Chapter Notes: Current Electricity - Class 12 Physics Notes


Notes for Current Electricity chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

 

The motion or dynamics of charges gives rise to new effects, which we shall study in this Chapter. Charges in motion makes an electric current.
Time rate of flow of charge through a cross-section is called electric current. Suppose, a charge \Delta q passes through a given cross section or area in a short time \Delta t, then the electric current is

I = {{\Delta q} \over {\Delta t}}

More precisely, the instantaneous current at a given time t is 
   I = \mathop {\lim }\limits_{\Delta \,t \to 0} {{\Delta q} \over {\Delta t}} = {{dq} \over {dt}}
If the current is steady (i.e., it does not change with time), then the charge q flowing through is just proportional to time t, and we have

    I = {q \over t}             (for steady current)

The SI unit of current is ‘ampere’ [A]. It is one of the base SI units. One ‘ampere’ is defined on the basis of the force produced by the current carrying conductors on each other, as we shall see in the next Chapter.
The system of units is such that one ampere is equal to one coulomb per second1A = {{1C} \over {1s}}
Electric current being one of the base or fundamental quantities in SI units,
Dimension of current = [A]

Conventional Direction of Current

Conventionally, the direction of current is taken to be the direction of flow of positive charge, (i.e., the direction of the field).In a conductor (such as copper, aluminum, etc.) the current flows due to the motion of free electrons (negatively charged particles). Hence, the direction of electric current is opposite to the direction of flow of electrons.

Current in Different Situations

(a)  Due to Translatory Motion of Charges
(i)   If n particles, each having a charge q, pass through a given area in time t, the current is given by

I = {{\Delta Q} \over {\Delta t}} = {{nq} \over t}

(ii)  If n particles, each having a charge q, pass per second per unit area, the current associated with cross-sectional area S is
I = {{\Delta Q} \over {\Delta t}} = nqS

(iii) If there are n particles per unit volume, each having a charge q and moving with velocity v, the current through cross-sectional area S is
I = {{\Delta Q} \over {\Delta t}} = nqS{{\Delta x} \over {\Delta t}} = nqvS

(b)  Due to Rotatory Motion of Charge
If a point charge q is moving in a circle of radius r with speed v, then its time period T = (2\pi r/v).
So through a given cross-section (perpendicular to motion), the current is

I = {q \over t} = {q \over T} = {{qv} \over {2\pi r}}

Application  1
The current in a wire varies with time according to the relation i = a + bt2,  where current i is in ampere and time t is in seconda = 4A, b = 2 As-2.
(a) How many coulomb pass a cross-section of the wire in the time interval between t = 5 s and t = 10 s ?
(b) What constant current could transport the same charge in same time interval ?

Solution:

(a) \Delta q = \int\limits_5^{10} {i\,\,dt} = \int\limits_5^{10} {\left( {4 + 2{t^2}} \right)\,\,dt}
\left| {4t + {2 \over 3}{t^3}} \right|_5^{10} = 4\left( {10 - 5} \right) + {2 \over 3}\left( {1000 - 125} \right)  C

(b)  {I_e} = {{\Delta q} \over {\Delta t}} = {{603.33} \over {10 - 5}} = 120.67  A

Application  2
In the Bohr model of hydrogen atom, the electron is pictured to rotate in a circular orbit of radius 5 x 10-11 m, at a speed of 2.2 x 106 m/s. What is the current associated with electron motion ?

Solution:

The time taken to complete one rotation is
T = {{2\pi r} \over v}
Therefore, the current is
I = {q \over t} = {e \over T} = {{ev} \over {2\pi r}} = {{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6}} \over {2 \times 3.14 \times 5 \times {{10}^{ - 11}}}} = 1.12  mA

Current Density (\bar J)

Electric current may be distributed nonuniformly over the surface through which it passes. Hence, in order to characterize the current in greater detail, we introduce the concept of current density vector \bar J at a point, defined as follows :
(1)  The magnitude of \bar J is equal to the current per unit area surrounding that point and normal to the direction of charge flow. Thus,

               J = {{dI} \over {dS}}
(2)  The direction of \bar J is the same as the direction of velocity vector \bar v of the ordered motion of positive charge-carriers.
If the current is due to the motion of both positive and negative charges, the current density  is given by

                \bar J = {\rho _ + }{\vec u_ + } + {\rho _ - }{\vec u_ - }
where {\rho _ + }  and  {\rho _ - } are volume densities of positive and negative charge carriers, respectively, and {\vec u_ + } and {\vec u_ - } are their velocities.


In conductors, the charge is carried only by electrons. Therefore,

               \bar J = {\rho _ - }{\vec u_ - }
Here, the value of {\rho _ - } is negative. Hence the directions of \bar J and {\vec u_ - } are opposite to each other.

If at point P current \Delta I passes normally through area \Delta s as shown. Current density \bar J at P is  given by
\vec J = \mathop {lim}\limits_{\Delta s \to 0} {{\Delta I} \over {\Delta S}}\vec n       or                 \vec J = {{dI} \over {ds}}\vec n

In general, if the cross-sectional area is not normal to the current, the cross-sectional area normal to current will be dS cos q and so in this situation,

         J = {{dI} \over {dS\cos \theta }}
dI = JdScos\theta
or    dI = \vec J.d\vec S
or    I = \int {\vec J.d\vec S}

Thus, the electric current is the flux of current density.
Note the following important points about current density :
(1)  Both current I and current density \vec J have directions, by definition current density \vec J is a vector while current I is a scalar.
(2)  Though the charge density {\rho _c} is defined as charge per unit volume, it may have different values at different points, as
{\rho _c} = \mathop {\lim }\limits_{\Delta V \to 0} {{\Delta q} \over {\Delta V}}
(3)  For uniform flow of charge through a cross-section normal to it, we have
I = nqvS
\vec J = {I \over S}\vec n = \left( {nqv} \right)\vec n    (as nv = {\rho _c})
\vec J = nq{\rm{\vec v}} = {\rho _c}\vec v
(4)  For a conductor, as we shall see,
V = IR   and   R = \rho {L \over S}
If E is the electric field, we have
V = IR
or  EL = I\rho {L \over S}
or  J = {I \over S} = {1 \over \rho }E  or   \vec J = \sigma \vec E
[As conductivity, \sigma = {1 \over {resistivity,\,\,\rho }} ]

Thus in case of conductors, current density is proportional to electric field \vec E. This in turn implies that
(a)  Direction of current density \vec J is same as that of electric field \vec E
(b)  If electric field is uniform (i.e.,  \vec E = constant) current density will be constant (as  \sigma = constant)
(c)  If electric field is zero (as in electrostatics, inside a conductor), current density and hence current will be zero.

Drift Velocity   ({\vec v_d})

The drift velocity is the average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for current through it.

Free elecrons, under the influence of thermal agitation, move randomly  throughout the metallic conductor from one point to another in an irregular manner in all possible directions as shown in Fig. (A). However, if a source of potential difference is applied across the two ends of the conductor, the electrons gain velocities tending them to move from negative to positive side of the conductor. Average value of these velocities is termed as drift velocity. The drift velocity of the electrons is no doubt superimposed on their random velocity, but in fact this is the cause for net transportation of the charge along the conductor and hence results in a current flow.In absence of any electric field [Fig.(A)], the random motion of electrons does not contribute to any current.  The number of electrons crossing any plane from left to right is equal to the number of electrons crossing from right to left (otherwise metal will not remain equipotential).

When an electric field is applied, due to electric force the path of electrons in general becomes curved instead of straight lines and electrons drift opposite to the field [Fig.(B)].

Analogy with Wind

The molecules in air have random thermal velocities whose average magnitude is somewhat larger than the speed of sound (330 m/s). A pressure difference between two regions causes a net flow of molecules in one direction resulting in wind. The wind velocity (say, about 10 m/s) is much smaller than random velocity. Similarly, in a conductor, electrons have random velocities upto 106 m/s. When a potential difference is applied, electrons drift with a very small velocity (  \approx 10-4 m/s) (opposite to field) resulting in current.

Current in a Conductor

Consider a conductor of uniform cross-sectional area S. If n is the number of free electrons per unit volume, the charge per unit volume is ne. If vd is the drift velocity, electrons move a distance vd in one second. Therefore, the volume moved in one second is vdS. Thus, the current,

I = {{charge{\mkern 1mu} {\mkern 1mu} density{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\mkern 1mu} volume} \over {time}} = {{\left( {ne} \right) \times \left( {{v_d}S} \right)} \over 1} = ne{v_d}S
⸫   J = {I \over S} = ne{v_d}  or  {v_d} = {J \over {ne}}

Metals (conductors) have large number of free electrons per unit volume (  \approx 1028/m3). Therefore, the drift velocity is very small (  \approx 10-4 m/s) compared to random speed of electrons at room temperature (  \approx 106 m/s).

Application  3
The area of cross-section, length and density of a piece of a metal of atomic mass 60 gm/mole are 10-6 m2, 1 m and 5 x 103 kg/m3 respectively. Find the number of free electrons per unit volume if every atom contributes one free electron. Also find the drift velocity of electrons in the metal when a current of 16 A passes through it. Given that Avogadro’s number NA = 6 x 1023/mole and charge on an electron e = 1.6 x 10-19 C.

Solution:

According to Avogadro’s hypothesis,

{N \over {{N_A}}} = {m \over M}  so  n = {N \over V} = {N_A}{m \over {VM}} = {{{N_A}d} \over M}   (as d = {m \over V})
⸫   n = {{6 \times {{10}^{23}} \times \left( {5 \times {{10}^3}} \right)} \over {\left( {60 \times {{10}^{ - 3}}} \right)}} = 5 \times {10^{28}} /{m^3}

Now as each atom contributes one electron, the number of electrons per unit volume is also the same.

J = {I \over S} = {{16} \over {{{10}^{ - 6}}}} = 16 \times {10^6} A/{m^2}
{v_d} = {J \over {ne}} = {{16 \times {{10}^6}} \over {\left( {5 \times {{10}^{28}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}} \right)}} = 2 \times {10^{ - 3}} m/s



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