Question 43: EM waves

Question

A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $\mathrm{E}_y=9.3 \mathrm{Vm}^{-1}$. Then, the magnetic field vector of the wave at that point is

Accepted Answer

For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light, $$c \approx 3 	imes 10^8 \, \mathrm{m/s},$$ through the equation
$$ E = cB. $$
Here's how to determine the magnetic field:
We are given the electric field magnitude: $$E=9.3 \, \mathrm{V/m}.$$
Use the relationship to find the magnetic field magnitude:
$$    B = \frac{E}{c} = \frac{9.3}{3 	imes 10^8} \, \mathrm{T}.    $$
Calculating the value gives:
$$    B \approx 3.1 	imes 10^{-8} \, \mathrm{T}.    $$
Since the electric field is along the $$y$$ direction and the wave propagates along the $$+x$$ direction, by the right-hand rule, the magnetic field is oriented along the $$z$$ direction.
Thus, the magnetic field vector is
$$ \mathrm{B}_z = 3.1 	imes 10^{-8} \, \mathrm{T}, $$
which corresponds to Option C.

Answer

$\mathrm{B}_z=1.55 	imes 10^{-8} \mathrm{~T}$

Answer

$\mathrm{B}_z=6.2 	imes 10^{-8} \mathrm{~T}$

Answer

$\mathrm{B}_z=3.1 	imes 10^{-8} \mathrm{~T}$

Answer

$\mathrm{B}_z=9.3 	imes 10^{-8} \mathrm{~T}$

EM waves

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is \mathrm{E}_y=9.3 \mathrm{Vm}^{-1}. Then, the magnetic field vector of the wave at that point is

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