Electromagnetics — Joule heating power density in a conductor For a conducting medium with electrical conductivity σ subject to an electric field of magnitude E, what is the heat generated per unit volume per second (power density W, in W/m^3)?

Introduction / Context:
Joule heating quantifies the conversion of electrical energy to heat within a conductor. The local power density depends on the electric field and the material’s conductivity. This relation is fundamental in power electronics, electromagnetic heating, and safety calculations.

Given Data / Assumptions:

• Ohmic, linear, isotropic conductor.
• Current density follows J = σ E.
• Steady-state fields; displacement current effects neglected for heating.

Concept / Approach:
Electromagnetic power density delivered to charge carriers is W = J · E. For an Ohmic conductor, J = σ E, giving W = σ E^2. Units check: σ (S/m) times E^2 (V^2/m^2) yields W/m^3, consistent with a volumetric heating rate.

Step-by-Step Solution:
Start from J · E formulation for power density.Substitute J = σ E.Obtain W = σ E^2.

Verification / Alternative check:
At the macroscopic level, total heat Q̇ equals ∫ J · E dV. In a uniform field region, this integrates to the product of W and volume, matching circuit-level I^2 R losses (with appropriate geometry relations).

Why Other Options Are Wrong:
σ E lacks correct units (W/m^3). E^2/σ applies to fixed current density scenarios incorrectly transposed. (σ E)^{0.5} and σ^2 E are dimensionally inconsistent for power density.

Common Pitfalls:

• Confusing current-controlled (I^2 R / volume) and field-controlled expressions; both reduce to W = σ E^2 with Ohm’s law.
• Mixing SI units and cgs quantities.

Final Answer:
W = σ E^2