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In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \mathrm{~Hz}$ and amplitude $48 \mathrm{~V} \mathrm{~m}^{-1}$.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the $\mathbf{E}$ field equals the average energy density of the $B$ field. [c $\left.=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1} \cdot\right]$
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the $\mathbf{E}$ field equals the average energy density of the $B$ field. [c $\left.=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1} \cdot\right]$
Solution:
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Verified Answer
Here $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, v=2 \times 10^{10} \mathrm{~Hz}, \mathrm{E}_0=48 \mathrm{~V} / \mathrm{m}$ $\lambda=?, \mathrm{~B}_0=?, \overrightarrow{\mathrm{u}_{\mathrm{E}}}=?, \overrightarrow{\mathrm{u}_{\mathrm{B}}}=$ ?
(a) $\because \mathrm{c}=v \lambda$
$$
\Rightarrow \lambda=\frac{\mathrm{c}}{v}=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2} \mathrm{~m}
$$
(b) $\frac{\mathrm{E}_0}{\mathrm{~B}_0}=\mathrm{c} \Rightarrow \mathrm{B}_0=\frac{48}{3 \times 10^8}=16 \times 10^{-8} \mathrm{~T}$.
(c) Energy density due to electric field,
$$
\mathrm{u}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^2
$$
(i) and Energy density due to magneticfield,
$$
\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \mu_{\mathrm{o}} \mathrm{B}^2
$$
We have $E=c B$ and $C^2=\frac{1}{\mu_o \varepsilon_o}$
Sub: these values in eqn (i)
$$
\begin{aligned}
\mathrm{U}_{\mathrm{E}} &=\frac{1}{2} \varepsilon_{\mathrm{o}}(\mathrm{cB})^2=\frac{1}{2} \varepsilon_{\mathrm{o}}\left(\frac{1}{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}\right) \mathrm{B}^2 \\
&=\frac{1}{2 \mu_{\mathrm{o}}} \mathrm{B}^2=\mathrm{U}_{\mathrm{B}} .
\end{aligned}
$$
(a) $\because \mathrm{c}=v \lambda$
$$
\Rightarrow \lambda=\frac{\mathrm{c}}{v}=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2} \mathrm{~m}
$$
(b) $\frac{\mathrm{E}_0}{\mathrm{~B}_0}=\mathrm{c} \Rightarrow \mathrm{B}_0=\frac{48}{3 \times 10^8}=16 \times 10^{-8} \mathrm{~T}$.
(c) Energy density due to electric field,
$$
\mathrm{u}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^2
$$
(i) and Energy density due to magneticfield,
$$
\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \mu_{\mathrm{o}} \mathrm{B}^2
$$
We have $E=c B$ and $C^2=\frac{1}{\mu_o \varepsilon_o}$
Sub: these values in eqn (i)
$$
\begin{aligned}
\mathrm{U}_{\mathrm{E}} &=\frac{1}{2} \varepsilon_{\mathrm{o}}(\mathrm{cB})^2=\frac{1}{2} \varepsilon_{\mathrm{o}}\left(\frac{1}{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}\right) \mathrm{B}^2 \\
&=\frac{1}{2 \mu_{\mathrm{o}}} \mathrm{B}^2=\mathrm{U}_{\mathrm{B}} .
\end{aligned}
$$
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