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A plane EM wave travelling along z-direction is described by $E=E_{0} \sin (k z-\omega t) \hat{i}$ and $B=B_{0} \sin (k z-\omega t) \hat{j}$ Show that i) the average energy density of the wave is given by $u_{a v}=\frac{1}{4} \epsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$ ii) the time-averaged intensity of the wave is given by $I_{a v}=\frac{1}{2} c \epsilon_{0} E_{0}^{2}$
A plane EM wave travelling along z-direction is described by $E=E_{0} \sin (k z-\omega t) \hat{i}$ and $B=B_{0} \sin (k z-\omega t) \hat{j}$ Show that
i) the average energy density of the wave is given by $u_{a v}=\frac{1}{4} \epsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$
ii) the time-averaged intensity of the wave is given by $I_{a v}=\frac{1}{2} c \epsilon_{0} E_{0}^{2}$
CBSE | Class 12 | Electromagnetic Waves | NCERT Exemplar | Physics
A plane EM wave travelling along z-direction is described by $E=E_{0} \sin (k z-\omega t) \hat{i}$ and $B=B_{0} \sin (k z-\omega t) \hat{j}$ Show that
i) the average energy density of the wave is given by $u_{a v}=\frac{1}{4} \epsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$
ii) the time-averaged intensity of the wave is given by $I_{a v}=\frac{1}{2} c \epsilon_{0} E_{0}^{2}$

i) The energy density due to electric field $E$ is given as $\mathrm{uE}=1 / 2 \varepsilon_{0} \mathrm{E}^{2}$

The energy density due to magnetic field $B$ is givena as $\mathrm{uB}=1 / 2 \mathrm{~B}^{2} / \mu_{0}$

The average energy density of the wave is given by the relation,
$u_{a v}=\frac{1}{4} \epsilon_{0} E_{0}^{2}+\frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$

ii) We know that $c=1 / \sqrt{\mu}_{0} \varepsilon_{0}$

The time-averaged intensity of the wave is given as

$I_{a v}=\frac{1}{2} c \epsilon_{0} E_{0}^{2}$

i

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