Electromagnetic Waves

JEE Main level, single correct, medium

91 of 165

Question 91 of 165

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}\left(x, t\right)=\left[1.2\times {10}^{-7}\sin \left(0.5\times {10}^{3}x+1.5\times {10}^{11}t\right)\hat{k}\right]T.
The instantaneous electric field E⃗\vec{E} corresponding to B⃗\vec{B} is :

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Answer: (A) E⃗(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\vec{E}\left(x,t\right)=\left[-36\sin \left(0.5\times {10}^{3}x+1.5\times {10}^{11}t\right)\hat{j}\right]\dfrac{V}{m}

E0=B0×C=1.2×10−7×3×108=36{E}_{0}={B}_{0}\times C=1.2\times {10}^{-7}\times 3\times {10}^{8}=36
As the light is propagating in xx direction.
& (E^×B^)∣∣C^\left(\hat{E}\times \hat{B}\right)||\hat{C}
∴∴ E⃗\vec{E} should be in j^\hat{j} direction.
So electric field E⃗=E0sin⁡E⃗⋅(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\vec{E}={E}_{0}\sin \vec{E}\cdot (x,t)=\left[-36\sin \left(0.5\times {10}^{3}x+1.5\times {10}^{11}t\right)\hat{j}\right]\dfrac{V}{m}

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