Q66 / 90 · JEE Main 2024 (04 Apr Shift 1) · Physics

The electric field in an electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit) :

  • A
    $\overrightarrow{\mathrm{B}}=\hat{k} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$
  • B
    $\vec{B}=\hat{j} 40 \cos \omega(t-z / c)$
  • C
    $\overrightarrow{\mathrm{B}}=\hat{i} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$
  • Correct answer
    D
    $\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

Solution

$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$ $\overrightarrow{\mathrm{E}}$ is along $+\mathrm{x}$ direction $\overrightarrow{\mathrm{v}}$ is along $+\mathrm{z}$ direction So direction of $\vec{B}$ will be along +y and magnitude of $B$ will be $\frac{E}{c}$ So answer is $\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}$

Hence, the correct answer is option: 'D'.

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