Question

Suppose that the electric field amplitude of an electromagnetic wave propagating along x-direction is $E_0$ = 120 N $C^{-1}$ and that its frequency is $\upsilon$ = 50.0 MHz.

• A. The expression of electric field is $\overrightarrow{E}=120\sin (\frac{\pi }{3}x-100\pi \times {10}^{6}t)\hat{j}$
• B. The expression of electric field is $\overrightarrow{E}=60\sin (\frac{\pi }{3}x-100\pi \times {10}^{6}t)\hat{j}$
• C. The expression of magnetic field is $\overrightarrow{B}=40\times {10}^{-8}\sin (\frac{\pi }{3}x-\pi \times {10}^{8}t)\hat{k}$
• D. Both (a) and (c)

Answer 1

Answer: (D)

Both (a) and (c)

$E_0=120NC$

$\omega =2\pi f$

$\omega =2\pi \times 50\times {10}^6=100\pi \times {10}^6$

$\frac{\omega }{k}=c$

$\Rightarrow \frac{100\pi \times {10}^6}{k}=3\times {10}^8$

$\Rightarrow k=\frac{\pi }{3}$

The standard equaton for electric field is:

$E=E_{0}sin(kx-\omega t)$

$⟹E=120\sin (\frac{\pi }{3}x-100\pi \times {10}^{6}t)\hat{j}$

The magnetic field intensity is given by:

$B_0=\frac{E_0}{c}$

$\Rightarrow B_0=\frac{120}{3\times {10}^8}=40\times {10}^{-8}T$

The standard equaton for magnetic field is:

$B=B_{0}sin(kx-\omega t)$

$⟹B=40\times {10}^{-8}\sin (\frac{\pi }{3}x-\pi \times {10}^{8}t)\hat{k}$

Option $\left(D\right)$ is correct.