$\begin{aligned}
& \overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\right) 30 \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right] \
& \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{c}} \text { and } \mathrm{E}=\mathrm{B}_0 \mathrm{c}
\end{aligned}Here\overrightarrow{\mathrm{E}}\left(\frac{\sqrt{3}}{2}(-\hat{\mathrm{j}})+\frac{1}{2} \hat{\mathrm{i}}\right)\begin{aligned}
& \mathrm{E}_0=30 \mathrm{c} \
& \overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]
\end{aligned}$
The magnetic field of an E.M. wave is given by overrightarrow B= ( √3 2 i+ 1 2 j ) 30 sin [ω ( t- z ) ] (S.I. Units). The corresponding electric…
JEE Main 2025 — Physics Electromagnetism
The magnetic field of an E.M. wave is given by B=(23i^+21j^)30sin[ω(t−cz)] (S.I. Units). The corresponding electric field in S.I. units is :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
E=(21i^−23j^)30csin[ω(t−cz)]
- B
E=(43i^+41j^)30ccos[ω(t−cz)]
- C
E=(21i^+23j^)30csin[ω(t+cz)]
- D
E=(23i^−21j^)30csin[ω(t+cz)]
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