Explanation Given data: Mass of the copper ( m ) is 1.0 g. Resistance of the cooper wire ( R ) is 1.0 Ω . Density of the copper wire ( D ) is 8900 kg m 3 . Formula used: Write the expression for resistance of a material in terms of resistivity and dimensions. R = ρ L A (1) Here, ρ is the resistivity of material, L is the length of the material, and A is the cross-section area of the material. Write the expression for cross-section area of the material as follows. A = π 4 d 2 (2) Here, d is the diameter of the material. Write the expression for density of a material (copper wire). D = m v (3) Here, m is the mass of the copper, and v is the volume of the material. Write the expression for volume of the material (copper). v = A L (4) Explanation: From equation (4), substitute ( A L ) for v in equation (3), D = m A L From equation (2), substitute ( π 4 d 2 ) for A . D = m ( π 4 d 2 ) L = 4 m π d 2 L Rewrite the expression to find the diameter of the copper as follows. d 2 = 4 m π D L d = 4 m π D L Substitute 1.0 g for m , and 8900 kg m 3 for D , d = 4 ( 1.0 g ) π ( 8900 kg m 3 ) L = 4 ( 1.0 × 10 − 3 kg ) π ( 8900 kg m 3 ) L d = 0.14306 × 10 − 6 m 3 L (5) From equation (2), substitute ( π 4 d 2 ) for A in equation (1), R = ρ L ( π 4 d 2 ) = 4 ρ L π d 2 Rewrite the expression to find the diameter of the copper wire as follows. d 2 = 4 ρ L π R d = 4 ρ L π R (6) Refer to the Table 22.1 in the textbook for resistivity of various materials. From the table, the resistivity of the copper wire is ( 1.7 × 10 − 8 ) Ω ⋅ m . Substitute ( 1

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Resistivity

Resistivity is a characteristic property of the conductor’s material that measures the extent to which the material opposes the flow of current. It is denoted by the Greek letter “rho”. Electrical resistivity, also known as specific resistance, depends o…

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Chapter 22, Problem 63GP

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Chapter 22 Solutions

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