An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 5.50 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.
Question
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 5.50 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.
Solution
To solve this problem, we need to use the formula for current, which is I = nAvq, where:
- I is the current
- n is the number of charge carriers per unit volume
- A is the cross-sectional area of the wire
- v is the drift speed of the electrons
- q is the charge of an electron
We are given I, A, and q, and we need to find v. We also need to find n, which we can do using the density of aluminum and the atomic mass of aluminum.
First, let's find n. The density of aluminum is given as 2.70 g/cm^3, which we need to convert to kg/m^3. 1 g/cm^3 is equal to 1000 kg/m^3, so the density of aluminum is 2700 kg/m^3.
The atomic mass of aluminum is approximately 27 g/mol, which is 27 x 10^-3 kg/mol. We also know that 1 mol of any substance contains Avogadro's number (6.022 x 10^23) of atoms.
So, the number of atoms per unit volume (n) is the density divided by the atomic mass, multiplied by Avogadro's number:
n = (2700 kg/m^3) / (27 x 10^-3 kg/mol) x (6.022 x 10^23 atoms/mol) = 6.022 x 10^28 atoms/m^3
Since each aluminum atom supplies one conduction electron, this is also the number of charge carriers per unit volume.
Next, we can rearrange the formula for current to solve for v:
v = I / (nAq)
Substituting the given values and the ones we calculated:
v = (5.50 A) / [(6.022 x 10^28 electrons/m^3) x (2.20 x 10^-6 m^2) x (1.60 x 10^-19 C/electron)]
Solving this gives a drift speed of approximately 2.12 x 10^-5 m/s.
Similar Questions
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 5.50 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. The equation for the drift velocity includes the number of charge carriers per volume, which in this case is equal to the number of atoms per volume. How do you calculate that if you know the density and the atomic weight of aluminum? mm
An aluminium (Al) rod with area of cross-section 4×10−6 m2 has a current of 5 ampere, flowing through it. Find the drift velocity of electron in the rod. Density of Al=2.7×103 kg/m3 and Atomic wt. =27. Assume that each Al atom provides one electron
The number density of conduction electrons in a particular metal is 1029 m-3. A current of 4 A flows through a wire of this material, which has a length of 1m and cross-sectional area of 2(mm)2. The voltage across the wire is 2.70x10-3 V. What is the average speed (drift speed) of the electrons in units of ms-1 to 3 significant figures using scientific notation (e.g. x.xxExx)?
An 18 gauge copper wire with a diameter of 1.02 mm carries a constant current of 1.65 A to a 200 W lamp. The free electron density in the wire is 8.5 X 1028 per cubic meter. Find (a) the current density and (b) the drift speed.
An aluminium wire of diameter 0.24 cm is connected in series to a copper wire of diameter 0.16 cm. The wires carry an electric current of 10 A. Determine the current density in aluminium wire.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.