# Convection Current by hcj

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```									Steady Electric Currents There are two types of electric current caused by the motion of free charges : I) convection current II) conduction current. Some basics : What is current ? To make a current flow in a medium, you have to push (apply a force F) on the charges. How fast the charges move (i.e current) depends on the nature of the material medium. The current density J is proportional to the force per unit charge or E. This is Ohm's Law in point form J = E where E is the force per unit charge,  is a property of the material medium called conductivity, in S/m, where S is called Siemen. (Another term for conductivity is mho, ohm spells backward as it is the opposite of resistivity) A perfect dielectric is a material with  =0, comparing to perfect conductor where  =  NB that inside a perfect conductor, E = 0 (always) for the electrostatic situation. For a perfect conductor where a current is flowing, E is also 0 !! as E = J/. In both of these cases where E = 0, we treat the connecting wires in a circuit as equipotential.

For a normal conductor, where  is large, E is often assumed as negligible, when a current is flowing. Convection Current Current which occurs in vacuum, gas and does not satisfy Ohm’s Law Definition of current Current through a given area is defined as the electric charge passing through the area per unit time, I = dQ/dt One amp is equal to charge being transferred through the area at a rate of one coulomb per second

X -section area s m2

Electron flow

U t

I = Q/t , where Q = Nq s ut (N is the number of carriers per unit volume) = Nq s ut/t = vS uy

where uy is the velocity of the moving electric charge and v = Nq is the volume free charge density (free charge per unit volume) Define the current density J at a given point as the current through a unit normal area at that point J = I /S = vuy In general J = vu The current I is the convection current in ampere and J is the convection current density in amperes/m2. If the current density is not normal to the surface, then I = J. S So the total current through a prescribed surface S is given by  J.dS
S

Current I through S is the flux of the vector current density J Conduction Current (Current Flowing In A Conductor, One Type of Charge Carrier Flow) When an electric field E is applied to a metallic conductor, conduction current occurs due to drift motion of electrons (charge carriers).

As electrons move, they encounter resistance. The average drift velocity ud is proportional to the applied electric field intensity E ud = -eE where e is the electron mobility in m2/V.s J = -eud = -eeE (where e = Nq is a -ve quantity since q for electrons is –1.6 x 10-19 C and J is a positive quantity) Therefore, J = E  = siemens per meter (S/m). or mho/m

Where = -ee is a macroscopic constitutive parameter of the medium called conductivity (a positive quantity)

J = E is also known as Ohm’s Law in point form, as it is valid at every point in a space and  can be a function of space coordinates. J = conduction current density (A/m2) The current density J in a conductor due to a uniform E is a constant The reciprocal of conductivity is called resistivity in ohm-meters (.m) Conduction Current (Current Flowing In A Semiconductor, Two Types of Charge Carrier Flow) In a semiconductor, the current flow is due to the movement of both electrons and holes. The holes are positive charge carriers and therefore the hole drift velocity uh is in the same direction of E uh = +hE (m/s) where h is the hole mobility

J = huh = hhE where h = Nq is a +ve quantity since q for holes for holes is also 1.6 x 10-19 C Thus the total conduction current density is J = Je + Jh = -eeE + hhE J = -eeE + hhE

Or J = E where  (S/m) is the overall conductivity for the semiconductor medium with  = -ee+ hh. (NB e is the electrons charge density, a negative value) Reason for electrons moving in a constant drift velocity Note that the charges inside the conductor upon experiencing a force due to E should accelerate, and the current will keep on growing !! Why this is not happening ? as by Ohms ' Law, the current J is a constant instead ? The answer is that the charges in the conductor is experiencing frequent collision as they pass down the wire, and after each collision, their velocity is reduced. The electrons accelerate at first and then stop when it collide with an atom and then accelerate again. So we take the average velocity and therefore we can imagine that the electrons move with a constant drift velocity ud instead. This constant drift velocity also applies to electrons moving into a bend.!! Why ? The above argument applies to the case of holes flowing in a semiconductor medium. Our study here thus concerns with steady DC current only i.e where charges are flowing with in a constant drift velocity

Resistance of a conducting material (0) Resistance is defined as V/I Given a conductor of uniform cross section S. A voltage V is applied to the conductor at the 2 ends Direction of E – same as positive charge direction E is uniform, where E = V/l Since the conductor has uniform cross section, the total current I =  J.dS = JS
S

Voltage applied here V volts 0 volts

S l

J E

J = I/S Substituting in J = E, where J and E are in the same direction I/S =  v/l

R = V/I = l/S R = l/S  = resistivity (ohm/m) This formula R = l/S is applicable only for homogeneous conductor with uniform cross section. If the conductor is of non-uniform cross section, then
 E.dl R = V/I = 

 E.ds

General Procedure For Resistance Calculation a) Choose an appropriate coordinate system for the given geometry Assume a potential difference V0 between conductor terminals Find electric field intensity E within conductor If material is homogeneous, solve V using Laplace’s Equation 2V = 0, then obtain E using E = - V Find the total current I   J.ds   E.ds S is the x-sectional area where
S S

b)

c) d)

e)

I flows f) Find resistance R by taking the ratio V0/I

For the calculation of resistance, when do you use the l l  formula R  and when you cannot use it ? S S The main thing to look for is to find out whether the E and therefore J is uniform (constant) and having the same direction as the surface area. If E is uniform, then the l simple formula R  can be used. S How can one generally deduce if the E field is uniform ? Of course, one can calculate derive E using Laplace's Equation and see how it varies in space. Often one need to quickly deduce if E is uniform without using Laplace's Equation. One way to check is ask the following questions i) Are the ends of the conductor flat surfaces ? ii) Are the ends of the conductor parallel to each other ? iii) Put a plane parallel to the ends of the conductor and passing through the side conductor. Would the crosssectional area of the conductor be the same for all such planes ? If the answer is yes to all the three questions, then the l simple formula R  can be used. S Uniform Field E We mentioned that if a uniform field E exist, then the l formula R  can be used. Let's us now prove that for a S flat homogeneous conducting material which satisfy the 3

requirements mentioned above, the electric field intensity E is uniform

Apply V0 to flat surface

l

V=0 here

E = constant

Applyying Laplace's Equation

d 2V 0 2 dx V = C1 x + C2
Boundary Conditions When X = L, V = 0 When X = 0, V = V0  C2 = V0

dV  C1 dx

 0 = C1L + C2  V0 = C1 0 + C2  C1 = -V0/L

V

V0 x  V0 L

Therefore E = -dv/dx ax = V0/L ax , a constant l Example when E is a constant (Use R  ) S

Example when E is a function of distance r

Use first principle, R 

SE.ds

  E.dl
L

Resistance Calculation Based on Capacitance We recall that the capacitance C can be calculated based on field quantities i.e Q SD.ds SE.ds C   V   E.dl   E.dl
L L

When the dielectric medium is lossy, leakage current can occurs and this means that the dielectric is not a perfect insulator. The insulator will have a small but non zero conductivity. Ohm's Law J = E ensures that the field lines for J and E will be in the same direction in an isotropic medium. The resistance between the conductors is 1

E E V  L .dl  L .dl R   I  J.ds  E.ds
S S

where the line and surface integrals are taken over the same L and S as those for the capacitance calculation. If you multiply R and C you will get   E.dl SE.ds L R C   E.dl  E.ds
S L

  This equation holds if  and  are the properties of a homogeneous con ducting medium. This equation is useful because if the capacitance between any two conductors is known, the resistance can be obtained directly from the ratio /
RC 

Example

2 C/m b ln a b ln   a ohm/m Therefore, for a coaxial cable R  C 2

We know that C for a coaxial cable is

Try using the first principle; R  same answer.

SE.ds

  E.dl
L

you will get the

Joule’s Law From the physical point of view, we know that there is a heating effect due to the transformation of electric energy in to heat in every current field Let us now derive a useful expression known as Joule’s Law in point form, which can help us calculate power dissipation in a conductor quickly. Power in watts is defined as the rate of change of energy W (in joules) or Force times distance /time In a conductor, let charge carrier density be N (number of carriers per unit volume), and each carrier has a charge Q. Let their individual drift velocities be ud. The electric force F on each charge is QE. The work done by the electrostatic force when moving one charge during a time interval t is equal to w = F.dl = QE.ud t The work done W in moving all the N v charges inside a small volume v is W = QE.ud t x N v Since NQ = , the work done W in moving all the Nv charges inside a small volume v is W = E. udt v = E.J tv Joules (since J = ud )

If we divide this expression by t , we get the desired power in a small volume v i.e the electric power that is lost to heat in the small volume v . P = W/t = E.J v
rate of energy dissipation in volume v

If we further divide this expression by v , we get the desired power per unit volume i.e the electric power density that is lost to heat. Electric power density wp = P/v = E.J
The is a point function which is valid for every point in the conductor medium

Hence, total power dissipated in the whole volume of the conductor is Joules Law P   E.Jdv
V
P is in watts

where

wp =

E.J

= [E]2

For a conductor with uniform cross section, dv = dSdl P =  E.dl  J .dS  VI
L S

P = I2R, since V=IR

Joules’ law in circuit theory

As a result of the frequent collisions of the electrons, the work done by the electrical force is converted to heat in the conductor.

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