The electric field of an EM wave in free space is given by \( E = 10\cos(10^{7}t + kx)\,\hat{\jmath}\,\mathrm{V/m} \) where t and \( x \) are in sec and meter respectively. It can be inferred that :
(A) The wavelength \( \lambda \) is 188.4 m
(B) The wave number k is 0.33 rad/m
(C) The wave amplitude is 10 V/m
(D) The wave is propagating along \( +x \) direction
Which of the following pairs of statement are correct ?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C)
Compare the field with the standard form of a plane wave \( E = E_0\cos(\omega t + kx) \). In free space the wave travels at \( c = \omega/k \).
\( E_0 = 10\,\mathrm{V/m} \)
\( \omega = 10^{7}\,\mathrm{rad/s} \)
\( c = 3\times10^{8}\,\mathrm{m/s} \)
\( \displaystyle k = \frac{\omega}{c} = \frac{10^{7}}{3\times10^{8}} = 0.033\,\mathrm{rad/m} \)
\( \displaystyle \lambda = \frac{2\pi}{k} = \frac{2\pi c}{\omega} = 2 \times 3.14 \times 30\,\mathrm{m} = 188.4\,\mathrm{m} \)
(A) The wavelength \( \lambda \) is 188.4 m — Correct. This is \( 2\pi c/\omega \) worked out above.
(B) The wave number k is 0.33 rad/m — Incorrect. \( k = \omega/c = 0.033\,\mathrm{rad/m} \), ten times smaller.
(C) The wave amplitude is 10 V/m — Correct. The coefficient of the cosine is \( E_0 = 10\,\mathrm{V/m} \).
(D) The wave is propagating along \( +x \) direction — Incorrect. The phase \( \omega t + kx \) stays constant when \( x \) decreases as \( t \) increases, so the wave travels along the negative \( x \)-direction.
Statements (A) and (C) are correct, which is Option (4).