CBSE 2024 · Region 3 · Set 1 · Q25 · 3 marks
The electric field in an electromagnetic wave in vacuum is given by : \[\overrightarrow{\mathrm{E}}=(6 \cdot 3 \mathrm{~N} / \mathrm{C})\left[\cos (1 \cdot 5 \mathrm{rad} / \mathrm{m}) \mathrm{y}+\left(4 \cdot 5 \times 10^{8} \mathrm{rad} / \mathrm{s}\right) \mathrm{t}\right] \hat{\mathrm{i}} \](a)Find the wavelength and frequency of the wave.(b)What is the amplitude of the magnetic field of the wave ?(c)Write an expression for the magnetic field of this wave.
The electric field in an electromagnetic wave in vacuum is given by : \[\overrightarrow{\mathrm{E}}=(6 \cdot 3 \mathrm{~N} / \mathrm{C})\left[\cos (1 \cdot 5 \mathrm{rad} / \mathrm{m}) \mathrm{y}+\left(4 \cdot 5 \times 10^{8} \mathrm{rad} / \mathrm{s}\right) \mathrm{t}\right] \hat{\mathrm{i}} \]
(a)
Find the wavelength and frequency of the wave.
(b)
What is the amplitude of the magnetic field of the wave ?
(c)
Write an expression for the magnetic field of this wave.
Marking-scheme solution
(a) \(\displaystyle \mathrm{k}=\frac{2 \pi}{\lambda}\)
\[\begin{aligned}
& \lambda=\frac{2 \pi}{K}=\frac{4 \pi}{3} \mathrm{~m}=4.18 \mathrm{~m} \\
& \omega=2 \pi v
\end{aligned}
\]
\[v=\frac{\omega}{2 \pi}=\frac{4.5 \times 10^{8}}{2 \pi} \mathrm{~Hz}
\]
(b) \(\displaystyle B_{0}=\frac{\mathrm{E}_{0}}{c}\)
\[B_{0}=\frac{6.3}{3 \times 10^{8}}=2.1 \times 10^{-8} \mathrm{~T}
\]
(c) \(\displaystyle \vec{B}=2.1 \times 10^{-8}\left[(\cos 1.5 \mathrm{rad} / \mathrm{m}) \mathrm{y}+\left(4.5 \times 10^{8} \mathrm{rad} / \mathrm{s}\right) \mathrm{t}\right] \hat{\mathrm{k}} \mathrm{T}\)
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.