Electromagnetic Waves

JEE Main level, single correct, medium

62 of 165

Question 62 of 165

Light wave travelling in air along xx-direction is given by Ey=540sin⁡π×104(x−ct)Vm−1{E}_{y}=540\sin \pi \times {10}^{4}\left(x-ct\right)V{m}^{-1}. Then, the peak value of magnetic field of wave will be (Given c=3×108ms−1c=3\times {10}^{8}m{s}^{-1})

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Answer: (A) 18×10−7T18\times {10}^{-7}T

Given here, Ey=540sin⁡π×104(x−ct)Vm−1{E}_{y}=540\sin \pi \times {10}^{4}\left(x-ct\right)V{m}^{-1}
Here, peak value of electric field, E0=540Vm−1{E}_{0}=540V{m}^{-1}
Using relation between magnetic and electric field,
Peak value of magnetic field is B0=E0c=5403×108=18×10−7T{B}_{0}=\dfrac{{E}_{0}}{c}=\dfrac{540}{3\times {10}^{8}}=18\times {10}^{-7}T

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