JEE Main 20249 Apr, Evening shiftPhysicsHardElectromagnetic Waves120× since 2002The magnetic field in a plane electromagnetic wave is By=(3.5×10−7)sin(1.5×103x+0.5×1011t)T\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}By=(3.5×10−7)sin(1.5×103x+0.5×1011t)T. The corresponding electric field will be :AOption A. Ez=105sin(1.5×103x+0.5×1011t)Vm−1E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}Ez=105sin(1.5×103x+0.5×1011t)Vm−1BOption B. Ey=10.5sin(1.5×103x+0.5×1011t)Vm−1E_y=10.5 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}Ey=10.5sin(1.5×103x+0.5×1011t)Vm−1COption C. Ey=1.17sin(1.5×103x+0.5×1011t)Vm−1E_y=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}Ey=1.17sin(1.5×103x+0.5×1011t)Vm−1DOption D. Ez=1.17sin(1.5×103x+0.5×1011t)Vm−1E_z=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}Ez=1.17sin(1.5×103x+0.5×1011t)Vm−1CheckSkip · show solution
