PhysicsMedium120×since 2002Q6761For a plane electromagnetic wave, the magnetic field at a point x and time t is B→(x,t)\overrightarrow B \left( {x,t} \right)B(x,t) = [1.2×10−7sin(0.5×103x+1.5×1011t)k^]\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right][1.2×10−7sin(0.5×103x+1.5×1011t)k] T The instantaneous electric field E→\overrightarrow EE corresponding to B→\overrightarrow BB is : (speed of light c = 3 × 10⁸ ms^–1)AE→(x,t)=[36sin(1×103x+1.5×1011t)i^]\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat i} \right]E(x,t)=[36sin(1×103x+1.5×1011t)i] Vm{V \over m}mVBE→(x,t)=[36sin(0.5×103x+1.5×1011t)k^]Vm\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]{V \over m}E(x,t)=[36sin(0.5×103x+1.5×1011t)k]mVCE→(x,t)=[36sin(1×103x+0.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}E(x,t)=[36sin(1×103x+0.5×1011t)j]mVDE→(x,t)=[−36sin(0.5×103x+1.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}E(x,t)=[−36sin(0.5×103x+1.5×1011t)j]mVCheck answerSkip