Problem — Dimensional analysis of the Poynting vector

Problem

Verify dimensionally that the Poynting vector $\vv I=\dfrac{\vv E\times\vv B}{\mu_0}$ has the dimensions of a power per unit surface (W/m²). Use $[E]=$ V/m, $[B]=$ T, $[\mu_0]=$ T·m/A.

Solution

Setup. We substitute the units of measure into the formula and simplify, remembering that $1$ V·A $=1$ W (power = voltage times current).

Step 1 — insert the units.

$[I]=\frac{[E]\,[B]}{[\mu_0]}=\frac{(\text{V/m})\,(\text{T})}{\text{T·m/A}}$

Step 2 — the tesla cancels.

$[I]=\frac{\text{V/m}\cdot\text{A}}{\text{m}}=\frac{\text{V·A}}{\text{m}^2}$

Step 3 — use $1$ V·A $=1$ W.

$\ev{[I]=\frac{\text{V·A}}{\text{m}^2}=\frac{\text{W}}{\text{m}^2}}$

The dimensions are indeed those of an intensity (power per surface).

Links

Topics: Onde elettromagnetiche Concepts: Vettore di Poynting Skills: Analisi dimensionale