JEE Main 202523 Jan, Morning shiftPhysicsEasyElectromagnetic Waves120× since 2002The electric field of an electromagnetic wave in free space is E⃗=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^) N/C\vec E=57\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](4\hat i-3\hat j)\ \mathrm{N/C}E=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^) N/C. The associated magnetic field in Tesla isAOption A. B⃗=−573×108cos[7.5×106t−5×10−3(3x+4y)](k^)\vec B=-\dfrac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](\hat k)B=−3×10857cos[7.5×106t−5×10−3(3x+4y)](k^)BOption B. B⃗=573×108cos[7.5×106t−5×10−3(3x+4y)](k^)\vec B=\dfrac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](\hat k)B=3×10857cos[7.5×106t−5×10−3(3x+4y)](k^)COption C. B⃗=573×108cos[7.5×106t−5×10−3(3x+4y)](5k^)\vec B=\dfrac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](5\hat k)B=3×10857cos[7.5×106t−5×10−3(3x+4y)](5k^)DOption D. B⃗=−573×108cos[7.5×106t−5×10−3(3x+4y)](5k^)\vec B=-\dfrac{57}{3\times10^8}\cos\left[7.5\times10^6t-5\times10^{-3}(3x+4y)\right](5\hat k)B=−3×10857cos[7.5×106t−5×10−3(3x+4y)](5k^)CheckSkip · show solution
