Physics

Physics

Problem

For a plane electromagnetic wave, the magnetic field at a point xx and time tt is B⃗(x,t)=[1.2×10−7sin⁡(0.5×103x+1.5×1011t)k^]T\vec{B}(x, t) = [1.2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{k}] \text{T} The instantaneous electric field E⃗\vec{E} corresponding to B⃗\vec{B} is: (speed of light c=3×108 ms−1c = 3 \times 10^8 \text{ ms}^{-1} ) [Main Sep. 06, 2020 (II)] a. E⃗(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\vec{E}(x, t) = [-36 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{j}] \frac{\text{V}}{\text{m}} b. E⃗(x,t)=[36sin⁡(1×103x+0.5×1011t)j^]Vm\vec{E}(x, t) = [36 \sin(1 \times 10^3 x + 0.5 \times 10^{11} t) \hat{j}] \frac{\text{V}}{\text{m}} c. E⃗(x,t)=[36sin⁡(0.5×103x+1.5×1011t)k^]Vm\vec{E}(x, t) = [36 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{k}] \frac{\text{V}}{\text{m}} d. E⃗(x,t)=[36sin⁡(1×103x+1.5×1011t)i^]Vm\vec{E}(x, t) = [36 \sin(1 \times 10^3 x + 1.5 \times 10^{11} t) \hat{i}] \frac{\text{V}}{\text{m}}

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