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Chapter 15: Problem 2165

An electric charge oscillating with a frequency of 1 kilo cycles/s can radiates electromagnetic waves of wavelength (A) \(100 \mathrm{~km}\) (B) \(200 \mathrm{~km}\) (C) \(300 \mathrm{~km}\) (D) \(400 \mathrm{~km}\)

Short Answer

Expert verified
The wavelength of the radiated electromagnetic waves emitted by an electric charge oscillating with a frequency of 1 kilo cycles/s can be found using the formula \(c = f \cdot \lambda\), where 'c' is the speed of light, 'f' is the frequency, and '\(\lambda\)' is the wavelength. By substituting the given values, we find the wavelength to be \(300~km\). Thus, the correct answer is (C) 300 km.

Step by step solution

01

Write down the given information and formula

We are given the frequency (f) of the oscillating electric charge as 1 kilo cycles/s or 1000 cycles/s. So, \(f = 1000~Hz\) Also, we know that the speed of light (c) is approximately \(3 \cdot 10^8~m/s\). The formula that relates frequency, wavelength, and the speed of light is: \(c = f \cdot \lambda\) where 'c' is the speed of light, 'f' is the frequency, and '\(\lambda\)' is the wavelength.
02

Solve for wavelength

We know the frequency (f) and the speed of light (c), so we can find the wavelength by rearranging the formula: \(\lambda = \frac{c}{f}\) Now, plug in the values of 'c' and 'f': \(\lambda = \frac{3 \cdot 10^8~m/s}{1000~Hz}\)
03

Calculate the wavelength

Now, calculate the wavelength: \(\lambda = \frac{3 \cdot 10^8}{1000}\) \(\lambda = 3 \cdot 10^5~m\) Since 1 km = 1000 m, \(\lambda = \frac{3 \cdot 10^5~m}{1000} = 300~km\)
04

Compare with the options and choose the correct one

Comparing the calculated value of the wavelength with the given options, we can see that the correct answer is: (C) 300 km

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Most popular questions from this chapter

Electromagnetic wave is produced by oscillating electric and magnetic fields \(E^{-}\) and \(B^{-}\). Choose only the incorrect statement from the following (A) \(\mathrm{E}^{-}\) is perpendicular to \(\mathrm{B}^{-}\). (B) \(E^{-}\) is perpendicular to the direction of propagation of the wave (C) \(\mathrm{B}^{-}\) is perpendicular to the direction of propagation of the wave (D) \(E^{-}\) is parallel to \(\mathrm{B}^{-}\)

What is the wave length of range of electromagnetic waves? (A) \(10^{-8} \mathrm{~m}\) to \(10^{15} \mathrm{~m}\) (B) \(10^{-15} \mathrm{~m}\) to \(10^{8} \mathrm{~m}\) (C) \(10^{-15} \mathrm{~m}\) to \(10^{15} \mathrm{~m}\) (D) \(10^{8} \mathrm{~m}\) to \(10^{15} \mathrm{~m}\)

A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum \(p\) and energy \(E\), then (A) \(p=0, E=0\) (B) \(p \neq 0, E \neq 0\) (C) \(p \neq 0, E=0\) (D) \(p=0, E \neq 0\)

An electromagnetic wave going through vacuum is described by $E=E_{0} \sin (k x-\omega t)\( then \)B=B_{0} \sin (k x-\omega t)$ then (A) \(E_{0} B_{0}=\operatorname{cok}\) (B) \(E_{0} k=B_{0} \omega\) (C) \(\mathrm{E}_{0} \mathrm{~m}=\mathrm{B}_{0} \mathrm{k}\) (D) none of these

The amplitude of the magnetic field part of an electromagnetic wave in vacuum is \(\mathrm{Bm}=510 \mathrm{nT}\). Then the amplitude of the electric part of the wave is (A) \(1.53 \times 10^{11} \mathrm{~V} / \mathrm{m}\) (B) \(1.53 \mathrm{~V} / \mathrm{m}\) (C) \(1.53 \times 10^{2} \mathrm{~V} / \mathrm{m}\) (D) \(1.53 \times 10^{8} \mathrm{~V} / \mathrm{m}\)
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