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The electric field of an electromagnetic wave in free space is E=57 [7.5 × 10^6 t-5 × 10^-3(3 x+4 y) ](4 i-3 j) N / C. The associated magnetic field in Tesla is

B=-(57)/(3 × 10^8) [7.5 × 10^6 t-5 × 10^-3(3 x+4 y) ](5 k)

B=(57)/(3 × 10^8) [7.5 × 10^6 t-5 × 10^-3(3 x+4 y) ](k)

B=-(57)/(3 × 10^8) [7.5 × 10^6 t-5 × 10^-3(3 x+4 y) ](k)

B=(57)/(3 × 10^8) [7.5 × 10^6 t-5 × 10^-3(3 x+4 y) ](5 k)

Electromagnetic Waves

JEE Main level, single correct, medium

27 of 165

Question 27 of 165

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

Show answer and solution

Answer: (C) $\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$

\begin{aligned} & \overrightarrow{\mathrm{K}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \\ & \hat{\mathrm{K}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{5} \\ & \hat{\mathrm{E}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}}{5} \\ & \hat{\mathrm{~B}}=\hat{\mathrm{K}} \times \hat{\mathrm{E}} \\ & \hat{\mathrm{B}}=-\hat{\mathrm{Z}} \\ & \mathrm{B}_0=\frac{\mathrm{E}_0}{\mathrm{C}}=\frac{57}{3 \times 10^8}\end{aligned}

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