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Electromagnetic Waves question

2020 · 6 Sep · Shift 2 · Q52
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  5. /2020 · 6 Sep · Shift 2 · Q52

Electromagnetic Waves question

2020 · 6 Sep · Shift 2 · Q52

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
For 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)
  1. A
    E→(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}mV​
  2. B
    E→(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]mV​
  3. C
    E→(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​]mV​
  4. D
    E→(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​]mV​
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