PhysicsHard120×since 2002Q6826A plane EM wave is propagating along xxx direction. It has a wavelength of 4 mm4 \mathrm{~mm}4 mm. If electric field is in yyy direction with the maximum magnitude of 60 Vm−160 \mathrm{~Vm}^{-1}60 Vm−1, the equation for magnetic field is :ABz=2×10−7sin[π2(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}Bz=2×10−7sin[2π(x−3×108t)]k^TBBz=2×10−7sin[π2×103(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}Bz=2×10−7sin[2π×103(x−3×108t)]k^TCBz=60sin[π2(x−3×108t)]k^T\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}Bz=60sin[2π(x−3×108t)]k^TDBx=60sin[π2(x−3×108t)]i^T\mathrm{B}_x=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{i}} \mathrm{T}Bx=60sin[2π(x−3×108t)]i^TCheck answerSkip