Question 46: EM waves

Question

A plane EM wave is propagating along $$x$$ direction. It has a wavelength of $$4 \mathrm{~mm}$$. If electric field is in $$y$$ direction with the maximum magnitude of $$60 \mathrm{~Vm}^{-1}$$, the equation for magnetic field is :

Accepted Answer

To find the correct equation for the magnetic field of the plane electromagnetic wave given its parameters, we can use a couple of known relationships from electromagnetism.
Firstly, the wavelength ($ \lambda $) of the wave is given as $4 \mathrm{~mm} = 4 	imes 10^{-3} \mathrm{~m}$. The speed of light (and all electromagnetic waves in vacuum) is $ c = 3 	imes 10^8 \mathrm{~m/s} $. Using these values, we can find the frequency ($ f $) of the wave using the relationship:
$ c = \lambda f $
$ f = \frac{c}{\lambda} = \frac{3 	imes 10^8}{4 	imes 10^{-3}} = 75 	imes 10^9 \mathrm{~Hz} $
The angular frequency ($ \omega $) which appears in wave equations is related to the frequency by $ \omega = 2\pi f $. However, in this context, what we need is the wave vector ($ k $), which defines how the phase of the wave changes with space. The wave vector $ k = \frac{2\pi}{\lambda} $. So, for this wave, $ k = \frac{2\pi}{4 	imes 10^{-3}} = \frac{\pi}{2} 	imes 10^3 \, \mathrm{m}^{-1} $.
Knowing that the electric field ($ \mathbf{E} $) and magnetic field ($ \mathbf{B} $) are related as $ E = cB $ in a vacuum, where $ E $ and $ B $ are the magnitudes of the electric and magnetic fields, respectively, we can calculate the magnitude of the magnetic field using the provided maximum electric field magnitude ($ E = 60 \, \mathrm{Vm}^{-1} $).
$ B = \frac{E}{c} = \frac{60}{3 	imes 10^8} = 2 	imes 10^{-7} \mathrm{~T} $
Therefore, the wave equation for the magnetic field, considering it propagates in the $ x $-direction and oscillates in a direction perpendicular to both the $ x $-direction and the direction of the electric field (thus, in the $ z $-direction if $ \mathbf{E} $ is in the $ y $-direction), is:
$ \mathbf{B} = B \sin(kx - \omega t) \hat{\mathbf{k}} $
$ \mathbf{B} = 2 	imes 10^{-7} \sin \left(\frac{\pi}{2} 	imes 10^3(x - 3 	imes 10^8t)ight) \hat{\mathbf{k}} \mathrm{T} $
This is represented by Option B:
$ \mathrm{B}_z=2 	imes 10^{-7} \sin \left[\frac{\pi}{2} 	imes 10^3\left(x-3 	imes 10^8 \mathrm{t}ight)ight] \hat{\mathrm{k}} \mathrm{T} $
Therefore, the correct answer is Option B.

Answer

$$\mathrm{B}_z=2 	imes 10^{-7} \sin \left[\frac{\pi}{2}\left(x-3 	imes 10^8 \mathrm{t}ight)ight] \hat{\mathrm{k}} \mathrm{T}$$

Answer

$$\mathrm{B}_z=2 	imes 10^{-7} \sin \left[\frac{\pi}{2} 	imes 10^3\left(x-3 	imes 10^8 \mathrm{t}ight)ight] \hat{\mathrm{k}} \mathrm{T}$$

Answer

$$\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(x-3 	imes 10^8 \mathrm{t}ight)ight] \hat{\mathrm{k}} \mathrm{T}$$

Answer

$$\mathrm{B}_x=60 \sin \left[\frac{\pi}{2}\left(x-3 	imes 10^8 \mathrm{t}ight)ight] \hat{\mathrm{i}} \mathrm{T}$$

EM waves

A plane EM wave is propagating along $x direction. It has a wavelength of 4 \mathrm{~mm}. If electric field is in y direction with the maximum magnitude of 60 \mathrm{~Vm}^{-1}$, the equation for magnetic field is :

Choose the correct answer: