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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

The oscillating electric and magnetic field vectors of an electromagnetic waves far away from source are oriented along (A) Mutually perpendicular direction and differ in phase by \(90^{\circ}\) (B) Mutually perpendicular and in same phase (C) In same direction and in same phase (D) In same direction and differ in phase by \(90^{\circ}\)

If the relative permeability and dielectric constant of a given medium are equal to \(\mu_{\mathrm{r}}\) and \(\mathrm{K}\) respectively, then the refractive index of the medium is equal to (A) \(\sqrt{\left(\mu_{\mathrm{T}} \mathrm{K}\right)}\) (B) \(\sqrt{\left(\mu_{1} E_{0}\right)}\)

Unit of \(\mu_{0} \mathrm{C}\) is same as that of (A) current (B) resistance (C) electric charge (D) velocity

The frequency of light wave of wavelength \(5000 \mathrm{~A}\) is \(\mathrm{Hz}\) (A) \(6 \times 10^{14}\) (B) \(1.5 \times 10^{-2}\) (C) \(1.5\) (D) \(6 \times 10^{1}\)

The wavelength of \(\mathrm{x}\) rays is of the order of (A) \(10^{-3} \mathrm{~m}\) (B) \(10^{-5} \mathrm{~m}\) (C) \(10^{-10} \mathrm{~m}\) (D) \(10^{-12} \mathrm{~m}\)
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