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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

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

A plane EM wave is propagating along x direction. It has a...

A plane EM wave is propagating along xx direction. It has a wavelength of 4 mm4 \mathrm{~mm}. If electric field is in yy direction with the maximum magnitude of 60Vm−160 \mathrm{Vm}^{-1}, the equation for magnetic field is :

Options

  1. ABz=2×10−7sin⁡[π2×103(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}
  2. BBz=60sin⁡[π2(x−3×108t)]k^T\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}
  3. CBx=60sin⁡[π2(x−3×108t)]i T^\mathrm{B}_x=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{i} ~ T}
  4. DBz=2×10−7sin⁡[π2(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}

Correct answer

A. Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}

Step-by-step solution

E=BC⇒60=B×3×108⇒B=2×10−7\mathrm{E}=\mathrm{BC} \Rightarrow 60=\mathrm{B} \times 3 \times 10^8 \Rightarrow \mathrm{B}=2 \times 10^{-7} Also \quad\mathrm{C}=\mathrm{f}λ\lambda⇒3×108=f×4×10−3⇒f=34×1011⇒ω=2πf=34×2π×1011⇒ω=π2×103C\begin{aligned} & \Rightarrow 3 \times 10^8=\mathrm{f} \times 4 \times 10^{-3} \\ & \Rightarrow \mathrm{f}=\frac{3}{4} \times 10^{11} \\ & \Rightarrow \omega=2 \pi \mathrm{f}=\frac{3}{4} \times 2 \pi \times 10^{11} \\ & \Rightarrow \omega=\frac{\pi}{2} \times 10^3 \mathrm{C}\end{aligned}⇒\Rightarrow \quad Electric field ⇒y\Rightarrow y direction Propagation ⇒x\Rightarrow \mathrm{x} direction Magnetic field ⇒z\Rightarrow \mathrm{z}-direction
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