Physics - EM Waves Question with Solution | TestHub

When undefined

Then$B = B_0\mathit{sin}$$\left(kx-\omega t\right)$

Of light in travelling along$\widehat{i}$then$\overrightarrow{B}$in either along$\widehat{j}$or$\widehat{k}$.

$\therefore$speed of light$C= \frac{E_0}{B_0} \Rightarrow B_0= \frac{E_0}{C}$

$\Rightarrow B_0= \frac{27}{3\times {10}^8}=9 \times {10}^{-8} T$

also,$\omega =2\pi \mathrm{f}=2\mathrm{\pi }\mathrm{ }\times 2\mathrm{ }\times {10}^{14}=4 \pi \times {10}^{14}$

$k=\frac{\omega }{c}=\frac{4\pi \times {10}^{14}}{3\times {10}^8}$

$=\left(\frac{2\pi }{1.5\times {10}^{-6}}\right)$

Looking into the option the correct

Answer is$\overrightarrow{B}=9\times {10}^{-8}\sin 2\pi \left(\frac{x}{1.5\times {10}^{-6}}-2\times {10}^{14}t\right)\widehat{k}$