Question

An electromagnetic wave in a vacuum has a magnetic field with a magnitude of 1.7 x 10^-6 T. What is the intensity S of the wave?

Answer 1

In plane waves

In a propagating sinusoidal linearly polarized electromagnetic plane wave of a fixed frequency, the Poynting vector always points in the direction of propagation while oscillating in magnitude. The time-averaged magnitude of the Poynting vector is

\langle S \rangle = \frac{1}{2 \mu_0 c} E_0^2 = \frac{\varepsilon_0 c}{2} E_0^2,

where \ E_0 is the maximum amplitude of the electric field and \ c is the speed of light in free space. This time-averaged value is also called the irradiance or intensity I.
[edit] Derivation

In an electromagnetic plane wave, \mathbf{E} and \mathbf{B} are always perpendicular to each other and the direction of propagation. Moreover, their amplitudes are related according to

B_0 = \frac{1}{c}E_0,

and their time and position dependences are

E\left(t,{\mathbf r}\right) = E_0\,\cos\left(\omega\,t- {\mathbf k} \cdot {\mathbf r} \right),
B\left(t,{\mathbf r}\right) = B_0\,\cos\left(\omega\,t- {\mathbf k} \cdot {\mathbf r} \right),

where \ \omega is the frequency of the wave and \mathbf{k} is wave vector. The time-dependent and position magnitude of the Poynting vector is then

S(t) = \frac{1}{\mu_0} E_0\,B_0\,\cos^2\left(\omega t-{\mathbf k} \cdot {\mathbf r}\right) = \frac{1}{\mu_0 c} E_0^2 \cos^2\left(\omega t-{\mathbf k} \cdot {\mathbf r} \right) = \varepsilon_0 c E_0^2 \cos^2\left(\omega t-{\mathbf k} \cdot {\mathbf r} \right).

In the last step, we used the equality \varepsilon_0\,\mu_0 = {c}^{-2} . Since the time- or space-average of \cos^2\left(\omega\,t-{\mathbf k} \cdot {\mathbf r}\right) is \textstyle\frac{1}{2}, it follows that

\left\langle S \right\rangle = \frac{\varepsilon_0 c}{2} E_0^2.

E= speedlight*B

The time averaged Poynting vector is equal to epsilonfreespace*speedlight*E^2/2

but if E= speedlight*B
S= epsilonFreespace*speedlight^3*B^2 /2

check this out: http://en.wikipedia.org/wiki/Poynting_vector