If σ = 2 &mho;/m, E = 10 V/m, the conduction current density is | Electrical MCQ | Examoogle

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ElectricalElectromagnetics Field Theory

If σ = 2 ℧/m, E = 10 V/m, the conduction current density is

$5\text{ A/m}^2$

$20\text{ A/m}^2$

$40\text{ A/m}^2$

$30\text{ A/m}^2$

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalElectromagnetics Field Theory

Option B

$20\text{ A/m}^2$

Quick Summary:

Given: Conductivity σ = 2 mho/m, Electric field intensity E = 10 V/m

📐MAMath SolutionDirect Formula

📋 Given

Conductivity σ = 2 mho/m, Electric field intensity E = 10 V/m

🔢 Formula Used

Jc=σEJ_c = \sigma EJc​=σE

🔢 Step-by-Step Solution

Identify the given parameters

The problem provides conductivity $\sigma = 2\text{ mho/m}$ and electric field intensity $E = 10\text{ V/m}$ .

σ=2,E=10\sigma = 2, E = 10σ=2,E=10

State the conduction current density formula

According to Ohm's law in point form, the conduction current density $J_c$ is given by the product of conductivity and electric field intensity.

Jc=σEJ_c = \sigma EJc​=σE

Calculate the current density

Substitute the given values into the formula to find $J_c$ .

Jc=2×10=20 A/m2J_c = 2 \times 10 = 20\text{ A/m}^2Jc​=2×10=20 A/m2

✅

B is correct because multiplying the conductivity by the electric field intensity yields a conduction current density of 20 A/ $m^2$ .

Core Concepts Used

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💡 EXAM TIP

In electromagnetics, remember that $J = \sigma E$ is the microscopic form of Ohm's law, similar to $V = IR$ in circuit analysis.

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