An electromagnetic wave passing through vacuum is described by the equations. $E = E_{0} sin (k x - ω t)$ and $B = B_{0} sin (k x - ω t)$ . Then.

E_{0} k = B_{0} ω

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E_{0} ω = B_{0} k

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E_{0} B_{0} = ω k

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E_{0} = B_{0}

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Solution

The correct option is A $E_{0} k = B_{0} ω$
Equations of an electromagnetic wave $E = E_{0} sin (k x - ω t)$ and $B = B_{0} sin (k x - ω t)$
The relation between electric field $(E_{0})$ and magnetic field $(B_{0})$ is $E_{0} = c B_{0}$ ........(i)
Velocity of light $c = v λ = \frac{ω}{2 π} λ = \frac{ω}{k}$ (where $k = 2 π / λ$ ).
Substituting this value of $c$ in equation (i), we get
$E_{0} = \frac{ω}{k} B_{0}$
or $E_{0} k = B_{0} ω$ .

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E = E_{0} s i n (k x - ω t); B = B_{0} s i n (k x - ω t)

E = E_{0} sin (k x - ω t)

B = B_{0} sin (k x - ω t)

E = E_{0} sin (k x - ω t)

E = E_{0} s i n (K x - ω t) .

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