Question

Chapter 26, Problem 018 A wire 5.66 m long and 7.36 mm in diameter has a resistance of 13.3 m2. A potential difference of 26.2 V is applied between the ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made. (a) Number Units Units (b) Number (c) Number Units

Answer 1

Chapter 26, Prooblem 018 Length of wire (2) = 5.66m; m. Diameter of wire = 7.36 mm = 7.36x 10 radins iso - 7.36 x 103 m = 3.68 x 10 m, cross sectional area f A AP2 - Tisz. A = 3.14x (3.68x103) me 4. 25x10-5 m? now, resistance of wire (R) = 13.3 mit = 13.3 x 103 di potential difference (v) = 2612 Volt; 2 to calculate current (I), we know, V=IR, 1 Putting 26.2 as 0, I= R - 13.3x10 VR 1.9699x103 Amp. i I = 1.9699 x 103 Amp in the current in the wire. 6 we know, current density 3 = 2 / 2 1.9699 x103 J: 4.25 X10-5 a = 46.35 x 106 A/m2 the values of
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© We know, R = 3 2 1 / 2 where 'f' is the resistivity of wire material: R = 8h or, 8 = 22 (13,3 x103) * (4.25X105) 5.66 22-m -9.987 X10-82m 2.m. i resistivity of mire material is 87=9.987 X103 (a) Number 11.9699x103) Unit Camperas (6) Number 146, 35x108 / unit [ Amp / m² (e) Number 19.987810-8 unit com.
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