An em wave going through vacuum is described by `E=E_0sin(kx-omegat)`
`B=B_0sin(kx-omegat)`

Options

A. `E_0B_0=k`
B. `E_0 omega=B_0k`
C. `E_0k=B_0omega`
D. None of these

Answer

c

Explanation

Suppose a plane em wave is
propagating along x-axis. Cosider a rectangular
path ABCD in x-y plane, fig. The electric
field `vecE` is parallel to the y-axis and magnetic field
`vecB` is parallel to z-axis.

Here, `oint vecE.vec(dl)=int_a^bvecE vec(dl)+int_b^cvecE vec(dl)`
`+int_c^dvecE vec(dl)+int_d^avecE vec(dl)`
`=0+E_(x_2)l+0E_(x_1)(-l)`
`=E_0l[sin(kx_2-omegat)-sin(kx_1-omegat)].......(i)`
Magnetic flux linked with rectangle abcd will be
`phi_B=intdphi_B=int_(x_1)^(x_2) B_(x) ldx`
`=int_(x_1)^(x_2) B_0[sin(kx-omegat)]l dx`
`=(B_0l)/k[cos(kx_2-omegat)-cos(kx_1-omegat)]`
`:. (dphi_B)/(dt)=-(B_0I)/k`
`x[-sin(kx_2-omegat)(-omega)+sin(kx_1-omegat)(-omega)]`
`=-(B_0lomega)/k[sin(kx_2-omegat)-sin(kx_1-omegat)]......(ii)`
Using Faraday's law of electromagnetic induction,
we have `oint vecE.vec(dl)=-(dphi_B)/(dt)`
Putting the values from (i) and (ii), we get
`E_0=B_0omega//k or kE_0=B_0omega`