Questions: Poynting Vector and Electromagnetic Energy Flow

Question 1

A resistive wire carries a steady DC current. According to Poynting vector analysis, where does the electromagnetic energy actually flow to deliver power to the resistor?

• A. Through the wire, carried by the conduction electrons moving along it
• B. Radially inward from the surrounding field into the wire, where it is converted to Joule heat
• C. Outward from the wire in all directions, dissipating into free space
• D. Along the surface of the wire in the direction of current flow

Answer: (B)

In a current-carrying wire, E points along the wire (driving the current) and B circles the wire (from Ampère's law). The Poynting vector S = (1/μ₀)E × B therefore points radially inward — energy flows from the surrounding electromagnetic field into the wire, where it is deposited as Joule heat. The wire is the sink, not the channel. Option A is the intuitive but wrong answer: conduction electrons carry charge, not energy. The energy delivery is through the external field.

Question 2

A plane electromagnetic wave travels in the +z direction with E pointing in the +x direction and B in the +y direction. What is the direction of the Poynting vector?

• A. +x direction, along the electric field
• B. +y direction, along the magnetic field
• C. +z direction, along the wave propagation
• D. Radially outward from the wave source

Answer: (C)

S = (1/μ₀)E × B = (1/μ₀)(x̂ × ŷ) = (1/μ₀)ẑ. The cross product of x̂ and ŷ is +ẑ, confirming that the Poynting vector points in the direction of wave propagation. This is a consistency check: electromagnetic energy must flow in the direction the wave travels. S is always perpendicular to both E and B, which is why it naturally aligns with propagation.

Question 3

The Poynting theorem, ∂u/∂t + ∇·S = −J·E, expresses a local conservation law for electromagnetic energy, analogous to a continuity equation.

• T. True
• F. False

Answer: True

The Poynting theorem has exactly the structure of a continuity equation: the rate of change of energy density (∂u/∂t) plus the divergence of energy flux (∇·S) equals the source/sink term (−J·E, the negative of work done on charges). If ∇·S > 0, energy is flowing out of the region. If J·E > 0, the field is doing work on charges and losing field energy. This is the electromagnetic analog of charge conservation ∂ρ/∂t + ∇·J = 0.

Question 4

In a DC circuit, the electric field inside a resistive wire points radially outward from the wire's axis, which is what drives the Poynting vector inward.

• T. True
• F. False

Answer: False

The electric field inside a resistive wire points along the wire in the direction of current flow (from high to low potential), not radially outward. It is this longitudinal E combined with the azimuthal B (which circles the wire per Ampère's law) that produces a cross product pointing radially inward. A radially outward E would give a Poynting vector with no inward component — it would be circumferential, not inward.

Question 5

Explain why electromagnetic energy in a DC circuit flows through the empty space surrounding the wire rather than through the wire itself.

Model answer: The energy is carried by the electromagnetic field, not by the electrons. The battery establishes an electric field along the circuit and a magnetic field around the current-carrying wires. The Poynting vector S = (1/μ₀)E × B points radially inward toward the wire everywhere along its length, meaning the field delivers energy from the surrounding space into the conductor where it is dissipated as heat. The electrons inside the wire are the mechanism for completing the circuit and maintaining the fields, but the energy transport is in the external field.

This result, though counterintuitive, is required by the Poynting theorem. Circuit theory gives the right answer for total power (P = IV) but says nothing about where the energy travels spatially. Field theory shows the energy flows in the field region, not in the wires. This reconciles with Joule heating: the inward-flowing field energy is converted to thermal energy inside the resistor at exactly the rate P = I²R predicted by circuit analysis.