 # A sinusoidal electromagnetic wave is propagating in vacuum in the + William Collins 2021-12-15 Answered
A sinusoidal electromagnetic wave is propagating in vacuum in the $+z-direction$. If at a particular instant and at a certain point in space the electric field is in the $+x-direction$ and has magnitude $4.00\frac{V}{m}$, what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?
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The Poynting vector of the energy rate in the electromagnetic wave in vacuum is given by equation (32.28) in the form
1) $\stackrel{\to }{S}=\frac{1}{{\mu }_{0}}\stackrel{\to }{E}×\stackrel{\to }{B}$
Equation(1) shows the vector product between the electric field and the magnetic field. So, to determine the direction of the wave we apply the right-hand rule where the wave comes from the opposite direction. The electric field and the magnetic field are perpendicular to each other and to the direction on the propagation.
Your index finger represents the direction of the electric field $+x$ - direction and your thump is the direction of the propagation $+z$ - direction, so your middle finger that represents the direction of magnetic field is in the direction of $+y$ - axis.
$+y-direction$
The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation (32.4) in the form
${B}_{max}=\frac{{E}_{max}}{c}$
Where c is the speed of the light. Now, plug the values for ${E}_{max}$ and c to get ${B}_{max}$
${B}_{max}=\frac{{E}_{max}}{c}=\frac{4.0\frac{V}{m}}{3×{10}^{8}\frac{m}{s}}=1.33×{10}^{-8}T$
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Magnitude of magnetic field
$B=\frac{E}{c}$
$=\frac{4.00\frac{V}{m}}{3×{10}^{8}\frac{m}{s}}$
$B=1.33×{10}^{-8}$ in the $+y$ direction.