Answer
$B_x = (1.67\times 10^{-8}~T)~sin[(1.00\times 10^6~m^{-1})z+(3.00\times 10^{14}~s^{-1}) t]$
Work Step by Step
$E_y = (5.00~V/m)~sin[(1.00\times 10^6~m^{-1})z+\omega t]$
We can find $B_m$:
$B_m = \frac{E_m}{c} = \frac{5.00~V/m}{3.0\times 10^8~m/s} = 1.67\times 10^{-8}~T$
We can find $\omega$:
$\omega = k~c = (1.00\times 10^6~m^{-1})(3.0\times 10^8~m/s) = 3.00\times 10^{14}~s^{-1}$
We can write an expression for the magnetic field component:
$B_x = (1.67\times 10^{-8}~T)~sin[(1.00\times 10^6~m^{-1})z+(3.00\times 10^{14}~s^{-1}) t]$
