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5. Consider two (1 and 2) wires both having constant circular cross sectional areas: wire 1 has diameter = d, ; wire 2 has diameter = d. Potential differences are applied across each wire so that they both carry current. The current in wire 1 is I, and in wire 2 it is I, ; the the magnitude of the current density in wire 1 is J, and in wire 2 it is J,. If I,= I, and J, = (2) J, then wire diameters must be such that d, = (?) d. a. 2/2 b. V212 c. 2 d. V2 e. V214

5. Consider two (1 and 2) wires both having constant circular cross sectional areas: wire 1 has diameter = d, ; wire 2 has diameter = d. Potential differences are applied across each wire so that they both carry current. The current in wire 1 is I, and in wire 2 it is I, ; the the magnitude of the current density in wire 1 is J, and in wire 2 it is J,. If I,= I, and J, = (2) J, then wire diameters must be such that d, = (?) d. a. 2/2 b. V212 c. 2 d. V2 e. V214

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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5. Consider two (1 and 2) wires both having constant circular cross sectional areas: wire 1 has diameter = d, ; wire 2 has diameter = d. Potential differences are applied across each wire so that they both carry current. The current in wire 1 is I, and in wire 2 it is I, ; the the magnitude of the current density in wire 1 is J, and in wire 2 it is J,. If I,= I, and J, = (2) J, then wire diameters must be such that d, = (?) d. a. 2/2 b. V212 c. 2 d. V2 e. V214
Transcribed Image Text:5. Consider two (1 and 2) wires both having constant circular cross sectional areas: wire 1 has diameter = d, ; wire 2 has diameter = d. Potential differences are applied across each wire so that they both carry current. The current in wire 1 is I, and in wire 2 it is I, ; the the magnitude of the current density in wire 1 is J, and in wire 2 it is J,. If I,= I, and J, = (2) J, then wire diameters must be such that d, = (?) d. a. 2/2 b. V212 c. 2 d. V2 e. V214
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