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Electromagnetic Waves question

2025 · Q173
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Electromagnetic Waves question

2025 · Q173

NEETPhysicsElectromagnetic WavesMCQ+4 / −1

The electric field in a plane electromagnetic wave is given by Ez=60cos⁡(5x+1.5×109t)V/mE_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{V} / \mathrm{m}Ez​=60cos(5x+1.5×109t)V/m Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :

  1. A
    Bz=60cos⁡(5x+1.5×109t)TB z=60 \cos \left(5 x+1.5 \times 10^9 t\right) TBz=60cos(5x+1.5×109t)T
  2. B
    By=60sin⁡(5x+1.5×109t)TB_y=60 \sin \left(5 x+1.5 \times 10^9 t\right) TBy​=60sin(5x+1.5×109t)T
  3. C
    By=2×10−7cos⁡(5x+1.5×109t)TB_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) TBy​=2×10−7cos(5x+1.5×109t)T
  4. D
    Bx=2×10−7cos⁡(5x+1.5×109t)TB_x=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) TBx​=2×10−7cos(5x+1.5×109t)T
View written solutionFree

Correct answer: C

In electromagnetic wave, $E$ and $B$ are in same phase and $B_0=\frac{E_0}{c}$; their planes are perpendicular to each other.

$$\begin{aligned} & \therefore B_y=\frac{60}{c} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \\ & =\frac{60}{3 \times 10^8} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \\ & B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \end{aligned}$$

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