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Electromagnetic Waves question

2025 · 28 Jan · Shift 2 · Q56
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  5. /2025 · 28 Jan · Shift 2 · Q56

Electromagnetic Waves question

2025 · 28 Jan · Shift 2 · Q56

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The magnetic field of an E.M. wave is given by B⃗=(32i^+12j^)30sin⁡[ω(t−zc)]\vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]B=(23​​i^+21​j^​)30sin[ω(t−cz​)] (S.I. Units). The corresponding electric field in S.I. units is:
  1. A
    E→=(12i^+32j^)30csin⁡[ω(t+zc)]\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{z}{\mathrm{c}}\right)\right]E=(21​i^+23​​j^​)30csin[ω(t+cz​)]
  2. B
    E→=(12i^−32j^)30csin⁡[ω(t−zc)]\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]E=(21​i^−23​​j^​)30csin[ω(t−cz​)]
  3. C
    E→=(32i^−12j^)30csin⁡[ω(t+zc)]\overrightarrow{\mathrm{E}}=\left(\frac{\sqrt{3}}{2} \hat{i}-\frac{1}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{z}{\mathrm{c}}\right)\right]E=(23​​i^−21​j^​)30csin[ω(t+cz​)]
  4. D
    E→=(34i^+14j^)30ccos⁡[ω(t−zc)]\overrightarrow{\mathrm{E}}=\left(\frac{3}{4} \hat{i}+\frac{1}{4} \hat{j}\right) 30 \mathrm{c} \cos \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]E=(43​i^+41​j^​)30ccos[ω(t−cz​)]
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