Question:
Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.
(a) Find the total current crossing the surface z = 0.1 m in the az direction.
(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.
(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.
Answer:
a. -39.8μA
b. -15.81mC/m³
c. 29.05m/s
Explanation:
Given
Density = J = −10^6z^1.5az A/m²
Region: 0 ≤ ρ ≤ 20 µm
ρ ≥ 20 µm
J = 0.
a. Total current is calculated by.
J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.
Where J = Density = -10^6 * z^1.5
ρ1 = Upper bound of ρ = 20
ρ0 = Lower bound of ρ = 0
π = 22/7
φdza = 10^-6
z = 0.1
Total current
= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6
= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6
= −39.7543477278310
= -39.8μA
b. Calculating velocity charge density at (ρv)
Density (J) = ρv * V
Where J = Density = -10^6 * z^1.5
V = 2 * 10^6
z = 0.1
Substitute the above values
-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6
ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)
ρv = -0.1^1.5/(2)
ρv = -0.015811388300841
ρv = -0.01581 --------- Approximated
ρv = -15.81mC/m³
c. Calculating Velocity
Velocity = J/V
Where Velocity Charge Density = -2000 C/m3
Where J = -10^6 * z^1.5
z = 0.15
J = -10^6 * 0.15^1.5
J = -58094.75019311125
Velocity = -58094.75019311125/-2000
Velocity = 29.047375096555625m/s
Velocity = 29.05m/s
