Electromagnetic Waves

JEE Main level, single correct, medium

112 of 165

Question 112 of 165

If the magnetic field of a plane electromagnetic wave is given by (The speed of light =3×108m/s=3\times {10}^{8}m/s)

B=100×10−6sin⁡⁡[2π×2×1015(t−xc)]B=100\times {10}^{-6}\sin ⁡\left[2\pi \times 2\times {10}^{15}\left(t-\dfrac{x}{c}\right)\right] then the maximum electric field associated with it is:

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Answer: (A) 3×104NC−13\times {10}^{4}N{C}^{-1}

We know that relation between electric field E and magnetic field B is given by:
EB=c⇒E=Bc=100×10−6×3×108Vm−1=3×104Vm−1\dfrac{E}{B}=c \Rightarrow E=Bc =100\times {10}^{-6}\times 3\times {10}^{8}V{m}^{-1} =3\times {10}^{4}V{m}^{-1}

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