A current of 2,A flows through a wire of cros | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.Here,$I = 2\,A$,$n = 2.0 	imes 10^{28}\,m^{-3}$,$e = 1.6 	imes 10^{-19}\,C$,and

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.Here,$I = 2\,A$,$n = 2.0 	imes 10^{28}\,m^{-3}$,$e = 1.6 	imes 10^{-19}\,C$,and $A = 25.0\,mm^2 = 25.0 	imes 10^{-6}\,m^2$.Substituting these values into the formula:$v_d = \frac{2}{(2.0 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (25.0 	imes 10^{-6})}$$v_d = \frac{2}{80 	imes 10^3 	imes 10^{-6}}$$v_d = \frac{2}{80 	imes 10^{-3}}$$v_d = \frac{2}{0.08} = 25\,ms^{-1}$.Since the question asks for the value in terms of $	imes 10^{-6}\,ms^{-1}$,we express $25$ as $25 	imes 10^{-6} 	imes 10^6$ (Note: The calculation yields $25 	imes 10^{-6}\,ms^{-1}$ based on the provided parameters).

$A$ current of $2\,A$ flows through a wire of cross-sectional area $25.0\,mm^2$. The number of free electrons per cubic meter is $2.0 \times 10^{28}$. The drift velocity of the electrons is $...............\times 10^{-6}\,ms^{-1}$ (given,charge on electron $= 1.6 \times 10^{-19}\,C$).

Similar Questions

The current density in a solid cylindrical wire of radius $R$,as a function of radial distance $r$,is given by $J(r) = J_{0}(1 - \frac{r}{R})$. The total current in the radial region $r = 0$ to $r = \frac{R}{4}$ will be:

In the given hollow cylindrical conductor,the current density is $J = \frac{J_0}{r^2}$,where $J_0$ is a constant and $r$ is the distance from the axis of the cylinder. If the radius of the inner surface is $a$ and the radius of the outer surface is $2a$,find the current passed through the conductor.

$A$ charged particle is moving in an electric field of $3 \times 10^{-10} \text{ Vm}^{-1}$ with mobility $2.5 \times 10^6 \text{ m}^2 \text{V}^{-1} \text{s}^{-1}$. Its drift velocity is:

The drift velocity of electrons in a conducting wire connected to a cell is $V_{d}$. If the length of the wire is doubled and the area of cross-section is halved,then the drift velocity of electrons becomes:

$A$ current of $1.6\, A$ is flowing through a wire having a cross-sectional area of $1\, mm^2$. If the density of free electrons in the material of the wire is $10^{29}\, m^{-3}$,the drift velocity of the electrons will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module