Tag: electric current, potential difference and resistance

Questions Related to electric current, potential difference and resistance

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


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:

  1. Is doubled

  2. Is halved

  3. Remain same

  4. Becomes zero


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.

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


Correct Option: B
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 $

There is current of 40 amperes in a wire of $10^{-6}m^{2}$ 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. $25.0\times 10^{-3}m/s$

  4. $250\times 10^{-3}m/s$


Correct Option: A

Drift speed of conduction electrons in the wire is

  1. $5.5cm/s$

  2. $0.55 mm/s$

  3. $0.22 mm/s$

  4. $1.1 mm/s$


Correct Option: B

In a wire of cross section radius r, free electrons travel with drift velocity V when a current a $I$ flows throught the wire. What is the current in another wire of half the radius and of the same material when the drift velocity is $2V$ ?

  1. $2I$

  2. $I$

  3. $I/2$

  4. $I/4$


Correct Option: B

The mean free path of electrons in a metal is $44 \times 10 ^ { - 8 } \mathrm { m }$ . Theelectric field which can give on an average 2$e \mathrm { V }$ energy to an electron in the metal will be in units of VIm 

  1. $8 \times 10 ^ { 7 }$

  2. $5 \times 10 ^ { - 11 }$

  3. $8 \times 10 ^ { - 11 }$

  4. $5 \times 10 ^ { 7 }$


Correct Option: D

When 3 V potential difference is applied a wire of length 0.1 m. having resistivity $1.6 \times 10^{-5}$ $\Omega m$, the electrons started moving. If the electron density in the wire is $6 \times 10^{10} m^{-1}$, the drift speed of electrons is  

  1. $1.94 \times 10^{-6}\ ms^{-1}$

  2. $1.94 \times 10^{-5}\ ms^{-1}$

  3. $1.94 \times 10^{-8}\ ms^{-1}$

  4. $1.94 \times 10^{-7}\ ms^{-1}$


Correct Option: C

The drift velocity of free electron in a metal wire of a given potential gradient along it is $ V _d $ if this potential gradient is doubled the new drift velocity will be 

  1. $ V _d $

  2. $ 2V _d $

  3. $ \frac {V _d}{2} $

  4. $4V _d $


Correct Option: B

When a potential difference $V $  is applied across a conductor at a temperature $T,$  the drift velocity of electrons is proportional to

  1. $\sqrt{V}$

  2. $V$

  3. $\sqrt{T}$

  4. $T$


Correct Option: B
Explanation:

We know that Drift velocity $v _d = \displaystyle \dfrac{eE}{m} \tau = \dfrac{e}{m} \tau \left ( \dfrac{V}{l} \right ) $ ($\because E = \dfrac{V}{l})$

so for a particular conductor of a particular length the drift velocity will directly depend upon voltage
Hence $v _d \propto V$. option B is correct.