JEE Main 2020 (6 Sep, Shift 2) Physics Q16: For a plane…

Question

For a plane electromagnetic wave, the magnetic field at a point $x$ and time t is $\vec{B}(x,t)=\left[1.2	imes10^{-7}\sin(0.5	imes10^{3}x+1.5	imes10^{11}t)\hat{k}ight]\mathrm{T}$
The instantaneous electric field $\vec{E}$ corresponding to $\vec{B}$ is :
(speed of light $c=3	imes10^{8}\,\mathrm{ms^{-1}}$)

Accepted Answer

Option 2: $\vec{E}(x,t)=\left[-36\sin(0.5	imes10^{3}x+1.5	imes10^{11}t)\hat{j}ight]\frac{\mathrm{V}}{\mathrm{m}}$

Answer

$\vec{E}(x,t)=\left[36\sin(1	imes10^{3}x+0.5	imes10^{11}t)\hat{j}ight]\frac{\mathrm{V}}{\mathrm{m}}$

Answer

$\vec{E}(x,t)=\left[-36\sin(0.5	imes10^{3}x+1.5	imes10^{11}t)\hat{j}ight]\frac{\mathrm{V}}{\mathrm{m}}$

Answer

$\vec{E}(x,t)=\left[36\sin(1	imes10^{3}x+1.5	imes10^{11}t)\hat{i}ight]\frac{\mathrm{V}}{\mathrm{m}}$

Answer

$\vec{E}(x,t)=\left[36\sin(0.5	imes10^{3}x+1.5	imes10^{11}t)\hat{k}ight]\frac{\mathrm{V}}{\mathrm{m}}$

JEE Main 6 September 2020, Shift 2, Physics Q16