Question

Consider a wire of length 4m and cross-sectional area Imm? carrying a current 2A. If each ci
metre of the material contains 1029 free electrons then the average time taken by an electron to a
the length of the wire is
[
a) 3.2 x 104 s
b) 1.2 x 104 s
c) 6.2 x 104 s
d) 3.4 x 10's​

Answer 1

Given :-

• Length of the wire = 4 m

• Area of cross section = 1 mm ² = 10 ^-6

• Current = 2 A

• current density = 10 ²⁹
• charge on an electron = 1.6 x 10 -¹⁹

To find :-

• Average time taken

Solution:-

According to the formula ,

$\implies\mathtt{ I = n e A V_d}$

here ,

• I = current
• n = current density
• e = charge on an electron
• A = area of cross section
• Vd = drift velocity

On rearranging ,

$\implies\mathtt{ V_d= \dfrac{I}{neA}}$

Now let's substitute the values in the above equation ,

$\implies\mathtt{ V_d= \dfrac{2}{10^{29} \times 1.6\times 10^{-19} \times 10^{-6} }}$

$\implies\mathtt{ V_d= \dfrac{2}{10^{29} \times 1.6\times 10^{-25} }}$

$\implies\mathtt{ V_d= \dfrac{1.25}{10^{4} }}$

$\implies\mathtt{ V_d= 1.25 \times 10^{-4} }}$

we know that,

$\implies\mathtt{ V_d= \dfrac{l}{t}}$

here ,

• Vd = drift velocity
• l = length of wire
• t = time taken

On rearranging

$\implies\mathtt{ t= \dfrac{l}{V_d}}$

$\implies\mathtt{ t= \dfrac{4}{1.25 \times 10^{-4} }}$

$\implies\mathtt{ t= 3.2 \times 10^{4} }$

The average time taken by an electron is 3.2 x 10 ^4 s