A copper wire has a diameter \(d_{\mathrm{Cu}}=0.0500 \mathrm{~cm}\) is \(3.00 \mathrm{~m}\) long, and has a density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). As shown in the figure, the copper wire is attached to an equal length of aluminum wire with a diameter \(d_{\mathrm{A} \mathrm{I}}=0.0100 \mathrm{~cm}\) and density of charge carriers of \(6.02 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). A current of 0.400 A flows through the copper wire. a) What is the ratio of the current densities in the two wires, \(J_{\mathrm{Cu}} / J_{\mathrm{Al}} ?\) b) What is the ratio of the drift velocities in the two wires, \(v_{\mathrm{d}-\mathrm{Cu}} / v_{\mathrm{d}-\mathrm{Al}} ?\)

Step by step solution

01

Define the given parameters for the copper and aluminum wires.

We are given the following: For Copper wire: \(d_{Cu} = 0.0500\ cm\) Length = \(3.00\ m\) Density of charge carriers = \(8.50 \times 10^{28}\ electrons/m^3\) For Aluminum wire: \(d_{Al} = 0.0100\ cm\) Length = \(3.00\ m\) Density of charge carriers = \(6.02 \times 10^{28}\ electrons/m^3\) Current flowing through copper wire = \(I_{Cu} = 0.400\ A\)

02

Calculate cross-sectional areas for the copper and aluminum wires.

To find the cross-sectional area of each wire, we use the following formula: \( A_{wire} = \pi r^2 \) For Copper wire: \(A_{Cu} = \pi (\frac{d_{Cu}}{2})^2 = \pi (\frac{0.0500\ cm}{2})^2 = \pi (\frac{0.000500\ m}{2})^2 = \pi (0.000250)^2 m^2\) For Aluminum wire: \(A_{Al} = \pi (\frac{d_{Al}}{2})^2 = \pi (\frac{0.0100\ cm}{2})^2 = \pi (\frac{0.000100\ m}{2})^2 = \pi (0.000050)^2 m^2\)

03

Calculate current densities for the copper and aluminum wires.

The current density (J) is given by: \( J = \frac{I}{A} \) For Copper wire: \(J_{Cu} = \frac{I_{Cu}}{A_{Cu}} = \frac{0.400\ A}{\pi (0.000250)^2 m^2}\) For Aluminum wire: The current flowing through the Aluminum is the same as the current in the copper wire, hence \(I_{Al} = 0.400\ A\) \(J_{Al} = \frac{I_{Al}}{A_{Al}} = \frac{0.400\ A}{\pi (0.000050)^2 m^2}\)

04

Calculate the ratio of the current densities in the two wires.

Now, we can find the ratio of the current densities: \(J_{Cu}/J_{Al} = \frac{J_{Cu}}{J_{Al}} = \frac{\frac{0.400\ A}{\pi (0.000250)^2 m^2}}{\frac{0.400\ A}{\pi (0.000050)^2 m^2}} = \frac{(0.000050)^2}{(0.000250)^2} = \frac{0.00000250}{0.0000625} = 0.04\)

05

Calculate the drift velocities for the copper and aluminum wires.

The drift velocity (v) is given by: \( v = \frac{I}{nqA} \) For Copper wire: \(v_{d-Cu} = \frac{I_{Cu}}{n_{Cu}qA_{Cu}} = \frac{0.400\ A}{(8.50 \times 10^{28}\ electrons/m^3)(1.6 \times 10^{-19}\ C)(\pi (0.000250)^2 m^2)}\) For Aluminum wire: \(v_{d-Al} = \frac{I_{Al}}{n_{Al}qA_{Al}} = \frac{0.400\ A}{(6.02 \times 10^{28}\ electrons/m^3)(1.6 \times 10^{-19}\ C)(\pi (0.000050)^2 m^2)}\)

06

Calculate the ratio of the drift velocities in the two wires.

Now, we can find the ratio of the drift velocities: \(v_{d-Cu}/v_{d-Al} = \frac{v_{d-Cu}}{v_{d-Al}} = \frac{\frac{0.400\ A}{(8.50 \times 10^{28}\ electrons/m^3)(1.6 \times 10^{-19}\ C)(\pi (0.000250)^2 m^2)}}{\frac{0.400\ A}{(6.02 \times 10^{28}\ electrons/m^3)(1.6 \times 10^{-19}\ C)(\pi (0.000050)^2 m^2)}}\) \(v_{d-Cu}/v_{d-Al} = \frac{(0.000050)^2(6.02 \times 10^{28}\ electrons/m^3)}{(0.000250)^2(8.50 \times 10^{28}\ electrons/m^3)} = 5\) So, the ratio of the current densities in the two wires is \(0.04\) and the ratio of the drift velocities in the two wires is \(5\).

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