KCET · Physics · Electromagnetic Waves
Suppose that the electric field amplitude of electromagnetic wave is \(E_{0}=120 \mathrm{NC}^{-1}\) and its frequency \(f=50 \mathrm{MHz}\). Then, which of the following value is incorrectly computed?
- A Magnetic field amplitude is \(400 \mathrm{nT}\).
- B Angular frequency of EM wave is \(\pi \times 10^{8} \mathrm{rad} / \mathrm{s}\).
- C Propagation constant (angular wave number) is \(2.1 \mathrm{rad} / \mathrm{m}\).
- D Wavelength of EM wave is \(6 \mathrm{~m}\).
Answer & Solution
Correct Answer
(C) Propagation constant (angular wave number) is \(2.1 \mathrm{rad} / \mathrm{m}\).
Step-by-step Solution
Detailed explanation
Given, \(E_{0}=120 \mathrm{NC}^{-1}\)
\(f=50 \mathrm{MHz}=50 \times 10^{6} \mathrm{~Hz}\)
(a) magnetic field amplitude,
\(\begin{aligned}
B_{0} &=\frac{E_{0}}{c}=\frac{120}{3 \times 10^{8}}=40 \times 10^{-8} \mathrm{~T} \\
&=400 \mathrm{n} \mathrm{T}
\end{aligned}\)
(b) Angular frequency, \(\omega=2 \pi f=2 \pi \times 50 \times 10^{6}\)
\(=\pi \times 10^{8} \mathrm{rad} / \mathrm{s}\)
(c) Propagation constant,
\(k=\frac{\omega}{c}=\frac{\pi \times 10^{8}}{3 \times 10^{8}}=1.047 \mathrm{rad} / \mathrm{m}\)
(d) Wavelength, \(\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{50 \times 10^{6}}=\frac{30}{5}=6 \mathrm{~m}\)
\(f=50 \mathrm{MHz}=50 \times 10^{6} \mathrm{~Hz}\)
(a) magnetic field amplitude,
\(\begin{aligned}
B_{0} &=\frac{E_{0}}{c}=\frac{120}{3 \times 10^{8}}=40 \times 10^{-8} \mathrm{~T} \\
&=400 \mathrm{n} \mathrm{T}
\end{aligned}\)
(b) Angular frequency, \(\omega=2 \pi f=2 \pi \times 50 \times 10^{6}\)
\(=\pi \times 10^{8} \mathrm{rad} / \mathrm{s}\)
(c) Propagation constant,
\(k=\frac{\omega}{c}=\frac{\pi \times 10^{8}}{3 \times 10^{8}}=1.047 \mathrm{rad} / \mathrm{m}\)
(d) Wavelength, \(\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{50 \times 10^{6}}=\frac{30}{5}=6 \mathrm{~m}\)
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