Question

Calculate the free electron concentration, mobility and drift velocity of
electrons in aluminum wire of length of 5 m and resistance 0.06 12
carrying a current of 15 A, assuming that each aluminum atom contributes
3 free elec-trons for conduction.
Given: Resistivity for aluminum = 2.7x 10-8 Q-m.
Atomic weight = 26.98
Density = 2.7 x 103 kg/m3​

Answer 1

Explanation:

kgm–3

 

Resistivity = 1.73 × 108    m

 

iii)Atomic weight = 63.5 kg

 

Calculate the mobility and the average time collision of electrons in copper obeying classical laws



Answer 2

The question is we have to calculate the free electron concentration, mobility and drift velocity of electrons in aluminium wire.

The wire is of length 5 m

resistance is 0.06

carries a current of 15 A

let assume each atom contributes 3 free electrons for conduction.

The given values are. :

Resistivity for aluminium=

$2.7 \times {10}^{ - 8} {q}^{ - m}$

Atomic weight os =26.98

Density =

$2.7 \times {10}^{3} \frac{kg}{ {m}^{3} }$

The drift velocity is the mean velocity of electrons within the conductor when it is exposed to the electric field.

J=neVd

sigma E= neVd.

Vd is the drift velocity of electrons of an aluminium.

n= number of electrons per unit volume

$n = \frac{density}{molar \: mass \: of \: aluminium} \times$ x Vd

$n = \frac{2.7 \times {10}^{3} }{26.98} \times$*Vd

$n = 100074 \times {m}^{ - 2}$xVd

$e = 1.6 \times {10}^{ - 19}$

sigma = conductivity=1/rho =1/resistivity.

$= \frac{1}{2.7 \times {10}^{ - 8} }$

The voltage is obtained by the formula

E=IR

$15 \times 0.06 = 0.9$

sibstitute the values in the formula

$\frac{1}{2.62 \times {10}^{ - 8} } \times 0.9 = 100074 \times 1.6 \times {10}^{ - 19 } \times$x Vd

$0.343 \times {10}^{8} = 160118 \times {10}^{ - 19} \times$Vd

$\frac{0.343 \times {10}^{8} }{160118 \times {10}^{ - 19} } =$Vd

$2.142 \times {10}^{ - 6} \times {10}^{8} \times {10}^{ 19} =$Vd

$2.142 \times {10}^{21}$

Mobility= u= drift velocity/E

substituting values we get

$\frac{2.142 \times {10}^{27} }{0.9} = 2.38 \times {10}^{27} \frac{ {m}^{2} }{v.s}$

# spj2

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