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: Understanding the Concept:
In an electromagnetic (EM) wave, the electric field (E) and magnetic field (B) are mutually perpendicular and also perpendicular to the direction of wave propagation. Their magnitudes are related by the speed of the wave.

Step 2: Key Formula or Approach:
1. For an EM wave propagating in vacuum, the ratio of the magnitudes of the electric field and magnetic field at any instant is equal to the speed of light, \(c\). \[ \frac{E}{B} = c \] This relationship also holds for their amplitudes: \[ \frac{E_0}{B_0} = c \] 2. For any wave described by an equation of the form \(\sin(kx - \omega t)\), the speed of the wave (\(v\)) is given by the ratio of the angular frequency (\(\omega\)) to the wave number (\(k\)): \[ v = \frac{\omega}{k} \] For an electromagnetic wave in vacuum, this speed is \(c\).

Step 3: Detailed Explanation:
From the first principle, we have: \[ \frac{E_0}{B_0} = c \] From the second principle, the speed of this EM wave is: \[ c = \frac{\omega}{k} \] By substituting the expression for \(c\) into the first equation, we get the relationship between the amplitudes and the wave parameters: \[ \frac{E_0}{B_0} = \frac{\omega}{k} \]

Step 4: Final Answer:
The correct relationship is \(\frac{E_0}{B_0} = \frac{\omega}{k}\).