Question

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

• A. The expression of elctric field is $\overrightarrow{E}=120sin(\frac{\pi }{3}x-100\pi \times {10}^{6}t)\hat{j}$
• B. The expression of electric field is $\overrightarrow{E}=60sin(\frac{\pi }{3}x-100\pi \times {10}^{6}t)\hat{j}$
• 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

The correct option is D Both (a) and (c)

$E_{0}max=120\omega =2\pi f=2\pi \times 50\times {10}^6=100\pi \times {10}^{6}K=\frac{\omega }{V}=\frac{100\pi \times {10}^6}{3\times {10}^8}=\frac{\pi }{3}\frac{E_{max}}{B_{max}}=C\frac{120}{B_{max}}=3\times {10}^8{B}_{max}=40\times {10}^{-8}$

Accordingly puttingthe equation.

$E=120\sin (Kx-\omega t)=120\sin (\frac{\pi }{3}x-100\pi \times {10}^{6}t)B=40\times {10}^{-8}\sin (Kx-\omega t)=40\times {10}^{-8}\sin (\frac{\pi }{3}-100\pi \times {10}^{-6}t)$