The electric field of an electromagnetic wave in free space isE→=57cos⁡7.5×106t−5×10−3(3x+4y)(4i^−3j^)N/CThe associated magnetic field in Tesla is-

K→=3i^+4j^K^=3i^+4j^5E^=4i^−3j^5B^=K^×E^B^=−Z^B0=E0C=573×108

The electric field of an electromagnetic wave in free space isE→=57cos⁡7.5×106t−5×10−3(3x+4y)(4i^−3j^)N/CThe associated magnetic field in Tesla is-

K→=3i^+4j^K^=3i^+4j^5E^=4i^−3j^5B^=K^×E^B^=−Z^B0=E0C=573×108

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The electric field of an electromagnetic wave in free space is
$\vec{E} = 57 \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (4 \hat{i} - 3 \hat{j}) N / C$
The associated magnetic field in Tesla is-

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$\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$

$\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$

$\vec{B} = - \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$

$\vec{B} = - \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$

answer is C.

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Detailed Solution

$\begin{array}{l} \vec{K} = 3 \hat{i} + 4 \hat{j} \\ \hat{K} = \frac{3 \hat{i} + 4 \hat{j}}{5} \\ \hat{E} = \frac{4 \hat{i} - 3 \hat{j}}{5} \\ \hat{B} = \hat{K} \times \hat{E} \\ \hat{B} = - \hat{Z} \\ B_{0} = \frac{E_{0}}{C} = \frac{57}{3 \times 10^{8}} \end{array}$