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Jaccobtw

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- Homework Statement
- A cylindrical copper wire has a radius of 1.2∗10^−3 m. It carries a constant current of 8.00A. What is the drift speed of the electrons in the wire in m/s? Assume each copper atom contributes one free electron to the current. The density of copper is 8.92g/cm^3.

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- Relevant Equations
- $$v_d = \frac{I}{neA}$$

We need to find each variable. ##I## is already given to us as 8 amps. The charge of an electron is 1.6 x 10^-19 coulombs. The cross sectional area will just be ##\pi(1.2∗10^−3)^2## m^2. Now we need to find the free electron density. We are given the density of of copper and can use dimensional analysis to find free electron density. Assume one free electron per copper atom:

$$\frac{8.92g}{cm^3} \times \frac{1 mol}{63.55g} \times \frac{6.022 \times 10^{23}atoms}{1mol} \times \frac{1 electron}{1 atom} = 8.45 \times 10^{22} \frac{electrons}{cm^3}$$

Plug in numbers

$$\frac{8.0 amps}{(\frac{8.45 \times10^{22}electrons}{cm^3})(1.6\times10^{-19}C)(\pi(1.2\times10^{-3})^2)}$$

I git 130.8 m/s but it was wrong. Can anyone help me find out why?

$$\frac{8.92g}{cm^3} \times \frac{1 mol}{63.55g} \times \frac{6.022 \times 10^{23}atoms}{1mol} \times \frac{1 electron}{1 atom} = 8.45 \times 10^{22} \frac{electrons}{cm^3}$$

Plug in numbers

$$\frac{8.0 amps}{(\frac{8.45 \times10^{22}electrons}{cm^3})(1.6\times10^{-19}C)(\pi(1.2\times10^{-3})^2)}$$

I git 130.8 m/s but it was wrong. Can anyone help me find out why?

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