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Physics · 2024 · 3 marks
CBSE 2024 · Region 3 · Set 1 · Q25
The electric field in an electromagnetic wave in vacuum is given by: E =(6 · 3 N / C )[ cos (1 · 5 rad / m ) y +(4 · 5 × 10^8 rad / s ) t ] i (a)…
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 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.