An aluminum wire carrying a current of 9.0 A has a cross-sectional area of 7.0X10^-6 m2. Find the drift speed of the electrons in the wire. The density of aluminum is 2.7 g/cm3. (Assume three electrons are supplied by each atom.)
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To find the drift speed of electrons in the aluminum wire, we can use the formula:
v_d = (I / (n * A * q)),
where:
v_d is the drift speed,
I is the current passing through the wire,
A is the cross-sectional area of the wire,
q is the charge of the electron, and
n is the number of charge carriers per unit volume.
First, we need to calculate n, the number of charge carriers per unit volume. Since aluminum has atomic mass of approximately 27 g/mol and a density of 2.7 g/cm^3, we can find the volume occupied by one atom of aluminum.
Using the relation: density = mass / volume,
we can rearrange it to find the volume:
volume = mass / density = (27 g/mol) / (2.7 g/cm^3).
We know that each aluminum atom contributes 3 electrons, so the number of electrons per unit volume (n) is given by:
n = (number of electrons per atom) * (number of atoms per volume),
= (3 electrons) * ( Avogadro's number / volume of one aluminum atom) ,
≈ (3) * (6.02214 x 10^23 / volume of one aluminum atom).
Now, we need to find the volume of one aluminum atom, which can be calculated using the atomic radius or other methods. Let's assume that volume of one aluminum atom is 4πr^3/3, where r is the atomic radius. The atomic radius of aluminum is around 1.43 Å (angstrom) or 1.43 x 10^-10 m.
Therefore, the volume of one aluminum atom is:
volume of one aluminum atom = 4π(1.43 x 10^-10 m)^3 / 3.
Now, we can substitute this value into our equation for n:
n = (3) * (6.02214 x 10^23) / (4π(1.43 x 10^-10 m)^3 / 3).
Next, we can find the drift speed (v_d) by plugging in the given values of current (I = 9.0 A), cross-sectional area (A = 7.0 x 10^-6 m^2), and charge of an electron (q = 1.6 x 10^-19 C) into the formula:
v_d = (9.0 A) / (n * 7.0 x 10^-6 m^2 * 1.6 x 10^-19 C).
Calculating all the values, we can determine the drift speed of electrons in the wire.