jangyasinniTeacher

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To find the electron density when the convection current density and the velocity are given, we can use the formula for current density which relates these quantities. The current density (J) (which here is given as the convection current density) can be related to the charge density (rho) and the velocity (v) by the equation:

[ J = rho v ]

where:

– (J) is the current density (in this case, 120 units, though the units need specification, e.g., A/m²),

– (rho) is the charge density (in C/m³, which is equivalent to the electron density times the fundamental charge, given that we are considering electron density),

– (v) is the drift velocity of the electrons (given as 5 m/s).

The question seems to aim to find the electron density, which we’ll denote as (n). Since (rho = n times e), where (e) is the elementary charge ((1.602 times 10^{-19}) C), the equation becomes:

[ J = n cdot e cdot v ]

Rearranging for (n), we have:

[ n = frac{J}{e cdot v} ]

Given:

– (J = 120) A/m² (assuming the “units” refers to Amperes per square meter for current density),

– (v = 5) m/s,

c

Explanation: The convection current density is given by J = ρe x v. To get ρe, put J =

120 and v = 5. ρe = 120/5 = 24 units