Drift speed - class-XII

A. Drift velocity only.

The drift velocity of electrons in a conductor is of the order of $10^{−4} m/s. It is very small compared to the thermal speed which is of the order of 105 m/s.

Drift velocity is given by v_d = (eE tau) / m, and thermal velocity is v_th = sqrt(3kT / m). Using the provided conductivity and density, the ratio is calculated as approximately 3.67 x 10^-6.

Resistance is a property of the material and its geometry (length and area). Bending a wire does not change its length or cross-sectional area, so the resistance remains constant. The reason provided is also physically incorrect as drift velocity is determined by the electric field and scattering, not the path geometry.

Given
Initial current through the wire is $I = 1A$
Initial drift velocity is, $v _d = 1.2 \times 10^{-4} ms^{-1}$
Increased current is, $I' = 5A$
The current, I through the wire is given by
$I = \mu _e.e.A.v _d$
where, $\mu _e$ is the free electron density, e is the charge on electron, A is the area of cross section of the wire and v_d is the drift velocity of electrons. 
Since,  the free electron density, the charge on electron and the area of cross section of the wire are constant, hence
$I \propto v _d$.................(1)
Now, current through the wire is increased to 5 A, if the new drift velocity of electrons is $v' _d$ then
$I' \propto v' _d$................(2)

$V _d = \mu E = \mu \dfrac{V}{\ell}$
$V _d = \dfrac{\mu. I R}{\ell} \dfrac{\mu. I _{\rho} \ell}{A \ell} = \dfrac{\mu . I _{\rho}}{A}$
$\mu = \dfrac{V _d . A}{I _{\rho}} = \dfrac{1.1 \times 10^{-3} \times \lambda \times 25 \times 10^{-6}}{5 \times 1.1 \times 10^{-8}}$
$\mu = 1.57 ,  m^2 $ volt/sec.

Considering law of conservation of momentum ,electrons possess drift velocity which is greater than velocity of ions.  Thus $K _1>K _2$. hence correct option is option C.

Mobility of free electrons, $\mu = \dfrac{q\tau}{m}$        

We know 
$I = \eta eAV$
$2 = 1 \times {10^{24}} \times 1.6 \times {10^{ - 24}} \times 1 \times v$
$\boxed{v = 1.25,m/s}$

We know,

Drift velocity v_d = I / (neA). In series, current I is the same, so v_d is inversely proportional to A. In parallel, voltage V is the same, so I = V/R, where R = rhoL/A. Thus v_d = V / (nerho*L). Calculating these ratios yields 3:2 for series and 2:1 for parallel.

Length ,$d=2m$

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 $0.36\,mm/s$ 

Drift velocity v_d = I / (neA). Given density, atomic weight, and current, calculate n = (density * Na) / atomic weight. Then solve for v_d = I / (ne * pi * r^2). The result is 0.1 mm/s.

Time t = length / v_d. First find v_d = I / (neA). With I=3, n=8.5e28, A=2e-6, v_d = 3 / (8.5e28 * 1.6e-19 * 2e-6) = 1.1e-4 m/s. Then t = 3 / 1.1e-4 = 2.7e4 s.

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

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  $

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} $

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 $

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.

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 $

Current is given by I = n e A v, where A = pi r^2. For the new wire, radius is r/2, so area becomes A/4, and drift velocity is 2v. Therefore, new current I' = n e (A/4) (2v) = (1/2) n e A v = I/2.

We know, for a conductor carrying current,
Drift speed $= neI =ne \dfrac{V}{R}$
Resistivity $= \dfrac {RA}{L}$
where,
$n$ is no. of electrons, 
$e$  is charge on electrons,
$V$  is applied voltage,
$R$  is resistance and 
$A$  and $L$  are area of cross-section and length of resistor.
From above equations it is clear that drift speed, resistivityand resistance of resistor will be affected due to heating of resistor

Mobility mu = v_d / E. E = V / L = 200 / 2 = 100 V/m. mu = 0.5 / 100 = 0.005 = 5 x 10^-3 m^2/Vs.

The electron concentration is much greater than the hole concentration, indicating an n-type semiconductor. Conductivity sigma = e(n * mu_n + p * mu_p). Computing sigma and taking its reciprocal gives the resistivity, which comes out to approximately 0.34 ohm-meter.

Drift velocity v_d = mu * E. If the potential gradient E is doubled, v_d becomes 2 * v_d.

Mobility    $\mu = \dfrac{v _d}{E}$

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

With decrease in temperature, resistance of metal conductor decreases. But  $R \propto  \dfrac{1}{\tau}$, therefore time of relation $\tau$ increases. that's why the correct answer is option A.

Drift velocity $\displaystyle v _d=\frac{I}{neA}=\frac{V/R}{neA}=\frac{V}{neA(\frac{\rho L}{A})}=\frac{V}{ne\rho L}$
Thus, drift velocity is independent of diameter so it does not change.

Drift velocity $v _d= \frac{eEt}{m}$ so drift velocity is directly proportional  to the electric filed. When switch is on, the filed is quick set up wire  and produce current through wire instantaneously.   

The drift velocity is $v _d=\frac{eVt}{2ml}$
The relaxation time t changes with change of temperature. Actually it decreases with rise of temperature. Thus at a given temperature drift velocity  is directly proportional to the potential difference, inversely proportional to length and is independent of cross-sectional area.
So when drift velocity is doubled , V becomes double and l is half.

Current flows when there in P.D.

Current in the conductor $I = An e v _d$

We know that drift velocity is inversely proportional to density of charge carriers.

The flow of current current is nothing but drift velocity of electrons. Since current is directly proportional to voltage applied, drift velocity of electrons will also be directly proportional to the voltage applied. And it follows Ohm's law as V = I/R. So, if voltage is doubled, current is doubled and hence the drift velocity is doubled.

As current is flowing through it , it means charges are flowing along its length therefore heir must be some electric field parallel to the length of the wire. Hence correct option is B.

$v=\dfrac{I}{ne}=\dfrac{1}{7.4\times 10^{-7}}\times 8.4\times 10^{28}$
$=\dfrac{10^{-2}}{8.4\times 1.6\times 7.4}=10^-3$.

Current , $I$ = $neAv _{d} $

v_d = I / (neA). I = 10, n = 9e28, A = 1e-4, e = 1.6e-19. v_d = 10 / (9e28 * 1.6e-19 * 1e-4) = 10 / (14.4e5) = 10 / 1440000 = 6.94e-6 m/s.

Relaxation time is the average times between two successive collision between the free electrons and the atoms(kernel) of atoms .

v_d = I / (neA) = I / (ne * pi * r^2). If I becomes 2I and r becomes 2r, then A becomes 4A. v_d' = (2I) / (ne * 4A) = (1/2) * v_d.

We assume the electrons in a copper wire of radius 0.815 mm and carrying current 1 A at room temp 300 K.
Drift velocity , $v _d=\frac{I}{neA}=\frac{1}{(8.47\times 10^{28})\times (1.6\times 10^{-19})\times \pi\times (0.815\times 10^{-3} )^2 } \sim 3.45\times 10^{-5} m/s$
and $v _{rms}=\sqrt{\frac{3k _BT}{m}}=\sqrt{\frac{3\times (1.38\times 10^{-23})\times 300}{9.1\times 10^{-31}}}=1.17\times 10^5  m/s $
thus, $\frac{v _d}{v _{rms}}=10^{-10}$

When no current pass through the conductor i.e. when there is no electric field is applied to the conductor, each electron moves along a straight path at constant speeds and collide with the lattice ions(positive). With each collision the direction of electrons is changing randomly. The resulting path of any electron over along period of time, covering many collisions is a random sequence of straight segments. Due to this, the average number of electrons crossing any small area is nearly equal(in any direction). Thus the average displacement of electrons along any direction during any long period of time is also zero. Hence, the average velocity component in any direction is zero.

In a series circuit (the wire), the electric current must be constant throughout the wire regardless of changes in cross-sectional area. Current density and electric field will vary inversely with the cross-sectional area.

$\mu=\dfrac{V _d}{E}$     $E=\rho J$

We know that 

Let I be the current flow through wire. 

Given that,