The electric field E, current density J and conductivity \$\sigma\$ of a conductor are related as

Let I current flow through a conductor when applied potential difference V volt. then, from ohm’s law,\$I=\dfrac VR\$But \$R=\rho\dfrac LA\$Therefore,\$I=\dfrac V{\rho\dfrac LA} \\\implies \dfrac IA=\dfrac V{\rho L}\$but, \$\dfrac IA=J, \dfrac VL=E\$Hence, \$J=\sigma E\\\implies \sigma = \dfrac JE\$

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>>Class 12
>>Physics
>>Current Electricity
>>Drift Velocity & Mobility
>> $Interpretation of Current f...$

Interpretation of Current from Drift Velocity

Q: Drive an expression for drift velocity of free electrons in a conductor.

When conductor is subjected to an electric field \$E\$, each electron experience a force \$F= -eE\$ and acquires an acceleration of \$a= F/m= -\dfrac{-eE}{m}\$ ..... (1)Here \$m\$= mass of electron, \$e\$= charge , \$E\$= Electric Field.The average time difference between two consecutive collisions is known as relaxation time of electron.\$\vec {\tau}= \dfrac{{\tau}_1 +{\tau}_2 + {\tau}_3 + .....}{n}\$......(2)As \$v=u+at\$ then drift velocity \$v_d\$ is \$v_d= \dfrac{v_1+v_2+v_3+....+v_n}{n}\$\$v_d = \dfrac{(u_1+u_2+....+u_n) +a({\tau}_1+ {\tau}_2+....)}{n}\$\$v_d = 0+ \dfrac{a({\tau}_1+ {\tau}_2+ {\tau}_3+....+ {\tau}_n)}{n}\$ ( since initially velocity is zero)\$v_d= 0+ a \tau\$\$v_d= - \dfrac{eE}{m} \tau\$According to drift velocity expression, the relaxation time is the time interval between successive collisions of an electron on increasing temperature, the electrons move faster and more collisions occur quickly. Hence, relaxation time decreases with increase in temperature which implies that drift velocity also decreases with temperature.

Q: Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area \$\displaystyle 1.0\times { 10 }^{ -7 }{ m }^{ 2 }\$ carrying a current of \$1.5\ A\$. Assume that each copper atom contributes roughly one conduction electron. The density of copper is \$\displaystyle 9.0\times { 10 }^{ 3 }{ kgm }^{- 3 }\$ and its atomic mass is \$63.5\ amu\$.

\$\textbf{Step 1: No. of conduction electron per unit volume}\$Molar mass of copper \$M = 63.5\$ gram \$= 63.5 \times 10^{-3}\ kg \$ Density of copper \$ \rho = 9 \times 10^{3}\ Kg/m^3 \$\$\text{No. of copper atoms per unit volume:}\$ \$N=\$ No. of moles in unit volume \$\times\$ No. of atoms in 1 mole\$(N_A)\$ \$....(1)\$\$\text{No. of moles in unit volume}\$ \$=\dfrac{\text{Mass of unit volume}}{\text{Mass of one mole}}\$ \$=\dfrac{\text{Density}}{\text{Molar Mass}}\$ \$=\dfrac{\rho}{M}\$Therefore, Using Equation \$(1)\$, We get: \$N = \dfrac{\rho}{M} \times N_{A} \$ \$\text{Where }{N_A = 6.023 \times 10^{23}} \$ \$\Rightarrow\ \ N = \dfrac{9 \times 10^3 \times 6.023 \times 10^{23}}{63.5 \times 10^{-3}} \$ \$= 8.54 \times 10^{28} m^{-3}\$ According to question one copper atom contributes one conduction electron.So, No. of conduction electron per unit volume \$=\$ No. of copper atoms per unit volume \$\therefore \ \ n = N = 8.54 \times 10^{28} m^{-3}\$ \$\textbf{Step 2: Drift velocity} \$Given: Current \$I = 1.5\ A \$, Cross sectional area \$A = 1 \times 10^{-7}\ m^2 \$ We know that, \$I = neAv_{d} \$ So, Drift Velocity \$v_{d} = \dfrac{I}{neA} \$ \$\Rightarrow\ \ v_{d} = \dfrac{1.5\ A}{8.5 \times 10^{28} \times (1.6 \times 10^{-19}C) \times (10^{-7}m^{-2})} \$ \${= 1.1 \times 10^{-3}\ m/s} \$Hence, option B is correct.

Q: Explain the term 'drift velocity' of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of 'drift velocity'.

The average velocity of all the free electrons in the conductor is called the drift velocity of free electrons of the conductor. When a conductor is connected to a source of emf an electric field is established in the conductor, such that \$\displaystyle E=\frac{V}{L}\$When \$V\$= potential difference across the conductor and \$L\$=length of the conductor The electric field exerts an electrostatics force '\$-Ee\$' on each free electron in the conductor The acceleration of each electron is given by \$\displaystyle \bar{a}=-\frac{e\vec{E}}{m}\$Where, \$e\$=electric charge on the electron and \$m\$=mass of electron Acceleration and electric field are in opposite directions, so the electrons attain a velocity in addition to thermal velocity in the direction opposite to that of electric field. \$\displaystyle \vec{v}_{d}=\frac{c\vec{E}}{m}\tau\$...............(i)\$\displaystyle E=\frac{-V}{L}\$...........(ii)Where \$\tau\$=relaxation time between two successive collision Let \$n\$=number density of electrons in the conductor N0. of free electrons in the conductor =\$nAL\$Total charge on the conductor , \$q=nALe\$Time taken by this charge to cover the length \$L\$ of the conductor,\$\displaystyle t=\frac{L}{v_{d}}\$Current \$\displaystyle I=\frac{q}{t}\$\$\displaystyle =\frac{nALe}{L}\times v_{d}\$\$=nAev_{d}\$Using equation (i) and (ii), we get that \$\displaystyle I=nAe\times\left(-\frac{e(-V)}{mL}\tau\right)\$\$\displaystyle =\left(\frac{ne^{2}A}{mL}\tau\right)V\$

Q: A steady current flows in a metallic conductor of non-uniform cross-section. Which of the following quantities is constant along the conductor?

When a steady current flows through a metallic conductor of non-uniform cross-section, then drift velocity\$ V_d = \dfrac {I}{enA} \$ or \$ V_d \,\infty \,\dfrac {1} {A} \$\$ E = \dfrac {I}{ \sigma A } \$ or \$ E\, \infty\, \dfrac {1} {A} \$Both \$V_d\$ and \$E\$ change with \$A\$, only current \$I\$ remains constant.

Q: Two wires each of radius of cross section \$r\$ but of different materials are connected together end to end (in series). If the densities of charge carriers in the two wires are in the ration \$1:4\$, the drift velocity of electrons in the two wires will be in the ratio:

Let \$n_1\$ and \$n_2\$ be the densities of charge carriers in the two wires, \$V_{d_{1}}\$ and \$V_{d_{2}}\$ be the drift velocities of charge carriers in the two wires,\$A\$ be the area of cross section,\$e\$ is the charge on electron and\$I\$ is the current flowing through both the wiresHence, \$I=n_1eAV_{d_{1}}= n_1eAV_{d_{2}}\$\$\therefore n_1 V_{d_1} = n_2V_{d_2}\$\$\therefore \dfrac{V_{d_1}}{V_{d_2}} = \dfrac{n_2}{n_1}\$\$\therefore \dfrac{V_{d_1}}{V_{d_2}} = \dfrac{4}{1}\$

Q: Drift speed of electrons, when \$1.5 A\$ of current flows in a copper wire of cross section \$5 mm2\$, is \$v\$. If the electron density in copper is \$9 \times 10^{28}/m^3\$ the value of \$v\$ in \$mm/s\$ is close to (Take charge of electron to be \$=1.6 \times 10^{-19}C\$)

\$I = neAv_d\$\$\Rightarrow v_d = \dfrac{I}{neA} = \dfrac{1.5}{9\times 10^{28} \times 1.6\times 10^{-19} \times 5 \times 10^{-6}}\$\$=0.02mm/s\$

Q: Two wires \$A\$ and \$B\$ of the same material, having radii in the ratio \$1 : 2\$ and carry currents in the ratio \$4 : 1\$. The ratio of drift speed of electrons in \$A\$ and \$B\$ is

Current flowing through the conductor, \$I = n e v A\$. Hence\$\displaystyle \dfrac{4}{1} = \dfrac{nev_{d_1}\pi (1)^2}{nev_{d_2}(2)^2} \$or\$\displaystyle \dfrac{v_{d_1}}{v_{d_2}} = \dfrac{4 \times 4}{1} = \dfrac{16}{1}\$Hence correct answer is option A.

Q: There are \$8.4\times 10^{22}\$ free electrons per \$cm^{3}\$ in copper, the current in the wire is \$0.21\ A\$ (\$e=1.6\times 10^{-19}C\$). Then the drifts velocity of electrons in a copper wire of \$1\:mm^{2}\$ cross section will be:-

Number of free electrons \$n = 8.4\times 10^{22} \times 10^6 \$ per \$m^3 = 8.4\times 10^{28}\$ per \$m^3\$Current in the wire \$I = 0.21\$ ACross-section of the wire \$A = 1 mm^2 = 10^{-6} m^2\$Thus drift velocity of electrons \$v=\dfrac{I}{neA}=\dfrac{0.21}{(8.4\times 10^{28})\times (1.6\times 10^{-19})\times (10^{-6})}=1.56 \times 10^{-5} m/s\$

Physics

formula

Current density in terms of drift velocity

j = n q v_{d}

where

n

is the number density (number per unit volume) of charge carriers each of charge

q

, and

v_{d}

is the drift velocity of the charge carriers.

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There are $8.4 \times 1 0^{22}$ free electrons per $c m^{3}$ in copper, the current in the wire is $0.21 A$ ( $e = 1.6 \times 1 0^{- 19} C$ ). Then the drifts velocity of electrons in a copper wire of $1 m m^{2}$ cross section will be:-

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formula

Current density in terms of drift velocity

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Interpretation from Drift Velocity

Drive an expression for drift velocity of free electrons in a conductor.

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0×10−7m2 carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0×103kgm−3 and its atomic mass is 63.5 amu.

The electric field E, current density J and conductivity σ of a conductor are related as

Explain the term 'drift velocity' of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of 'drift velocity'.

A steady current flows in a metallic conductor of non-uniform cross-section. Which of the following quantities is constant along the conductor?

Two wires each of radius of cross section r but of different materials are connected together end to end (in series). If the densities of charge carriers in the two wires are in the ration 1:4, the drift velocity of electrons in the two wires will be in the ratio:

Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires

Drift speed of electrons, when 1.5A of current flows in a copper wire of cross section 5mm2, is v. If the electron density in copper is 9×1028/m3 the value of v in mm/s is close to (Take charge of electron to be =1.6×10−19C)

Two wires A and B of the same material, having radii in the ratio 1:2 and carry currents in the ratio 4:1. The ratio of drift speed of electrons in A and B is

There are 8.4×1022 free electrons per cm3 in copper, the current in the wire is 0.21 A (e=1.6×10−19C). Then the drifts velocity of electrons in a copper wire of 1mm2 cross section will be:-

The electric field E, current density J and conductivity $σ$ of a conductor are related as

Two wires each of radius of cross section $r$ but of different materials are connected together end to end (in series). If the densities of charge carriers in the two wires are in the ration $1 : 4$ , the drift velocity of electrons in the two wires will be in the ratio:

Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross section $5 m m 2$ , is $v$ . If the electron density in copper is $9 \times 1 0^{28} / m^{3}$ the value of $v$ in $m m / s$ is close to (Take charge of electron to be $= 1.6 \times 1 0^{- 19} C$ )

Two wires $A$ and $B$ of the same material, having radii in the ratio $1 : 2$ and carry currents in the ratio $4 : 1$ . The ratio of drift speed of electrons in $A$ and $B$ is

There are $8.4 \times 1 0^{22}$ free electrons per $c m^{3}$ in copper, the current in the wire is $0.21 A$ ( $e = 1.6 \times 1 0^{- 19} C$ ). Then the drifts velocity of electrons in a copper wire of $1 m m^{2}$ cross section will be:-