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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2024 · 9 April · Evening Shift · Physics · Electromagnetic Waves

The magnetic field in a plane electromagnetic wave is B_ y =(3.5 ×...

The magnetic field in a plane electromagnetic wave is By=(3.5×10−7)sin⁡(1.5×103x+0.5×1011t)T\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}. The corresponding electric field will be :

Options

  1. AEy=10.5sin⁡(1.5×103x+0.5×1011t)Vm−1E_y=10.5 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}
  2. BEz=1.17sin⁡(1.5×103x+0.5×1011t)Vm−1E_z=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}
  3. CEy=1.17sin⁡(1.5×103x+0.5×1011t)Vm−1E_y=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}
  4. DEz=105sin⁡(1.5×103x+0.5×1011t)Vm−1E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

Correct answer

D. Ez=105sin⁡(1.5×103x+0.5×1011t)Vm−1E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

Step-by-step solution

E0=B0C&E0=3×108×(3.5×10−7)sin⁡(1.5×103x+0.5×1011t)&E0=105sin⁡(1.5×103x+0.5×1011t)Vm−1\begin{aligned}& \mathrm{E}_0=\mathrm{B}_0 \mathrm{C} \& \mathrm{E}_0=3 \times 10^8 \times\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \& \mathrm{E}_0=105 \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1} \end{aligned} Data inconsistent while calculating speed of wave. You can challenge for data.
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