The magnetic field in a plane electromagnetic wave is given by
By = (2 × 10−7) sin (0.5 × 103 x + 1.5 × 1011 t) T
The expression for the electric field will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Ez = 60 sin (5 × 102 x + 1.5 × 1011 t) V/m
In a plane electromagnetic wave, \( \vec{E} \), \( \vec{B} \) and the direction of travel are mutually perpendicular, the two fields oscillate in phase (same k and \( \omega \)), and their amplitudes are related by \( E_0 = cB_0 \).
Amplitude:
\( \displaystyle E_0 = cB_0 = (3\times10^{8}\,\mathrm{m\,s^{-1}})(2\times10^{-7}\,\mathrm{T}) \)
\( \displaystyle E_0 = 60\,\mathrm{V\,m^{-1}} \)
Check of the speed from the phase:
\( \displaystyle v = \frac{\omega}{k} = \frac{1.5\times10^{11}\,\mathrm{rad\,s^{-1}}}{0.5\times10^{3}\,\mathrm{rad\,m^{-1}}} = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
Direction: the phase \( (kx + \omega t) \) means the wave travels along the x-axis (towards −x). \( \vec{B} \) is along y, so \( \vec{E} \) must be perpendicular to both x and y, that is along z. The phase stays the same, and \( 5\times10^{2} \) is the same number as \( 0.5\times10^{3} \).
\( \displaystyle E_z = 60\sin(5\times10^{2}\,x + 1.5\times10^{11}\,t)\,\mathrm{V\,m^{-1}} \)