JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 10)
The electric field of an electromagnetic wave in free space is represented as $$\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}$$. The corresponding magnetic induction vector will be :
$$\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}$$
$$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}$$
$$\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}$$
$$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}$$
Explanation
$$\begin{aligned}
& \text { Given } \vec{E}=E_0 \cos (\omega t-k z) \hat{i} \\
& \vec{B}=\frac{E_0}{C} \cos (\omega t-k z) \hat{j} \\
& \hat{C}=\hat{E} \times \hat{B}
\end{aligned}$$
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