The magnetic field of a plane electromagnetic wave is given by $\overrightarrow B = 3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat j$, then the associated electric field will be :

Question

$9\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat k$ V/m

$3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat i$ V/m

$3 \times {10^{ - 8}}\sin (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat i$ V/m

$9\sin (1.6 \times {10^3}x - 48 \times {10^{10}}t)\widehat k$ V/m

Solution

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