Question

A plane electromagnetic wave having a frequency \(\nu = 23.9\ \mathrm{GHz}\) propagates along the positive z-direction in free space. The peak value of the Electric Field is \(60\ \mathrm{V/m}\). Which among the following is the acceptable magnetic field component in the electromagnetic wave ?

• A. \(\vec{B} = 2\times 10^{-7}\sin(1.5\times 10^{2}\,x + 0.5\times 10^{11}\,t)\,\hat{j}\)
• B. \(\vec{B} = 60\sin(0.5\times 10^{3}\,x + 0.5\times 10^{11}\,t)\,\hat{k}\)
• C. \(\vec{B} = 2\times 10^{-7}\sin(0.5\times 10^{3}\,z + 1.5\times 10^{11}\,t)\,\hat{i}\)
• D. \(\vec{B} = 2\times 10^{-7}\sin(0.5\times 10^{3}\,z - 1.5\times 10^{11}\,t)\,\hat{i}\)

Hint:

Find \(B_0 = E_0/c\), \(\omega = 2\pi\nu\) and \(k = \omega/c\), then pick the option travelling along +z, i.e. of the form \(\sin(kz-\omega t)\).

Answer 1

The Correct Option is D

Peak magnetic field: \(B_0 = E_0/c = 60/(3\times 10^{8}) = 2\times 10^{-7}\ \mathrm{T}\).
Angular frequency: \(\omega = 2\pi\nu = 2\pi\times 23.9\times 10^{9} \approx 1.5\times 10^{11}\ \mathrm{rad/s}\).
Wave number: \(k = \omega/c = 1.5\times 10^{11}/(3\times 10^{8}) = 500 = 0.5\times 10^{3}\ \mathrm{m^{-1}}\).
A wave moving along +z has argument \(kz - \omega t\), so the function must contain \(z\) with a minus sign before \(\omega t\).
Options (1) and (2) contain x, so they are not for propagation along z; option (2) also has amplitude 60 (that is E, not B). Option (3) has \(kz+\omega t\) which travels along -z. Option (4) has the correct amplitude, k, \(\omega\) and the sign \(kz-\omega t\).
Answer: (4) \(\vec{B} = 2\times 10^{-7}\sin(0.5\times 10^{3}\,z - 1.5\times 10^{11}\,t)\,\hat{i}\)