Tag: physics

Questions Related to physics

Current is flowing with a current density $J=480\ amp/cm^{2}$ in a copper wire. Assuming that each copper atom contribution one free electron and gives that  Avogadro number$=6.0\times 10^{23}\ atoms/mole$  Density of copper $=9.0\ g/cm^{3}$ .Atomic weight of copper $=64\ g/mole$ Electronic charge $=1.6\times 10^{-19}$ coulomb. The drift velocity of electrons is:

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

  1. $1\ mm/s$

  2. $2\ mm/s$

  3. $0.5\ mm/s$

  4. $0.36\ mm/s$

Show answer
Correct Option: D
Explanation:

Given that,

 Current density $J=480\,amp/c{{m}^{2}}$

Avogadro number $N=6.0\times {{10}^{23}}\,atoms/mole$

Density of copper $\rho =9.0\,g/c{{m}^{3}}$
Atomic weight of copper $m=64\,g/mole$

Electronic charge $e=1.6\times {{10}^{-19}}\,coulomb$

We know that

Current density $J=\dfrac{I}{A}$

Now, the drift velocity is

${{v} _{d}}=\dfrac{J}{ne}....(I)$

We know that,

$n={{N} _{A}}\times \dfrac{1}{m}\times \rho $

Now, put the value of n in equation (I)

  $ {{v} _{d}}=\dfrac{Jm}{{{N} _{A}}e\rho } $

 $ {{v} _{d}}=\dfrac{480\times 64}{6.0\times {{10}^{23}}\times 1.6\times {{10}^{-19}}\times 9} $

 $ {{v} _{d}}=\dfrac{30720}{86.4\times {{10}^{4}}} $

 $ {{v} _{d}}=355.5\times {{10}^{-4}}\,cm/s $

 $ {{v} _{d}}=0.36\,mm/s $

Hence, the drift velocity is $0.36\,mm/s$ 

Assume that each atom of copper contributes one free electron. The density of copper is $9g cm^{-3}$ and atomic weight of copper is $63$. If the current flowing through a copper wire of $1mm$ diameter is $1.1 $ ampere, the drift velocity of electrons will be:- 

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

  1. $0.01 mm s^{-1}$

  2. $0.02 mm s^{-1}$

  3. $0.2 mm s^{-1}$

  4. $0.1 mm s^{-1}$

Show answer
Correct Option: D

Find the time an electron takes to drift from one end of a uniform wire $3m$ long to its other end if the wire is $2$ x ${ 10 }^{ -6 }{ m }^{ 2 }$ in cross section and carries a current $3A$.The density of free electrons in a copper conductor is $8.5$ x ${ 10 }^{ 28 }{ m }^{ 3 }$.

physics electric current drift velocity and mobility drift speed drift velocity & mobility

  1. $2.7$ x ${ 10 }^{ 4 }s$

  2. $1.1$ x ${ 10 }^{ -4 }s$

  3. $0.9$ x ${ 10 }^{ -4 }s$

  4. $1.414$ x ${ 10 }^{ 4 }s$

Show answer
Correct Option: A

How many electrons should be removed from a coil of mass 1.6 gram so that it may float in an electric field of intensity $10^9 NC^-1$ directed upwards ?

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

  1. $10^6$

  2. $10^7$

  3. $10^8$

  4. $10^9$

Show answer
Correct Option: C

How many electrons should be removed from a coin of mass 1.6 gram, so that it may float in an electric field of intensity $10^9 NC^-1$ directed upwards?

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

  1. $10^6$

  2. $10^7$

  3. $10^9$

  4. $10^8$

Show answer
Correct Option: A

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 $9000kgm^{2})$

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

  1. $0.74 mms^{-1}$

  2. $7.4 mms^{-1}$

  3. $74 mms^{-1}$

  4. $0.074 mms^{-1}$

Show answer
Correct Option: D
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 $vd = \dfrac{i}{neA}$
$=\dfrac{1}{8.2\times 10^{28}\times 1.602\times 10^{19}\times 10^{-3}}$
$=7.4\times 10^{-5}m/s$
$0.07mm/s$

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  

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

  1. 1.0 mm/sec

  2. 1.0 m / sec

  3. 0.1 mm/sec

  4. 0.01 mm / sec

Show answer
Correct Option: A
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  $

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 

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

  1. v/4

  2. v/2

  3. 2v

  4. 4 v

Show answer
Correct Option: A
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} $

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:

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

  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. $25\times 10^{6}m/s$

Show answer
Correct Option: D
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 $

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

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

  1. Is doubled

  2. Is halved

  3. Remain same

  4. Becomes zero

Show answer
Correct Option: A
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.