JEE Main 202528 Jan, Evening shiftPhysicsMediumElectromagnetic Waves120× since 2002The magnetic field of an E.M. wave is given by B⃗=(32i^+12j^)30sin[ω(t−zc)]\vec B = \left(\frac{\sqrt3}{2}\hat i + \frac12\hat j\right)30\sin\left[\omega\left(t-\frac{z}{c}\right)\right]B=(23i^+21j^)30sin[ω(t−cz)] (S.I. Units). The corresponding electric field in S.I. units is :AOption A. E⃗=(12i^−32j^)30csin[ω(t−zc)]\vec E = \left(\frac12\hat i - \frac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]E=(21i^−23j^)30csin[ω(t−cz)]BOption B. E⃗=(34i^+14j^)30ccos[ω(t−zc)]\vec E = \left(\frac34\hat i + \frac14\hat j\right)30c\cos\left[\omega\left(t-\frac{z}{c}\right)\right]E=(43i^+41j^)30ccos[ω(t−cz)]COption C. E⃗=(12i^+32j^)30csin[ω(t+zc)]\vec E = \left(\frac12\hat i + \frac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t+\frac{z}{c}\right)\right]E=(21i^+23j^)30csin[ω(t+cz)]DOption D. E⃗=(32i^−12j^)30csin[ω(t+zc)]\vec E = \left(\frac{\sqrt3}{2}\hat i - \frac12\hat j\right)30c\sin\left[\omega\left(t+\frac{z}{c}\right)\right]E=(23i^−21j^)30csin[ω(t+cz)]CheckSkip · show solution
