1.For a plane electromagnetic wave, the electric field is given by E⃗=90sin(0.5×103x+1.5×1011t)k^\vec{E} = 90 \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{k}E=90sin(0.5×103x+1.5×1011t)k^ V/m. The corresponding magnetic field B⃗\vec{B}B will bea.B⃗=3×10−7sin(0.5×103x+1.5×1011t)i^\vec{B} = 3 \times 10^{-7} \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{i}B=3×10−7sin(0.5×103x+1.5×1011t)i^ Tb.B⃗=3×10−7sin(0.5×103x+1.5×1011t)j^\vec{B} = 3 \times 10^{-7} \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{j}B=3×10−7sin(0.5×103x+1.5×1011t)j^ Tc.B⃗=27×109sin(0.5×103x+1.5×1011t)j^\vec{B} = 27 \times 10^9 \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{j}B=27×109sin(0.5×103x+1.5×1011t)j^ Td.B⃗=3×10−7sin(0.5×103x+1.5×1011t)k^\vec{B} = 3 \times 10^{-7} \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{k}B=3×10−7sin(0.5×103x+1.5×1011t)k^ TLogin to continueOnly logged in users canattempt or see the solution.
