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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2025 · 28 January · Evening Shift · Physics · Electromagnetic Waves

The magnetic field of an E.M. wave is given by B = ( 3 2 i +(1)/(2)...

The magnetic field of an E.M. wave is given by B→=\overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right) 30 \sin [ω(t−zc)]\left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right] (S.I. Units). The corresponding electric field in S.I. units is :

Options

  1. AE→=(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]
  2. BE→=(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]
  3. CE→=(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]
  4. DE→=(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]

Correct answer

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]

Step-by-step solution

B→=(32i^+12j^)30sin⁡[ω(t−zc)]E→=B→×c→ and  E=B0c\begin{aligned} & \overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\right) 30 \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right] \\ & \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{c}} \text{\text{ and } } \mathrm{E}=\mathrm{B}_0 \mathrm{c} \end{aligned} Here E→(32(−j^)+12i^)\overrightarrow{\mathrm{E}}\left(\frac{\sqrt{3}}{2}(-\hat{\mathrm{j}})+\frac{1}{2} \hat{\mathrm{i}}\right)E0=30cE→=(12i^−32j^)30csin⁡[ω(t−zc)]\begin{aligned} & \mathrm{E}_0=30 \mathrm{c} \\ & \overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right] \end{aligned}$$
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