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Electromagnetic Waves question

2024 · 4 Apr · Shift 1 · Q75
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  5. /2024 · 4 Apr · Shift 1 · Q75

Electromagnetic Waves question

2024 · 4 Apr · Shift 1 · Q75

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field in an electromagnetic wave is given by E→=i^40cos⁡ω(t−z/c)NC−1\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}E=i^40cosω(t−z/c)NC−1. The magnetic field induction of this wave is (in SI unit) :
  1. A
    B→=j^40ccos⁡ω(t−z/c)\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})B=j^​c40​cosω(t−z/c)
  2. B
    B→=i^40ccos⁡ω(t−z/c)\overrightarrow{\mathrm{B}}=\hat{i} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})B=i^c40​cosω(t−z/c)
  3. C
    B⃗=j^40cos⁡ω(t−z/c)\vec{B}=\hat{j} 40 \cos \omega(t-z / c)B=j^​40cosω(t−z/c)
  4. D
    B→=k^40ccos⁡ω(t−z/c)\overrightarrow{\mathrm{B}}=\hat{k} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})B=k^c40​cosω(t−z/c)
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