Question
Derive the equation of energy density for electromagnetic wave.
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Answer
→Energy density : "Energy stored per unit volume is called Energy density." An electromagnetic wave contains Electric field and magnetic field both.
→Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
→Energy-density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
→Total Energy-density associated with EM wave,
$\varrho =\varrho_{ E }+\varrho_{B}
\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
→But the magnitudes of electric field and magnetic field change as per sine or cosine functions in an EM wave.Hence, in the equation, the rms value of electric field and magnetic field is considered.
→Energy-density $ \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{ B _{\text {rms }}^2}{2 \mu_0}$
But $c ^2=\frac{1}{\mu_0 \varepsilon_0} \therefore \mu_0=\frac{1}{ c ^2 \varepsilon_0}$
and $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
→Substituting both these values in eq. (1),
$\begin{array}{l}\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)} \\\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2 \\\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2\end{array}$
→In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.
→Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
→Energy-density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
→Total Energy-density associated with EM wave,
$\varrho =\varrho_{ E }+\varrho_{B}
\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
→But the magnitudes of electric field and magnetic field change as per sine or cosine functions in an EM wave.Hence, in the equation, the rms value of electric field and magnetic field is considered.
→Energy-density $ \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{ B _{\text {rms }}^2}{2 \mu_0}$
But $c ^2=\frac{1}{\mu_0 \varepsilon_0} \therefore \mu_0=\frac{1}{ c ^2 \varepsilon_0}$
and $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
→Substituting both these values in eq. (1),
$\begin{array}{l}\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)} \\\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2 \\\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2\end{array}$
→In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.
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