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 in mm-s is close to -Take charge of electron to be -1-6-10-19C-

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Toppr Admin · Accepted Answer

The correct option is D 0-02I-neAvd-x21D2-vd-IneA-1-59-xD7-1028-xD7-1-6-xD7-10-x2212-19-xD7-5-xD7-10-x2212-6-0-02mm-s

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 $= 1.6 \times 10^{- 19} C$ )

The correct option is D $0.02$
$I = n e A v_{d}$
$\Rightarrow v_{d} = \frac{I}{n e A} = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6}}$
$= 0.02 m m / s$

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)

The correct option is D 0.02I=neAvd⇒vd=IneA=1.59×1028×1.6×10−19×5×10−6=0.02mm/s

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 $= 1.6 \times 10^{- 19} C$ )

The correct option is D $0.02$
$I = n e A v_{d}$
$\Rightarrow v_{d} = \frac{I}{n e A} = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6}}$
$= 0.02 m m / s$