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
Question
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
Solution
To find the drift velocity of the electron in the rod, we need to use the formula for current, which is given by:
I = nAev
where: I = current n = number of free electrons per unit volume A = cross-sectional area e = charge of an electron v = drift velocity
We know that the current I is 5 A, the cross-sectional area A is 4×10−6 m², and the charge of an electron e is 1.6×10^-19 C.
We need to find the number of free electrons per unit volume n. We can find this by using the density of aluminium and its atomic weight.
The density of aluminium is 2.7×10³ kg/m³ and its atomic weight is 27.
The number of moles of aluminium per cubic meter is given by the density divided by the atomic weight.
So, the number of moles = 2.7×10³ kg/m³ / 27 kg/mol = 100 mol/m³
Since each mole of aluminium provides one electron, the number of free electrons per unit volume n is given by the number of moles times Avogadro's number (6.022×10²³ per mol).
So, n = 100 mol/m³ * 6.022×10²³ per mol = 6.022×10²5 per m³
Now we can substitute the values into the formula for current to find the drift velocity v.
5 A = (6.022×10²5 per m³)(4×10−6 m²)(1.6×10^-19 C)(v)
Solving for v, we get:
v = 5 A / [(6.022×10²5 per m³)(4×10−6 m²)(1.6×10^-19 C)] = 0.52 m/s
So, the drift velocity of the electron in the rod is approximately 0.52 m/s.
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