SATHEE: Current Electricity 1 Question 11

Current Electricity 1 Question 11

11. Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross-section $5 m m^{2}$ is $v$ . If the electron density in copper is $9 \times 10^{28} / m^{3}$ , the value of $v$ in $m m / s$ is close to (Take, charge of electron to be $= 1.6 \times 10^{- 19} C$ )

(201 9Main, 9 Jan I)

(a) 0.02

(b) 0.2

(d) 3

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Solution:

Relation between current $(I)$ flowing through a conducting wire and drift velocity of electrons $(v_{d})$ is given as

$I = n e A v_{d}$

where, $n$ is the electron density and $A$ is the area of cross-section of wire.

$\Rightarrow v_{d} = \frac{I}{n e A}$

Substituting the given values, we get

$\begin{aligned} v = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6}} \\ v = \frac{1.5 \times 10^{- 3}}{72} m / s = 0.2 \times 10^{- 4} m / s \\ v = 0.02 m m / s \end{aligned}$

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Current Electricity 1 Question 11

11. Drift speed of electrons, when 1.5A of current flows in a copper wire of cross-section 5mm2 is v. If the electron density in copper is 9×1028/m3, the value of v in mm/s is close to (Take, charge of electron to be =1.6×10−19C )

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11. Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross-section $5 m m^{2}$ is $v$ . If the electron density in copper is $9 \times 10^{28} / m^{3}$ , the value of $v$ in $m m / s$ is close to (Take, charge of electron to be $= 1.6 \times 10^{- 19} C$ )