ELECTRIC CURRENT AND OHM'S LAW

Electron Drift Velocity
Suppose that in a conductor, the number of free electrons available per m3 of the conductor
material is n and let their axial drift velocity be v metres/second. In time dt, distance travelled would
be v x dt. If A.is area of cross-section of the conductor, then the volume is vAdt and the number of
electrons contained in this volume is vA dt. Obviously, all these electrons will cross the conductor
cross-section in time dt. If e is the charge of each electron, then total charge which crosses the
section in time dt is dq =nAev dt.
Since current is the rate of flow of charge, it is given as
= -dq = nAev dt :. I.=nAev
dt dt
Current density J = ilA =ne v amperelmetre2
Assuming a normal current density J =1.55 X 106 Alm2, n = 1029for a copper conductor
6 -19
and e =1. x 10 coulomb, we get
1.55 x 106 = 1029x 1.6X 10-19x v :. v = 9.7 X 10-5m/s = 0.58 cm/min
It is seen that contrary to the common but mistaken view, the electron drift velocity is rather
very slow and is independent of the current flowing and the area of the conductor.
N.H.Current density i.e., the current per unit area. is a vector quantity. It is denoted by the symbol J .
-->
Therefore. in vector notation, the relationship between current I and J is:
-->--> -->
I = J. a [where a is the vector notation for area 'a']
For extending the scope of the above relationship. so that it becomes applicable for area of any shape, we
write:
f
--> -->
J = J .d a
Themagnitudeof the currentdensitycan. therefore.be writtenas J.a.
Example 1.1. A conductor material has a free-electron density of J{j4 electrons per metrl.
When a voltage is applied, a constant drift velocity of 1.5 X ]0-2 metre/second is attained by the
electrons. If the cross-sectional area of the material is 1 cm2,calculate the magnitude of the current.
Electronic charge is 1.6 x ]0-19coulomb. (Electrical Engg. Aligarh Muslim University 1981)
Solution. The magnitude of the current is
= nAev amperes
n = 1024;A = 1cm2 = 10-4m2
e = 1.6 x 10-19C ; v = 1.5X 10-2m/s
= 1024X 10-4x 1.6 X 10-19x 1.5X 10-2=0.24 A
Here,
..
1.2. Charge Velocity and Velocity of Field Propagation
The speed with which charge drifts in a conductor is called the velocity of charge. As seen fromabove. its value is quite low, typically fraction of a metre per second.
However. the speed with which the effect of e.m.f. is experienced at all parts of the conductor
resulting in the flow of current is called the velocit.v of propagation of electri8calfield. It is independent
of current and voltage and has high but constant value of nearly 3 x 10 mls.
Example 1.2. Find the velocity of charge leading to 1 A current which flows in a copper
conductor of cross-section 1 em2and length 10 km. Free electron dellSityof copper =8.5.x 1028per
m3. How long will it take the electric charge to travelfrom one end of the conductor to the other.
Solution. i =neAv or v =ih,('A
.. v = 1/(85 y I02X) x 1.6 x 10 1'1X (I X 10-4) = 7.35 X 10-7 mls =0.735 Jlmls
Time taken by the charge 10lravd conductor length of 10 kIn is
t = distance _ 10x 103 =1.36 x 1010S
velocity 7.35 x 10-7
Now. I year::: 365 x 24 x 3600 = 31,536.000 s
10
t = 1.36 x 10 /31.536.000 = 431 years
1.3. The Idea of Electric Potential
In Fig. 1.1is shown a simple voltaic cell. It consists of copper plate (known as anode) and a zinc
rod (I.e. cathode) immersed in dilute sulphuric acid (H2S04) contained in a suitable vessel. The
chemical action taking place within the cell causes the electrons to be removed from Cu plate and to
be deposited on the zinc rod at the same time. This transfer of electrons is accomplished through the
agency of the diluted H2S04 which is known as the electrolyte. The result is that zinc rod becomes
negative due to the deposition of electrons on it and the Cu plate becomes positive due to the removal
of electrons from it. The large number of electrons collected on the zinc rod is being attracted by
anode but is prevented from returning to it by the force set up by the chemical action within the cell.
Conventional Direction
of Current
--0 --0 --0. Directionof '"
E-l-e-ctronFlow0- .'"
1 1
Water
,'-:' -- - --- --- -- - -- - - -- - - - - :-,
r
Direction
of Flow
1
Cu Zn
Fig. 1.1. Fig. 1.2
But if the two electrodes are joined by a wire externally. then electrons rush to the anode thereby
equalizing the charges of the two electrodes. However. due to the continuity of chemical action. a
continuous difference in the number of electrons on the two electrodes is maintained which keeps up
a continuous flow of current through the external circuit. The action of an electric cell is similar to
that of a water pump which. while working. maintains a continuous flow of water i.e. water current
through the pipe (Fig. 1.2).