Question

The electric field E associated with a progressive electromagnetic wave is given by E = E\(_0\)sin(kx - \(\omega\)t). If B\(_0\) is the amplitude of the magnetic field associated with the wave, then

• A. \(\frac{E_0}{B_0} = \frac{\omega}{k}\)
• B. \(\frac{E_0}{B_0} = \frac{\omega^2}{k^2}\)
• C. \(\frac{E_0}{B_0} = \frac{k}{\omega}\)
• D. \(\frac{E_0}{B_0} = \frac{k^2}{\omega^2}\)

Hint:

A simple way to remember the core relationships for EM waves is \(E = cB\) and \(c = f\lambda\). Remembering that \(\omega = 2\pi f\) and \(k = 2\pi/\lambda\) allows you to quickly derive \(c = \omega/k\). Combining these gives the answer.

Answer 1

The Correct Option is A

Step 1: Concept Definition:
In an electromagnetic (EM) wave, the electric (E) and magnetic (B) fields are mutually perpendicular. They are also perpendicular to the direction the wave travels. The magnitudes of these fields are linked by the wave's speed.

Step 2: Core Principles:
1. In a vacuum, the ratio of the electric field magnitude (E) to the magnetic field magnitude (B) at any moment equals the speed of light (c): \[ \frac{E}{B} = c \] This also applies to their amplitudes: \[ \frac{E_0}{B_0} = c \] 2. For a wave defined by \(\sin(kx - \omega t)\), its speed (v) is the angular frequency (\(\omega\)) divided by the wave number (\(k\)): \[ v = \frac{\omega}{k} \] For an EM wave in a vacuum, \(v = c\).

Step 3: Detailed Derivation:
From the first principle: \[ \frac{E_0}{B_0} = c \] From the second principle, the wave speed is: \[ c = \frac{\omega}{k} \] Substituting the expression for \(c\) into the first equation establishes the link between amplitudes and wave parameters: \[ \frac{E_0}{B_0} = \frac{\omega}{k} \]

Step 4: Conclusion:
The governing relationship is \(\frac{E_0}{B_0} = \frac{\omega}{k}\).