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(C) The electric field is given by $E_{y} = 900 \sin \omega(t - x/c)$,so the amplitude of the electric field is $E_{0} = 900 \, V/m$.The electric forc

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$10: 1$

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(C) The electric field is given by $E_{y} = 900 \sin \omega(t - x/c)$,so the amplitude of the electric field is $E_{0} = 900 \, V/m$.The electric force on a charge $q$ is $F_{E} = qE_{0}$.The magnetic force on a charge $q$ moving with velocity $v$ is $F_{B} = qvB_{0}$,where $B_{0}$ is the amplitude of the magnetic field.In an electromagnetic wave,the relationship between the amplitudes of the electric and magnetic fields is $E_{0} = cB_{0}$,which implies $B_{0} = E_{0}/c$.Substituting $B_{0}$ into the expression for magnetic force: $F_{B} = qv(E_{0}/c)$.The ratio of electric force to magnetic force is:$\frac{F_{E}}{F_{B}} = \frac{qE_{0}}{qv(E_{0}/c)} = \frac{c}{v}$.Given $c = 3 	imes 10^{8} \, m/s$ and $v = 3 	imes 10^{7} \, m/s$,the ratio is:$\frac{F_{E}}{F_{B}} = \frac{3 	imes 10^{8}}{3 	imes 10^{7}} = 10$.Thus,the ratio is $10: 1$.

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