Electromagnetic Waves

JEE Main level, single correct, medium

106 of 165

Question 106 of 165

The magnetic field of an electromagnetic wave is given by:
B⃗=1.6×10−6cos⁡⁡(2×107z+6×1015t)(2i^+j^)Wbm2\vec{B}=1.6\times {10}^{-6}\cos ⁡\left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(2\hat{i}+\hat{j}\right)\dfrac{Wb}{{m}^{2}}
The associated electric field will be:

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Answer: (C) E⃗=4.8×102cos(2×107z+6×1015t)(−i^+2j^)Vm\vec{E}=4.8\times {10}^{2}cos\left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(-\hat{i}+2\hat{j}\right)\dfrac{V}{m}

Given magnetic field, B⃗=1.6×10−6cos(2×107z+6×1015t)(2i^+j^)Wbm2\vec{B}=1.6\times {10}^{-6}cos\left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(2\hat{i}+\hat{j}\right)\dfrac{Wb}{{m}^{2}}
The relation between magnetic field and electric field is E=cB=(3×108)×(1.6×10−6×5)=4.8×102×5E=cB=(3\times {10}^{8})\times (1.6\times {10}^{-6}\times \sqrt{5})=4.8\times {10}^{2}\times \sqrt{5}
The direction of light propagation will be along E⃗×B⃗\vec{E}\times \vec{B} .
E⃗=4.8×102cos(2×107z+6×1015t)(−i^+2j^)Vm\vec{E}=4.8\times {10}^{2}cos\left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(-\hat{i}+2\hat{j}\right)\dfrac{V}{m}

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