Question

A plane electromagnetic wave propagates through a medium of a refractive index n = 1.5. The Electric field vector associated with this wave is given by $\overrightarrow{E}$ (x, t)= 4sin(wt - kx)$\hat{j}$, The magnetic field corresponding with this wave is [All terms are in Std. Sl units]

• A. 2 × 10^{-8} sin(wt - kx) $\hat{i}$ T
• B. 2 × 10^{-8} sin(wt - kx) $\hat{j}$ T
• C. 2 × 10^{-8} sin(wt - kx) $\hat{k}$ T
• D. None of these

Answer 1

Answer: (C)

2 × 10^{-8} sin(wt - kx) $\hat{k}$ T

Answer 2

The electric field $\overrightarrow{E}(x, t) = 4 \sin(\omega t - kx) \hat{j}$ indicates propagation in the +x direction and oscillation along the +y direction. For an electromagnetic wave, the electric field, magnetic field, and propagation direction are mutually perpendicular and form a right-handed system ($\overrightarrow{k}_{prop} = \overrightarrow{E} \times \overrightarrow{B}$). Given propagation along +x and E-field along +y, the magnetic field must be along +z ($\hat{i} = \hat{j} \times \hat{k}$). The amplitude of the magnetic field $B_0$ is related to the electric field amplitude $E_0$ by $B_0 = E_0/v$, where $v$ is the speed of light in the medium. Since $v = c/n$, $B_0 = E_0 n/c$. Substituting $E_0=4$ V/m, $n=1.5$, and $c=3 \times 10^8$ m/s, we get $B_0 = (4 \times 1.5) / (3 \times 10^8) = 2 \times 10^{-8}$ T. The magnetic field oscillates with the same phase as the electric field. Thus, $\overrightarrow{B}(x, t) = 2 \times 10^{-8} \sin(\omega t - kx) \hat{k}$ T.