The electric field of an electromagnetic wave in free space is represented as vecE = E_0cos (omega t - kz)hati$\vec{E} = E_0\cos (\omega t - kz)\hat{i}$ The corresponding magnetic induction vector will be:

### Related Formula B_0 = fracE_0C$$B_0 = \frac{E_0}{C}$$ hatC = hatE times hatB$$\hat{C} = \hat{E} \times \hat{B}$$ ### Core Logic In an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by E_0 = c B_0$E_0 = c B_0$. Thus, B_0 = E_0 / C$B_0 = E_0 / C$. The direction of wave propagation is given by the cross product of the electric field and magnetic field vectors: hatC = hatE times hatB$\hat{C} = \hat{E} \times \hat{B}$. ### Step 1: Determine Wave Direction and Magnetic Field Direction Given the phase term (omega t - kz)$(\omega t - kz)$, the wave propagates in the +z$+z$ direction, so hatC = hatk$\hat{C} = \hat{k}$. The electric field oscillates in the +x$+x$ direction, so hatE = hati$\hat{E} = \hat{i}$. We know: hatk = hati times hatB$$\hat{k} = \hat{i} \times \hat{B}$$ Since hati times hatj = hatk$\hat{i} \times \hat{j} = \hat{k}$, the magnetic field must oscillate in the +y$+y$ direction (hatj$\hat{j}$). ### Step 2: Construct Final Vector The full magnetic field vector shares the same phase and applies the above amplitude and direction: vecB = fracE_0C cos(omega t - kz) hatj$$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$$ ### Pattern Recognition Phase remains identical. Amplitude scales by 1/c$1/c$. Direction satisfies the right-hand triad (vecE, vecB, vecv)$(\vec{E}, \vec{B}, \vec{v})$ where vecv = vecE times vecB$\vec{v} = \vec{E} \times \vec{B}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves