The electric field in a plane electromagnetic wave is given by \( E_z=60\sin\left(0.5\times10^{3}x+1.5\times10^{11}t\right)\,\mathrm{V/m} \). What is the wavelength and magnetic field amplitude of the wave?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.26\,\mathrm{cm},\ 2\times10^{-7}\,\mathrm{T} \)
Compare the given field with the standard form of a travelling wave, \( E = E_0\sin(kx + \omega t) \), where \( k = \frac{2\pi}{\lambda} \) is the wave number. Here \( E_0 = 60\,\mathrm{V\,m^{-1}} \), \( k = 0.5\times10^{3}\,\mathrm{m^{-1}} \) and \( \omega = 1.5\times10^{11}\,\mathrm{rad\,s^{-1}} \).
Wavelength
\( \displaystyle \lambda = \frac{2\pi}{k} = \frac{2\times3.14}{0.5\times10^{3}}\,\mathrm{m} = 1.26\times10^{-2}\,\mathrm{m} = 1.26\,\mathrm{cm} \)
Magnetic field amplitude, using \( B_0 = E_0/c \) for an electromagnetic wave in vacuum:
\( \displaystyle B_0 = \frac{E_0}{c} = \frac{60}{3\times10^{8}}\,\mathrm{T} = 2\times10^{-7}\,\mathrm{T} \)
Check: \( \omega/k = 1.5\times10^{11}/500 = 3\times10^{8}\,\mathrm{m\,s^{-1}} \), the speed of light.