Question

A plane electromagnetic wave of frequency 50MHz travels in free space along the +x direction. At a particular point in space $\vec{E}=7.2\hat{j}$ V/m . At this point, $\vec{B}$ is equal to;

• A. -2.4x10^{-8}\hat{k}T
• B. 2.4x10^{-8}\hat{j}T
• C. 7.4x10^{-6}\hat{i}T
• D. 2.4x10^{-8}\hat{j}T

Answer 1

Answer: (A)

(1) $-2.4\times10^{-8}\hat{k}T$

Answer 2

The magnitude of the magnetic field is $B = E/c$. Given $E = 7.2$ V/m and $c \approx 3 \times 10^8$ m/s, $B = \frac{7.2}{3 \times 10^8} = 2.4 \times 10^{-8}$ T. The wave travels along the +x direction ($\hat{i}$), and the electric field is along the +y direction ($\hat{j}$). The relation $\vec{E} \times \vec{B}$ points in the direction of propagation. Thus, $\hat{j} \times \vec{B}$ must be in the $+\hat{i}$ direction. For this to be true, $\vec{B}$ must be in the $-\hat{k}$ direction, because $\hat{j} \times (-\hat{k}) = -(\hat{j} \times \hat{k}) = -\hat{i}$. This implies that the wave is represented by $\vec{E} = E_0 \sin(kx-\omega t)\hat{j}$ and $\vec{B} = B_0 \sin(kx-\omega t)(-\hat{k})$. Therefore, $\vec{B} = -2.4 \times 10^{-8} \hat{k}$ T.