Question

The electric field \((E)\) and magnetic field \((B)\) of an electromagnetic wave passing through vacuum are given by
\[ E=E_0\sin(kx-\omega t) \] \[ B=B_0\sin(kx-\omega t) \] Then the correct statement among the following is

• A. \(E_0k=B_0\omega\)
• B. \(E_0\omega=B_0k\)
• C. \(E_0B_0=\omega k\)
• D. \(E_0B_0=\dfrac{\omega}{k}\)

Hint:

For electromagnetic waves in vacuum: \[ E_0=cB_0 \] and \[ c=\frac{\omega}{k} \] Combining both relations gives the required result.

Answer 1

The Correct Option is A

Step 1: Recall the field amplitude relation.
For an electromagnetic wave travelling in vacuum, the peak electric and magnetic fields are linked by \[ E_0 = c\, B_0 \] where $c$ is the speed of light.
Step 2: Express $c$ through the wave parameters.
The wave speed is also the ratio of angular frequency to wave number, \[ c = \frac{\omega}{k} \]
Step 3: Combine the two relations.
Substituting for $c$, \[ E_0 = \frac{\omega}{k}\, B_0 \]
Step 4: Clear the fraction.
Multiplying both sides by $k$, \[ E_0\, k = \omega\, B_0 \]
Step 5: Compare with the options.
This is exactly the relation $E_0 k = B_0 \omega$, while the other choices mismatch the way $\omega$ and $k$ pair with the amplitudes.
Step 6: State the correct statement.
Hence the correct relation is \[ \boxed{E_0 k = B_0 \omega} \]