The magnitude of the electric field of a plan | vedclass

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Question

(C) For a plane electromagnetic wave travelling in free space,the relationship between the magnitude of the electric field $E$ and the magnetic field

Vedclass · Accepted Answer

$E \sqrt{\mu_0 \varepsilon_0}$

Vedclass · Answer

(C) For a plane electromagnetic wave travelling in free space,the relationship between the magnitude of the electric field $E$ and the magnetic field $B$ is given by $E = cB$,where $c$ is the speed of light in free space.The speed of light $c$ in free space is related to the permeability $\mu_0$ and permittivity $\varepsilon_0$ by the formula $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.Substituting this expression for $c$ into the relationship $E = cB$,we get:$E = \left( \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \right) B$Rearranging the equation to solve for the magnitude of the magnetic field $B$:$B = E \sqrt{\mu_0 \varepsilon_0}$.

The magnitude of the electric field of a plane electromagnetic wave travelling in free space is $E$. If $\mu_0$ and $\varepsilon_0$ are respectively permeability and permittivity of the free space,then the magnitude of magnetic field of the wave is

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