Q.Show that the average value of radiant flux density $S$ over a single period $T$ is given by $S = \dfrac{1}{2c\mu_0}E_0^2$.

The key idea is that the instantaneous Poynting vector $S = \frac{1}{\mu_0} E \times B$ gives the power per unit area in an electromagnetic wave. For a plane wave, $E$ and $B$ are perpendicular and related by $B = E/c$.

1. For a sinusoidal wave, $E = E_0 \sin(\omega t)$ and $B = B_0 \sin(\omega t)$, with $B_0 = E_0 / c$. The instantaneous magnitude is: $S(t) = \frac{1}{\mu_0} E(t) B(t) = \frac{1}{\mu_0} E_0 B_0 \sin^2(\omega t) = \frac{E_0^2}{c \mu_0} \sin^2(\omega t).$ …

The key idea is that radiant flux density $S$ — the power per unit area carried by an electromagnetic wave — is given by the magnitude of the Poynting vector $\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{B})$. For a plane wave, $\mathbf{E}$ and $\mathbf{B}$ are perpendicular, in phase, and related by $E = cB$. Since both fields oscillate sinusoidally, the instantaneous flux density oscillates at twice the wave frequency. The average over a full period picks out the constant term, leaving half the peak value.

Let’s walk through it step by step.

1. Write the instantaneous fields. For a plane electromagnetic wave propagating along the $x$-axis, with the electric field along $y$ and magnetic field along $z$, we have:

The amplitudes are related by $E_0 = c B_0$, and $c = 1/\sqrt{\mu_0 \varepsilon_0}$.

2. Write the instantaneous Poynting vector. The Poynting vector is $\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{B})$. For our fields, $\mathbf{E} \times \mathbf{B}$ points along $x$ (the direction of propagation), and its magnitude is:

Using $B_0 = E_0 / c$, this becomes:

3. Average over one period. The time average of $\sin^2(\omega t)$ over a full period $T = 2\pi/\omega$ is $\frac{1}{2}$. This is a standard result: $\langle \sin^2(\omega t) \rangle = \frac{1}{T} \int_0^T \sin^2(\omega t)\, dt = \frac{1}{2}.$ …

Method: Deriving a Time-Averaged Wave Quantity from Its Instantaneous Form

Use this for any "show that the average value of [X] is [formula]" derivation involving a quantity built from oscillating $E$ and/or $B$ fields.

Steps

Step 1: Write the instantaneous form of the quantity first, without averaging anything yet.

For the Poynting vector, $S(t) = \frac{1}{\mu_0}E(t)B(t)$, with both fields sinusoidal and in phase.

Step 2: Reduce to a single field amplitude using the fixed $E_0$–$B_0$ relation.

Write the instantaneous expression purely in terms of one amplitude (pick $E_0$ or $B_0$ and stay consistent) times $\sin^2(\text{phase})$.

Step 3: Apply the standard time-average of $\sin^2$ over one full period.

$\langle \sin^2(\omega t)\rangle = \frac{1}{T}\int_0^T \sin^2(\omega t)\,dt = \frac{1}{2}$ …