# obtain an expression for the energy density of an electromagnetic wave .Inthe electromagnetic wave show that the average energy density of the electric field equals the average energy density of the magnetic field?

Dear student,

The energy in an electromagnetic wave is tied up in the electric and magnetic fields. In general, the energy per unit volume in an electric field is given by:

${U}_{E}=\frac{1}{2}{\epsilon }_{0}{E}^{2}$

In a magnetic field, the energy per unit volume is:

${U}_{B}=\frac{1}{2{\mu }_{0}}{B}^{2}$

An electromagnetic wave has both electric and magnetic fields, so the total energy density associated with an electromagnetic wave is:

U= $\frac{1}{2}{\epsilon }_{0}{E}^{2}+\frac{1}{2{\mu }_{0}}{B}^{2}$

It turns out that for an electromagnetic wave, the energy associated with the electric field is equal to the energy associated with the magnetic field, so the energy density can be written in terms of just one or the other:

$U={\epsilon }_{0}{E}^{2}=\frac{1}{{\mu }_{0}}{B}^{2}$

${U}_{E}=\frac{1}{2}{\epsilon }_{0}{E}^{2}$ and ${U}_{B}=\frac{1}{2{\mu }_{0}}{B}^{2}$
$E=cB\phantom{\rule{0ex}{0ex}}E=\frac{1}{\sqrt{{\epsilon }_{0}{\mu }_{0}}}B\phantom{\rule{0ex}{0ex}}{E}^{2}=\frac{1}{{\epsilon }_{0}{\mu }_{0}}{B}^{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2}{\epsilon }_{0}{E}^{2}=\frac{{B}^{2}}{2{\mu }_{0}}\phantom{\rule{0ex}{0ex}}$
Hence the two energies are equal

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