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Chapter 19: Problem 42

An electromagnetic wave going through vacuum is described by \(E=E_{0} \sin (k x-\omega t) ; B=B_{0} \sin (k x-\omega t)\) Which of the following equation is true? (A) \(E_{0} k=B_{0} \omega\) (B) \(E_{0} \omega=B_{0} k\) (C) \(E_{0} k_{0}=\omega k\) (D) None of these

Short Answer

Expert verified
The correct equation is (B) \(E_{0} \omega = B_{0} k\).

Step by step solution

01

Recall the relationship between E and B fields in vacuum

In a vacuum, the electric field (E) and magnetic field (B) magnitudes are related by the formula: \(c = \frac{E}{B}\) where c is the speed of light in vacuum.
02

Recall the relationship between wave number and wavelength

In any wave, the wave number (k) and wavelength (位) are related by the formula: \(k = \frac{2\pi}{\lambda}\)
03

Recall the relationship between angular frequency and speed of light

For any electromagnetic wave, the angular frequency (蠅) and speed of light (c) are related by the formula: \(\omega = k c\)
04

Substitute the relationships found in Steps 1-3 into the original equations

Using the relationships from Steps 1, 2, and 3, we can rewrite the given equations as follows: \(E = E_{0} \sin (kx - \omega t)\) \(B = B_{0} \sin (kx - \omega t)\) Using the relationship from step 1, we now have: \(c = \frac{E_{0}}{B_{0}}\)
05

Use the relationship from step 3 to find the correct equation

The relationship from step 3 tells us that \(\omega = k c\). Substituting this into the equation from step 4, we have: \(\omega = k \frac{E_{0}}{B_{0}}\) Now, rearrange this equation to get: \(E_{0} \omega = B_{0} k\) Thus the correct answer is choice (B), which states: \(E_{0} \omega=B_{0} k\)

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