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Physics Question on Electromagnetic induction

Current is flowing with a current density j=480Acm−2j = 480 Acm ^{-2} in a copper wire. Assuming that each copper atom contributes one free electron and given that Avogadro number =6.0×1023 atom mol−1= 6.0 \times 10^{23} \, atom \, mol^{-1} Density of copper =9.0 g cm−2 = 9.0 \, g \, cm^{-2} \, Electronic charge =1.6×10−19C= 1.6 \times 10^{-19} C Atomic weight of copper =64 g mol−1= 64 \, g \, mol^{-1} The drift velocity of electrons is

A

1mm s−11 mm \, s^{-1}

B

2mm s−12 mm \, s^{-1}

C

0.5mm s−10.5 mm \, s^{-1}

D

0.36mm s−10.36 mm \, s^{-1}

Answer

0.36mm s−10.36 mm \, s^{-1}

Explanation

Solution

Drift velocity is given by
        vd=InqA\, \, \, \, \, \, \, \, v_d = \frac {I}{nqA}
where I is current, n the number of electrons, A the area, q the charge.
Given IA=480Acm2 and q=1.6×10−19C \frac {I}{A} = \frac {480A}{cm^2} \, and \, q = 1.6 \times 10^{-19}C
     n=6×1023×964\, \, \, \, \, n = \frac {6 \times 10^{23}\times 9} {64}
∴     vd=480×646×1023×9×1.6×10−19\therefore \, \, \, \, \, v_d = 480 \times \frac {64}{6 \times 10^{23}\times 9 \times 1.6 \times 10^{-19}}
⇒    vd=480×646×9×1.6×10000cms−19\Rightarrow \, \, \, \, v_d = \frac {480 \times 64}{6 \times 9 \times 1.6 \times 10000}cms^{-19}
⇒    vd=32900cms−1\Rightarrow \, \, \, \, v_d = \frac {32}{900}cms^{-1}
     =32×10900cms−1\, \, \, \, \, = \frac {32 \times 10}{900}cms^{-1}
        =0.36mms−1\, \, \, \, \, \, \, \, = 0.36 mms^-1
⇒        vd=0.36 mms−1\Rightarrow \, \, \, \, \, \, \, \, v_d = 0.36 \, mms^-1