Physics · 2019
JEE Main · 10 January 2019, Shift 1 · Q22
If the magnetic field of a plane electromagnetic wave is given by (The speed of light =3 × 10^8 m / s ) B =100 × 10^-6 sin [2 π × 2 × 10^15( t -x/c)]…
If the magnetic field of a plane electromagnetic wave is given by (The speed of light $\displaystyle =3 \times 10^8 \mathrm{~m} / \mathrm{s}$ )
$$\mathrm{B}=100 \times 10^{-6} \sin \left[2 \pi \times 2 \times 10^{15}\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right)\right]
$$
then the maximum electric field associated with it is :
Official answer
(3)
$\displaystyle 3 \times 10^4 \mathrm{~N} / \mathrm{C}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.