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This is second of our many quizzes on Drift velocity. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Drift velocity Quiz 2

physics

 In a closed circuit, the current flows through a wire of radius $''r''$. If the charge carriers are protons, then the corresponding drift velocity will be ${{V}_{d1}}$. But if the charge carriers are electrons, then drift corresponding velocity is ${{V}_{d2}}$. Ratio of $\dfrac{{{V}_{d1}}}{{{V}_{d2}}}$ for the same set of working conditions would be  

$''I''$ related to $''{{V}_{d}}''$ as $I=neA{{V}_{d}}$ use this relation to solve the problem. 

  For a charge particle e, the relation between current and its drift velocity is $ \\ I=neA{{V}_{d}} \\ \Rightarrow {{V}_{d}}=\dfrac{I}{neA} \\ $ As working conditions are the same, ${{\text{I}}_{\Pr oton}}\text{ and }{{\text{I}}_{Electron}} $ will have same $n, e=1.6\times {{10}^{-19}}\text{ C, A}\text{.} \\ \dfrac{I}{neA}=\text{constant} \\ \therefore \dfrac{{{V}_{d1}}}{{{V}_{d2}}}=\dfrac{1}{1}$

A charge $''Q''$ is spread over a length  $''\ell ''$ pivoted at the point $''O''$. It’s rotated with an angular velocity $''W''$. Find the value of net current due to this arrangement.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8132410d-0251-4322-bedd-1e2ad35e2ccc7549826805567088354.png"/> 

$I=\dfrac{dq}{dt}$, where $''I''$ is current flow $''q''$ is charge flow. 

  $I=\dfrac{Q}{T}=\dfrac{Q}{\dfrac{2\pi }{\omega }}=\dfrac{\omega Q}{2\pi } \\ $ current flowing due to the arrangement is rate of flow of charge through a given area of cross-section, where $ \omega$ is the angular velocity. If the rod is made up of the same charge distribution, with each discrete elemental charge moving at the same angular velocity. T is the time period to complete one revolution.

A circular ring having resistance per unit length $''R''$ is connected to two points $A$ and $B$ as shown in figure. Find ${{R}_{AB}}=\left( ? \right)$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/81d18172-3ea6-49b5-bdc4-b861c62d1122522325831365888512.png"/> 

 If $''R''$ is resistance per unit length, then total resistance of circular wire is $\left( 2\pi r \right)\times R$. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/edff8a0d-9734-4f49-bd25-c59f34013d394959234309917913024.png"/>$ \\ \text{Total Resistance of the complete circle  = R}\times \text{2}\pi \text{r} \\ \therefore \,\,\text{Resistance of the arc which subtends }90{}^\circ \text{ at center} \\ \Rightarrow {{R}_{1}}=R\times 2\pi r\times \dfrac{90}{360}=\dfrac{\left( 2\pi r \right)\times R}{4}=\dfrac{\pi rR}{2} \\ \therefore \,\text{ Resistance of arc subtends }360{}^\circ -90{}^\circ =270{}^\circ \\ \text{ Angle at center}=R\times 2\pi r\times \dfrac{270{}^\circ }{360{}^\circ }=\left( 2\pi rR \right)\times \dfrac{3}{4}=\dfrac{3}{2}\pi rR \\ \text{These both arcs are parallel to each other because}\\ \text{they are connected across the same terminals A and B}\text{.} \\ \text{Their equivalent resistance is} \\{{R}_{AB}}=\dfrac{{{R}_{1}}{{R}_{2}}}{\left( {{R}_{1}}+{{R}_{2}} \right)} \\ \Rightarrow {{R}_{AB}}=\dfrac{\left( \dfrac{\pi rR}{2} \right)\times \dfrac{3}{2}\pi rR}{\dfrac{\pi rR}{2}+\dfrac{3}{2}\pi rR}=\left( \pi rR \right)\times \dfrac{3}{4\times 2} \\ \therefore {{R}_{AB}}= \dfrac{3}{8}\pi rR$

For a current carrying conductor having resistance $''R''$, voltage $''V''$  is applied across it and this results in an induced electric field $''E''$ then how ${{V}_{d}}$ and $''E''$ is related to each other?

 $V=IR$ and $V=E\times d$. Use both of these formulas to analyze the problem. Here, $E$ is an electric field across the conductor, $d$ is the length of the conductor.

 From Ohm’s law $V=IR \\ \Rightarrow I=\dfrac{V}{R} \\ $ As the induced electric field is uniform, if d is the length of the conductor, $ \\V=E\times d \\ \Rightarrow \,\,\,I=\dfrac{E\times d}{R} \\ $ Also, the relation of drift velocity of charges and current in a conductor is, $ \\I=neA{{V}_{d}} \\ \therefore \,\,\dfrac{E\times d}{R}=neA{{V}_{d}} \\As\text{ d, R, n, e, A are constants} \\{{V}_{d}}\propto E$

In a conductor, the charge is flowing with time as described by the equation $q=2t+3{{t}^{2}}-4{{t}^{3}}$ . At what $t=?$ does the current hit its maximum value??

 $I=\dfrac{dq}{dt}$, current is the rate of charge flow, and for ${{\operatorname{I}}_{max}}$, $\dfrac{dI}{dt}=0$, where ${{I}_{\max }}$ is the maximum current flow in the circuit.

 $As\,\,I=\dfrac{dq}{dt}=2+6t-12{{t}^{2}} \\ \text{For maxima of current, the condition is} \\ \dfrac{dI}{dt}=0 \\ \Rightarrow \,\,6-24t=0 \\ \therefore \,\,t=\dfrac{1}{4}\,\sec  \\As\,\,\dfrac{{{d}^{2}}I}{d{{t}^{2}}}=-24 \\ \text{The negative sign means that at t=}\dfrac{1}{4}\,\,\sec, \text{ we get the maximum value of current}\text{.}$ 