Question
Derive the equation of energy density for electromagnetic wave.
✓
Answer
$\rightarrow $ Energy density : "Energy stored per unit volume is called Energy density."
An electromagnetic wave contains Electric field and magnetic field both.
$\rightarrow $ Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
$\rightarrow $ Energy $-$ density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
$\rightarrow $ Total Energy $-$ density associated with $EM$ wave,
$\varrho =\varrho_{ E }+\varrho_{B}$
$\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
$\rightarrow $ 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.
$\rightarrow $ 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 $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
$\rightarrow $ Substituting both these values in eq. $(1),$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)}$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2$
$\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2$
$\rightarrow $ In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.
An electromagnetic wave contains Electric field and magnetic field both.
$\rightarrow $ Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
$\rightarrow $ Energy $-$ density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
$\rightarrow $ Total Energy $-$ density associated with $EM$ wave,
$\varrho =\varrho_{ E }+\varrho_{B}$
$\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
$\rightarrow $ 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.
$\rightarrow $ 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 $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
$\rightarrow $ Substituting both these values in eq. $(1),$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)}$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2$
$\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2$
$\rightarrow $ In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.
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