Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 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 × 106 m/s at z = 0.1 m, find rhoν there. (c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

A) The total current crossing the surface z = 0.1 m in the z^ direction is; I_tot ≈ -39.8μA

B) If the charge velocity is 2 × 10⁶ m/s, then ρv is; -15.81 mC/m³

C) If the volume charge density at z = 0.15 m is −2000 C/m³, the Charge velocity is; 29.05m/s

We are given;

Current Density; J = −10⁶z^(1.5) (z^) A/m²

Region: 0 ≤ ρ ≤ 20 µm

At ρ ≥ 20µm,  J = 0.

I_tot = J × ½((ρ1)² - (ρ0)²) × 2π × φdza.

Where;

J is current Density = −10⁶z^(1.5) A/m²

ρ1 is Upper bound of ρ = 20 µm

ρ0 is Lower bound of ρ = 0 µm

 φdza = 10⁻⁶

z = 0.1

Thus plugging in the relevant values, we have;

Total current;

I_tot = -10⁶ × 0.1^(1.5) * ½(20² - 0²) × 2 × π × 10⁻⁶

I_tot ≈ -39.8μA

ρv = J/V

where;

J is current density = −10⁶z^(1.5) A/m²

 V is charge velocity = 2 × 10⁶ m/s

z = 0.1

Thus;

 ρv = ( −10⁶ × 0.1^(1.5))/(2 × 10^6)

ρv = -¹/₂(0.1^(1.5))

ρv ≈  -0.01581 C/m³

Thus;

ρv = -15.81 mC/m³

Charge Velocity = J/V

Where;

J is current density = −10⁶z^(1.5) A/m²

V is volume charge density = -2000 C/m³

At z = 0.15;

Charge velocity = (−10⁶ × 0.15^(1.5))/(-2000)

Charge velocity ≈ 29.05m/s

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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.

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

answer:

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explanation: