Tag: drift velocity & mobility

Quizzes (0) Questions (60)

Questions Related to drift velocity & mobility

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A current of $1.0A$ exists in a copper wire of cross-section $1.0mm^2$.Assuming one free electron per atom

Calculate drift speed of the free electron in the wire _______(the density of copper is $9000kgm^{2})$

  1. $0.74 mms^{-1}$

  2. $7.4 mms^{-1}$

  3. $74 mms^{-1}$

  4. $0.074 mms^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$i=ne AV _d$
$n=\dfrac{6\times 10^{23}\times 9000}{63.5\times 10^{-3}}$


$=8.5\times 10^{28}m^{-3}$

Hence $vd = \dfrac{i}{neA}$
$=\dfrac{1}{8.2\times 10^{28}\times 1.602\times 10^{19}\times 10^{-3}}$
$=7.4\times 10^{-5}m/s$
$0.07mm/s$

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is $ 1 mm^2 $. If the number of free electrons per $ cm^3  is 8.4 \times 10^22 $, then the drift velocity would be  

  1. 1.0 mm/sec

  2. 1.0 m / sec

  3. 0.1 mm/sec

  4. 0.01 mm / sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Current $I=1.344\,A$

Area of cross-section, $A=1\,m{{m}^{2}}=0.01\,c{{m}^{2}}$

Number of free electrons, $n=8.4\times {{10}^{22}}$

Charge on electron $e=1.6\times {{10}^{-19}}\,$

Drift velocity is given by

$ {{v} _{d}}=\dfrac{i}{nAe} $

$ =\dfrac{1.344}{8.4\times {{10}^{22}}\times 0.01\times 1.6\times {{10}^{-19}}} $

$ =\dfrac{1.344}{{{10}^{3}}\times 0.1344} $

$ ={{10}^{-2}}\,cm/\sec  $

$ =1\,mm/\sec  $

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A current I flows through a uniform wire of diameter d when the electron drift velocity is V .The same current will flow through a wire of diameter d/2  made of the same material if the drift velocity of the electrons is 

  1. v/4

  2. v/2

  3. 2v

  4. 4 v
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relation between current and drift velocity is

$ I=neA{{v} _{d}} $

$ {{v} _{d}}=\dfrac{I}{neA} $

$ {{v} _{d}}=\dfrac{I}{ne\pi {{r}^{2}}} $

According to given data,

$ v _{d}^{'}=\dfrac{1}{ne\pi {{(2r)}^{2}}} $

$ v _{d}^{'}=\dfrac{1}{4}\dfrac{1}{ne\pi {{r}^{2}}} $

$ {{v} _{d}}'=\dfrac{{{v} _{d}}}{4} $

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

There is a current of 40 amperes in a wire of $10^{-16}m^{2}$ area of cross-section. If the number of free electrons per $m^{3}$ is $10^{29}$, then the drift velocity will be:

  1. $1.25\times 10^{3}$ m/s

  2. $2.50\times 10^{3}m/s$

  3. $2.0\times 10^{6}m/s$

  4. $25\times 10^{6}m/s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If L is the length of wire so velocity is given by $v=\dfrac{L}{t}$

Total number of free electrons in the wire, $Q=nqLA$

Current,

$ I=\dfrac{Q}{t} $

$ I=\dfrac{nqLA}{t} $

$ I=nqvA $

$ v=\dfrac{I}{nqA} $

Where, n is the number of electron, $n={{10}^{29}}$

q is the charge of an electron, $q=1.6\times {{10}^{-19\,}}C$

A is area, $A={{10}^{-16}}\,{{m}^{2}}$

I is current, $I=40\,A$

So, drift velocity,

$ v=\dfrac{40}{{{10}^{29}}\times 1.6\times {{10}^{-19}}\times {{10}^{-16}}} $

$ v=25\times {{10}^{6}}\,m/s $

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A potential difference $V$ is applied to a copper wire of length $l$ and thickness $d$. If $V$ is doubled, the drift velocity:

  1. Is doubled

  2. Is halved

  3. Remain same

  4. Becomes zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the first case, $E= V/l$ so $E$ is directly proportional to $V$ so $E$ will be doubled. Since $V= IR$ so, $E= IR/l$ so Resistance is doubled and $Vd$ is directly proportional to $E$ so $Vd$ is doubled. So in the case, everything is doubled.

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

The number of free electrons per $10$ mm ordinary copper wire is about $2\times 10^{21}$. The average drift speed of the electrons is $0.25$ mm current flowing is:

  1. $0.8$ A

  2. $8$ A

  3. $80$ A

  4. $5$ A
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

Number of electron, $n=2\times {{10}^{21}}$

Average drift speed, $0.25\,mm/s$

$ Q=ne $

$ Q=2\times {{10}^{21}}\times 1.6\times {{10}^{-19}} $

$ Q=320\,C $

Since,

$ s=\dfrac{D}{T} $

$ T=\dfrac{D}{s} $

$ T=\dfrac{10}{0.25}=40 $

So, current

$ I=\dfrac{Q}{T} $

$ I=\dfrac{320}{40} $

$ I=8\,A $