A copper wire has a diameter of d_Cu = 0.0400 cm,is 3.00 m long,has a resistivity of 1.72 * 10^-8 $\varOmega$m and has a carrier density of 8.50 * 10²⁸ electrons/m³.The copper wire is attached to an equal length of aluminum wire with a a carrier density of 6.02 * 10²⁸ electrons/m³.A current of 0.400 A flows through the copper and aluminum wires.If the ratio of the current densities the two wires,J_Cu/J_Al is 0.50,what is the diameter of the aluminum wire? A)0.0800 cm B)0.0283 cm C)0.0200 cm D)0.0565 cm E)0.0400 cm

Question

Quizplus · Accepted Answer

We can start by using the equation for current density: J=nevd, where J is current density, n is carrier density, e is charge of an electron, v is drift velocity, and d is the diameter of the wire. We can rearrange this equation to solve for the diameter: d=J/(nev)

The current density for the copper wire can be found using Ohm's law: J=V/R, where V is voltage and R is resistance. The voltage across the copper wire can be found using the length and resistivity: V = IRhoL/A, where A is cross-sectional area. Solving for the resistance of the copper wire using its resistivity and length: R = RhoL/A = 8.95 x 10^-3 ohms. So, the current density for the copper wire is Jc = V/R = (0.4 A)(3 m)/(8.95 x 10^-3 ohms)(pi/4)(0.0400 cm)^2 = 7.99 x 10^6 A/m^2.

The ratio of the current densities is given as Jc/JA = 0.50. We can use this to solve for the current density in the aluminum wire: JA = Jc/0.50 = 1.60 x 10^7 A/m^2. Using the carrier density and current density for the aluminum wire, we can solve for its diameter: dA = JA/(neA) = (1.60 x 10^7 A/m^2)/(6.02 x 10^28 m^-3)(1.60 x 10^-19 C/electron) = 2.83 x 10^-4 m = 0.0283 cm.

Therefore, the answer is B.

A Copper Wire Has a Diameter of DCu = 0 Ω\varOmegaΩ

A Copper Wire Has a Diameter of D_Cu = 0 $\varOmega$