Verified Solution • 2025–26 Session moderate

QJEEM25Ph2072: JEE Main 2025 Physics - Class 12 JEEM25Ph2072 Solution: The magnetic field of an E.M. wave is given by (S.I. Units).…

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] (S.I. Units).

The corresponding electric field in S.I. units is:

Multiple Choice Options

  • 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]
  • 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]
  • 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]
  • 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]

Step-by-Step Verified Solution

B→=(32i^+12j^)30sin⁡[ω(t−zc)]E→=B→×c→ and E=B0c Here E→(32(−j^)+12i^)E0=30cE→=(12i^−32j^)30csin⁡[ω(t−zc)]

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