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JEE MAIN 2019
08-04-2019 S2
Question
The magnetic field of an electromagnetic wave is given by: $$ \vec{B}=1.6 \times 10^{-6} \cos \left(2 \times 10^7 z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{W b}{m^2} $$ The associated electric field will be:
Select the correct option:
A
$\overrightarrow{\mathrm{E}}=4.8 \times 10^2 \cos \left(2 \times 10^7 z+6 \times 10^{15} t\right)(-\hat{i}-2 \hat{j}) \frac{V}{m}$
B
$\overrightarrow{\mathrm{E}}=4.8 \times 10^2 \cos \left(2 \times 10^7 z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{V}{m}$
C
$\vec{E}=4.8 \times 10^2 \cos \left(2 \times 10^7 z-6 \times 10^{15} t\right)(-2 \hat{i}+\hat{j}) \frac{V}{m}$
D
$\overrightarrow{\mathrm{E}}=4.8 \times 10^2 \cos \left(2 \times 10^7 z+6 \times 10^{15} t\right)(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \frac{\mathrm{V}}{\mathrm{m}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Physics
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