Related Formula
B₀ = (E₀)/(c)$$B_0 = \frac{E_0}{c}$$
c = E × B$$\hat{c} = \hat{E} \times \hat{B}$$
Core Logic
The phase of the wave is (0.6 × 10³ x - 1.8 × 10¹¹ t)$(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$. This indicates the wave propagates in the +x$+x$ direction, so c = i$\hat{c} = \hat{i}$.
The electric field oscillates along the y$y$-axis, so E = j$\hat{E} = \hat{j}$.
From B = c × E$\hat{B} = \hat{c} \times \hat{E}$, we have B = i × j = k$\hat{B} = \hat{i} \times \hat{j} = \hat{k}$.
So, the magnetic field is along the z$z$-axis (Bz$B_z$).
Step 1: Calculate Amplitude of B
Wave speed v = c = (ω)/(k) = 1.8 × 10¹¹0.6 × 10³ = 3 × 10⁸ m/s$v = c = \frac{\omega}{k} = \frac{1.8 \times 10^{11}}{0.6 \times 10^3} = 3 \times 10^8\text{ m/s}$.
The amplitude of the magnetic field is:
B₀ = (E₀)/(c) = (69)/(3 × 10⁸) = 23 × 10⁻⁸ = 2.3 × 10⁻⁷ T$$B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 23 \times 10^{-8} = 2.3 \times 10^{-7}\text{ T}$$
The phase remains exactly the same as the electric field:
Bz = 2.3 × 10⁻⁷ (0.6 × 10³ x - 1.8 × 10¹¹ t)$$B_z = 2.3 \times 10^{-7} \sin(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$$
Pattern Recognition
B₀ = E₀/c$B_0 = E_0/c$ gives the magnitude. The vector identity B = v × E$\hat{B} = \hat{v} \times \hat{E}$ gives the direction. Phase part never changes sign or terms between E and B equations.
Chapter Mix
Class 12 Physics: Electromagnetic Waves