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Thursday, 12 September 2019

Convection Current Density Derivation

Convection Equation Derivation:


When the e beam passes through the electric current, it induces a current. This current is known as Convection Current.

The electron velocity, charge density, current density and axial electric field are given by the equations.

v = v0 + v1 e(jωt-γz) ------------------- (a)
ρ = ρ0 + ρ1 e(jωt-γz) ------------------- (b)
J = -J0 + J1 e(jωt-γz) ------------------- (c)
Eγ= -E1e(jωt-γz) -------------------- (d)

For a small signal, the current density of electron is given by,
J = ρv

Substitute for ρ and v from eqn (a) and (b)


J = 0 + ρe(jωt-γz))(v0 + v1 e(jωt-γz))

Since the signal is very small, the term e(jωt-γz) is neglected.

Therefore, J = (ρ0 + ρ1)(v0 + v1)
J = ρ0 v0 + ρ0 v1+ ρ1v0 + ρ1 v1

Let ρ1 v1 = 0
Therefore, J1 = ρ0 v0 + ρ0 v1+ ρ1v0

Substitute for J from eqn (c), the above equation becomes,

-J0 + J1 e(jwt-rz) = ρ0 v0 + ρ0 v1+ ρ1v0

Let, ρ0 v0 = -J0 and

J1 = ρ0 v1+ ρ1v0 ---------------------- (e)

The axial electric field affects the velocity of electrons as,

dv/dt = -e/mE1 e(jwt-rz)  ----------- (1)

Substitute for v from eqn (a)

dv/dt = ∂/ ∂t(v0 + v1 e(jωt-γz)) + ∂z/ ∂t. ∂/∂z(v0 + v1 e(jωt-γz))
dv/dt = v1 e(jωt-γz)(jω - γ ∂z/ ∂t)

Let, ∂z/ ∂t = v0
Therefore, dv/dt = v1 e(jωt-γz)(jω - γ v0) -------------------- (2)

Sub eqn (2) in (1)

v1 e(jwt-γz)(jω - γ v0) = -e/mE1 e(jωt-γz) 

so, v1(jω - γ v0) = -e/mE1

v1= -E1e/m (jω - γ v0) -------------------- (3)

eqn (3) defines the change in electron velocity with respect to electric field, E1.



By law of conservation of electric charge, the continuity equation can be written as,

∇. J = -ρ/ ∂t ----------------------- (4)
∇ = ∂/ ∂x + ∂/ ∂y + ∂/ ∂z

Consider only the z component

∇ = ∂/ ∂z
∇. J = -J/ ∂z

Substitute for J from eq (c), we get

∇. J = - ∂/ ∂z(-J0 + J1 e(jωt-γz))
∇. J = - γJ1 e(jωt-γz) ------------------------- (5)

From eq (b)

-ρ/ ∂t = - ∂/ ∂t(ρ0 + ρe(jωt-γz))
-ρ/ ∂t = - ρ1 jω e(jωt-γz) ----------------------- (6)

Substitute eq (5) and (6) in eq (4)

∇. J = -ρ/ ∂t
- γJ1 e(jωt-γz) = - ρ1 jω e(jωt-γz)

Therefore - γJ1 = - ρ1 jω

ie, ρ1 = γJ1/ jω --------------------------------- (7)
eqn (7) defines the change in charge density ρ with respect to current density J.

Consider eqn (e)

J1 = ρ0v1 + ρ1v0

Sub. for v1 from eqn (3) and ρ1 from eqn (7)

J1 = ρ0(-E1e/m (jω - γ v0)) + (γJ1/ jω) v0

Thus, J1 = -ρ0eE1ω/m(jω - γv0)(ω + γv0j)

Multiplying Numerator and Denominator by v0

J1 = -ρ0eE1ω v0/m(jω - γv0)(ω + γv0j) v0

We have,

J0 = -ρ0 v0
Be = ω0 v0

Therefore,

J1 = J0eE1Be/m(jω - γv0)(ω + γv0j)

Multiplying numerator and denominator with j

J1 = J0eE1Bej/jm(jω - γv0)(ω + γv0j)
J1 = J0eE1Bej/m(-ω - γv0j)(ω + γv0j)
= J0eE1Bej/m((jω)2 - 2ω γv0j + (γv0)2)
= J0eE1Bej/m(jω - γv0)2
= J0eE1Bej/ v02m(jω/ v0 - γ) 2
= J0eE1Bej/ v02m(jBe - γ) 2    [Since ω/ v0 = Be]

The uniform velocity v0 is given by the eqn,

v0 = (2e v0/m)
v02 = 2e v0/m
e/ m v02 = 1/2 v0

Therefore J1 = J0E1Bej/ 2v0(jBe - γ) 2

When the electric field is uniform, J1 = i

Hence, i = J0E1Bej/ 2v0(jBe - γ) 2

This is the equation for convection current. This equation is also known as ‘electronic equation’ as it determines the current induced by the electric field.


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