Electric Current , Drift velocity & Deduction of Ohm’s Law

Electric Current :
Electric current is the rate of flow of electric charge across any cross-section of a conductor.

current flowing , $\large I = \frac{dq}{dt}$

Where dq is the quantity of electric charge flowing through cross-section A in time dt.

Mechanism of current flow in metallic conductors

When an external potential difference is applied across a metallic conductor then an electric field is set up within the conductor.

Applied electric field → Force on electrons → drift of electrons

Due to the externally applied electric field electrons drift with an average velocity called drift velocity.

Drift velocity  :

It is defined as the velocity with which the free electrons gets drifted towards the +ve terminal under the effect of applied electric field .

$\large \vec{v_d} = \vec{a}\tau$

$\large \vec{v_d} = -\frac{e\vec{E}}{m}\tau$

$\large v_d = \frac{e E}{m}\tau$

Where : vd = drift velocity , a = acceleration , τ  = Relaxation time , E = electric field , e = electronic charge , m = mass of electron

Relation between drift velocity & current :

Consider a conductor of length l & cross-sectional area A . If n be the number of free electrons per unit volume of the conductor .

Total charge on all the free electrons in the conductor ,

$\large q = n(Al)e$

Let constant Potential difference V be applied across the ends of conductor , the electric field set up across the conductor is $\large E = \frac{V}{l}$

Time taken by free electrons to cross the length l is

$\large t = \frac{l}{v_d}$

Current , $\large I = \frac{q}{t} = \frac{n A l e}{l/v_d}$

$\large I = n e A v_d$

Current density( $\vec{J}$):
Current density at a point, within a conductor, is the current through a unit area of the conductor, around that point, provided the area is perpendicular to the direction of flow of current at that point.

$\large J = \frac{I}{A} = n e v_d$

In vector form, $\large I = \vec{J}.\vec{A}$

Mobility (μ): Mobility of a charge carrier (like electron) is defined as the average drift velocity per unit applied electric field strength.

$\large \mu = \frac{v_d}{E}$

Electrical Resistance & Ohm’s Law

“The potential difference across a conductor is directly proportional to the current flowing through it at constant temperature”.

This fact is known as Ohm’s law. It is established by experiment.

Mathematically,

V ∝ I

$\large \frac{V}{I} = R = constant$

Obviously, this constant is what we call resistance (R)

Deduction of Ohm’s Law : 

As we know that $\large v_d = \frac{e E}{m}\tau$

$\large v_d = \frac{e V}{m l}\tau$ ,(since E = V/l)

Relation between current & drift velocity is

$\large I = n e A v_d$

$\large I = n e A (\frac{e V}{m l}\tau)$

$\large I = \frac{n e^2 A \tau}{m l}V$

$\large \frac{V}{I} =\frac{m l}{ne^2 A \tau} $

$\large \frac{m l}{ne^2 A \tau} = R = constant $ .It is known as resistance of conductor  .

$\large \frac{V}{I} =R $ . This is Ohm’s Law .

$\large R = \frac{m }{ne^2 \tau} \frac{l}{A}$

Here $\large \rho = \frac{m }{ne^2 \tau}$ is called resistivity of the material .

$ \displaystyle R = \rho \frac{l}{A} $

Where ρ is a constant which depends upon the material of the wire , it is called the resistivity of the material and has the unit ohm-meter.

Electrical Conductivity : The inverse of resistivity is known as conductivity (σ)

$ \large \sigma = \frac{1}{\rho} $ , which has units of mho/m

Conductance (G):  It is reciprocal of resistance .

$\large G = \frac{1}{R}$ , Unit is mho or siemen(S)

Relation between J , σ & E :

$\large I = n e A v_d$

$\large I = n e A (\frac{e E}{m}\tau)$

$\large I = \frac{n e^2 A \tau E}{m}$

$\large \frac{I}{A} = \frac{n e^2  \tau E}{m}$

$\large \frac{I}{A} = \frac{ E}{m/n e^2  \tau}$

$\large J= \frac{ E}{\rho}$

$\large J= \sigma E$

Variation of Resistance with temperature

$ \displaystyle R_T = R_0 [1 + \alpha (T – T_0 )] $ ; Where alpha (α) is temperature coefficient of resistance .

Solved Example : The resistance of a thin silver wire is 1.0Ω at 20°C. The wire is placed in a liquid bath and its resistance rises to 1.2 Ω. What is the temperature of the bath ? α for silver is 3.8 × 10-3/°C.

Sol : R(T) = R0 [1 + α(T-T0)]

Substituting the values, we have 1.2 = 1.0[1 + 3.8 × 10-3 (T – 20)]

0.2 = 3.8 × 10-3 (T – 20)

On Solving this, we get T = 72.6°C

Example : A wire carries a current of 2.0 A. What is the charge that has flowed through its cross-section in 1.0s ? How many electrons does this correspond to ?

Sol: q = I t = 2 × 1 = 2 C

$\large n = \frac{q}{e} = \frac{2}{1.6 \times 10^{-19}}$

= 1.25 × 1019

Solved Example : The current in a wire varies with time according to the relation i=(3.0A)+(2.0A/s)t
(a) How many coulombs of charge pass a cross-section of the wire in the time interval between t = 0 and t = 4.0s ?
(b) What constant current would transport the same charge in the same time interval?

Sol: $\large I = \frac{dq}{dt}$

$\large \int_{0}^{q}dq = \int_{0}^{4}I dt$

$\large q = \int_{0}^{4} (3 + 2 t) dt$

$\large q = [3t + 2t^2/2]_{0}^{4}$

$\large q = 12 + 16 = 28 C$

(b) $\large I = \frac{q}{t} = \frac{28}{4} = 7 A$

Solved Example : Two copper wires of the same length have got different diameters,

(a) which wire has greater resistance?

(b) greater specific resistance?

Sol: (a) For a given wire,

$\large R = \rho \frac{l}{A}$

$\large R \propto \frac{l}{A}$

So the thinner wire will have grater resistance.

(b) Specific resistance (ρ) is a material property. It does not depend on l or A.

So, both the wires will have same specific resistance.

Also Read :

→ Combination of Resistors
→ Grouping of Cells
→Concept of EMF & Terminal P.D
→Kirchhoff ‘ s Laws & Solved Examples
→Wheatstone Bridge Principle & Metre Bridge
→Ammeter and Voltmeter
→Potentiometer & its Applications
→Sensitivity of Potentiometer
→Charging of capacitors
→Discharging of Capacitors
→ Heating effect of currents

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