JEE Main 2019 — Physics solution

JEE Main2019Physics

The magnetic field of an electromagnetic wave is given by: $B = 1.6 \times 1 0^{- 6} cos (2 \times 10^{7} z + 6 \times 10^{15} t) (2 \hat{i} + \hat{j}) \frac{W b}{m ^{2}}$ The associated electric field will be:

A
$E = 4.8 \times 1 0^{2} cos (2 \times 10^{7} z + 6 \times 10^{15} t) (- \hat{i} + 2 \hat{j}) \frac{V}{m}$
B
$E = 4.8 \times 1 0^{2} cos (2 \times 10^{7} z - 6 \times 10^{15} t) (2 \hat{i} + 2 \hat{j}) \frac{V}{m}$
C
$E = 4.8 \times 1 0^{2} cos (2 \times 10^{7} z - 6 \times 10^{15} t) (- 2 \hat{i} + \hat{i}) \frac{V}{m}$
D
$E = 4.8 \times 1 0^{2} cos (2 \times 10^{7} z + 6 \times 10^{15} t) (\hat{i} - 2 \hat{i}) \frac{V}{m}$

Solution

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