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An aluminum wire having a cross-sectional area of 2.10 ✕ 10⁻⁶ m² carries a current of 6.00 A. The density of aluminum is 2.70 g/cm³. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.

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Safari File Edit View History Bookmarks Window Help 恸 令49%DEE1 Mon 3:12:37 PM 6.585*10*-5 X Your response is off by a multiple of ten. mA Submit Answer Save Progress Practice Another Version +-/10 points SerCP11 17.2.P.008 My Notes Ask YourT An aluminum wire having a cross-sectional area of 2.10 x 10-6 m carries a current of 6.00 A. The density of aluminum is 2.70 g/cm3 Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire. mm/s Need Help? LReaan -/10 points SerCP11 17.4.P.016 My Notes Ask Your T A wire of diameter 0.650 mm and length 40.0 m has a measured resistance of 1.40 Ω·What is the resistivity of the wire? Ω·m Need Help? Read It

Answer 1

The current is given by
  $I =6.00~A$
And the cross sectional area
$A =2.10\times 10^{-6}~m^2$
And so, the current density is
  $J=\frac{I}{A} =\frac{6}{2.10\times 10^{-6}}~A/m^2$
  $\Rightarrow J=2.857\times 10^{6}~A/m^2$
And in terms of the drift velocity, this is given by
  $J=\rho_n e v_{d}$
  $\Rightarrow v_d=\frac{J}{\rho_n e}$
where, $\rho_n$ is the number density of conduction electrons, and e is the electron charge. Now this is related to the mass density $\rho_m$ of the aluminium is given by
$\rho_n=\frac{\rho_mN_An}{M}$
where, M is the molar mass, N_A is the Avogadro constant, n is the number of conduction electrons per atom. So, putting the given values,
$\rho_m = 2.70~g/cm^3 =2700~kg/m^3$
  $M =27\times 10^{-3}~kg$
$N_A =6.023\times 10^{23}$
$n=1$
we get the number density of conduction electrons
  
  $\rho_n=\frac{2700\times 6.023\times 10^{23}\times 1}{27\times 10^{-3}}~m^3$
  $\Rightarrow \rho_n=6.023\times 10^{28}~m^3$
So, the drift velocity becomes
$\Rightarrow v_d=\frac{J}{\rho_n e}=\frac{2.857\times 10^{6}}{6.023\times 10^{28}\times 1.602\times 10^{-19}}~m/s$
  $\Rightarrow v_d=0.296\times 10^{-3}~m/s$
  $\Rightarrow v_d=0.296~mm/s$

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Resistivity and resistance are related via
$R=\rho \frac{l}{A}$
$\Rightarrow \rho= \frac{RA}{l}$
where, l is the length of the wire and A is cross sectional area of the wire. So, we get in terms of the diameter d as
  $A= \frac{\pi d^2}{4}$
So, we get

$\Rightarrow \rho= \frac{\pi d^2 R}{4l}$
Now putting the given values,
  $d=0.650~mm=0.650\times 10^{-3}~m$, $l=40.0~m$,  $R =1.40~\Omega$.
we get the resistivity of the wire as
$\Rightarrow \rho= \frac{\pi\times (0.650\times 10^{-3})^2\times 1.4}{4\times 40}~\Omega . m$
  $\Rightarrow \rho=1.161\times 10^{-8}~\Omega . m$