Suppose that the electric field amplitude of an electromagnetic wave us `E_0=120 N// C` and that its frequency is `50.0 MHz`.
(a) Determine` B_0,omega,k` and `lambda`,
(b) find expressions for `E` and `B`.

Options

A. The expression of electric field is `vec(E)=120 sin((pi)/(3)x-100pixx10^(6)t)hat(j)`
B. The expression of electric field is `vec(E)=60sin((pi)/(3)x -100pixx10^(6)t)hat(j)`
C. The expression of magnetic field si `vec(B)=40xx10^(-8)sin((pi)/(3)x-pixx10^(8)t)hat(k)`
D. Both (a) and (c)

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Given:
- Electric field amplitude, \( E_0 = 120 \, \text{N/C} \)
- Frequency, \( f = 50.0 \, \text{MHz} = 50 \times 10^6 \, \text{Hz} \)

### (a) Determine \( B_0, \omega, k, \) and \( \lambda \)

1. **Calculate \( B_0 \)**:
The relationship between the electric field amplitude \( E_0 \) and the magnetic field amplitude \( B_0 \) in an electromagnetic wave is given by:
\[ B_0 = \frac{E_0}{c} \]
where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ B_0 = \frac{120 \, \text{N/C}}{3 \times 10^8 \, \text{m/s}} = 40 \times 10^{-8} \, \text{T} = 4.0 \times 10^{-7} \, \text{T} \]

2. **Calculate \( \omega \)**:
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
Substituting the frequency:
\[ \omega = 2 \pi (50 \times 10^6) = 100 \pi \times 10^6 \, \text{rad/s} \approx 3.14 \times 10^8 \, \text{rad/s} \]

3. **Calculate \( k \)**:
The wave number \( k \) is given by:
\[ k = \frac{\omega}{c} \]
Substituting the values:
\[ k = \frac{100 \pi \times 10^6}{3 \times 10^8} = \frac{100 \pi}{3} \, \text{rad/m} \]

4. **Calculate \( \lambda \)**:
The wavelength \( \lambda \) is given by:
\[ \lambda = \frac{c}{f} \]
Substituting the values:
\[ \lambda = \frac{3 \times 10^8}{50 \times 10^6} = 6 \, \text{m} \]

### Summary of Part (a):
- \( B_0 = 4.0 \times 10^{-7} \, \text{T} \)
- \( \omega = 100 \pi \times 10^6 \, \text{rad/s} \)
- \( k = \frac{100 \pi}{3} \, \text{rad/m} \)
- \( \lambda = 6 \, \text{m} \)

### (b) Find expressions for \( E \) and \( B \)

1. **Expression for \( E \)**:
The electric field \( E \) in an electromagnetic wave can be expressed as:
\[ E(x, t) = E_0 \sin(kx - \omega t) \]
Substituting the values:
\[ E(x, t) = 120 \sin\left(\frac{100 \pi}{3} x - 100 \pi \times 10^6 t\right) \]

2. **Expression for \( B \)**:
The magnetic field \( B \) in an electromagnetic wave can be expressed as:
\[ B(x, t) = B_0 \sin(kx - \omega t) \]
Substituting the values:
\[ B(x, t) = 4.0 \times 10^{-7} \sin\left(\frac{100 \pi}{3} x - 100 \pi \times 10^6 t\right) \]

### Summary of Part (b):
- \( E(x, t) = 120 \sin\left(\frac{100 \pi}{3} x - 100 \pi \times 10^6 t\right) \, \text{N/C} \)
- \( B(x, t) = 4.0 \times 10^{-7} \sin\left(\frac{100 \pi}{3} x - 100 \pi \times 10^6 t\right) \, \text{T} \)