapter 26: Current and Resistance 26-3 Problem 13 What is the current in a wire of radius R = 3.40 mm if the magnitude of the current density is given by (a) J = J r/R and (b)...
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chapt26 - Chapter 26 Current and Resistance Problem 1...

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