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For a plane electromagnetic wave, the magnetic field at a point x and time t is : B(x, t)=[1.2× 10^-7 sin(0.5× 10^3x+1.5× 10^11t) k]T. The…

JEE Main 2020 — Physics Electromagnetism

2020mcqeasy

For a plane electromagnetic wave, the magnetic field at a point xxx and time ttt 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}\mathrm{sin}(0.5\times {10}^{3}x+1.5\times {10}^{11}t)\hat{k}]TB(x,t)=[1.2×10−7sin(0.5×103x+1.5×1011t)k^]T.

The instantaneous electric field E⃗\vec{E}E corresponding to B⃗\vec{B}B is :

Official previous-year question

Held on 6 Sept 2020 · Verified 6 Jul 2026.

Options

  1. A

    E⃗(x,t)=[−36sin(0.5×103x+1.5×1011t)j^]Vm\vec{E}(x,t)=[-36\mathrm{sin}(0.5\times {10}^{3}x+1.5\times {10}^{11}t)\hat{j}]\frac{V}{m}E(x,t)=[−36sin(0.5×103x+1.5×1011t)j^​]mV​

  2. B

    E⃗(x,t)=[36sin(1×103x+0.5×1011t)j^]Vm\vec{E}(x,t)=[36\mathrm{sin}(1\times {10}^{3}x+0.5\times {10}^{11}t)\hat{j}]\frac{V}{m}E(x,t)=[36sin(1×103x+0.5×1011t)j^​]mV​

  3. C

    E⃗(x,t)=[36sin(0.5×103x+1.5×1011t)k^]Vm\vec{E}(x,t)=[36\mathrm{sin}(0.5\times {10}^{3}x+1.5\times {10}^{11}t)\hat{k}]\frac{V}{m}E(x,t)=[36sin(0.5×103x+1.5×1011t)k^]mV​

  4. D

    E⃗(x,t)=[36sin(1×103x+15×1011t)i^]Vm\vec{E}(x,t)=[36\mathrm{sin}(1\times {10}^{3}x+15\times {10}^{11}t)\hat{i}]\frac{V}{m}E(x,t)=[36sin(1×103x+15×1011t)i^]mV​

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