CUET Preparation Today
The electric field vector oscillates according to the equation $\vec E = 4\sin[1.8y+(5.4×10^6)t]\hat k$. The equation of magnetic field vector is: |
$\vec B=1.33×10^{-8} \sin[1.8y +(5.4×10^6)t]\hat i$ $\vec B=4\sin[1.8y+(5.4×10^4)i]\hat j$ $\vec B=1.33×10^{-8} \sin[1.8y +(5.4×10^6)t]\hat k$ $\vec B=5\sin[1.8y+(5.4×10^4)i]\hat j$ |
$\vec B=1.33×10^{-8} \sin[1.8y +(5.4×10^6)t]\hat i$ |
The correct answer is Option (1) → $\vec B=1.33×10^{-8} \sin[1.8y +(5.4×10^6)t]\hat i$ The given vector field is, $\vec E = 4\sin(1.8y+(5.4×10^6)t)\hat k$ and, $B_0=\frac{E_0}{C}=\frac{4}{3×10^8}\sin(1.8y+(5.4×10^6)t)\hat i$ $=(1.33×10^{-8})\sin(1.8y +(5.4×10^6)t)\hat i$ |
