A beam of light travelling aling x-axis is described by the magnetic field, `B _(o) =5.2 xx 10 ^(-9) T sin omega (t - x//c).` Then,
(Charge on electron `=1. 6 xx 10 ^(-19) C)`

Options

A. the maximum electric field is `1.56 V m ^(-1).`
B. the maximum electric field is `2.85 V m ^(-1)`
C. the maximum electric force an alpha particle due to electric field is `2 xx 10 ^(-10) N`
D. the maximum electric force an alpha particle due to electric field is `5 xx 10 ^(-19)N.`

Explanation

To solve the problem, we need to determine the maximum electric field (E₀) from the given magnetic field (B₀) and then calculate the maximum electric force (F_max) on an alpha particle.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The magnetic field is given by:
\[ B_0 = 5.2 \times 10^{-9} \, \text{T} \]
- The speed of light (c) is:
\[ c = 3 \times 10^8 \, \text{m/s} \]
- The charge of an electron (e) is:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]
- The charge on an alpha particle is twice the charge of an electron:
\[ Q_{\text{alpha}} = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]

2. **Calculate the Maximum Electric Field (E₀):**
- The relationship between the maximum electric field (E₀) and the maximum magnetic field (B₀) in electromagnetic waves is given by:
\[ E_0 = B_0 \cdot c \]
- Substituting the values:
\[ E_0 = (5.2 \times 10^{-9} \, \text{T}) \cdot (3 \times 10^8 \, \text{m/s}) \]
- Calculating E₀:
\[ E_0 = 5.2 \times 3 \times 10^{-1} = 15.6 \times 10^{-1} = 1.56 \, \text{V/m} \]

3. **Calculate the Maximum Electric Force (F_max) on the Alpha Particle:**
- The electric force (F) on a charge in an electric field is given by:
\[ F = Q \cdot E \]
- For the alpha particle, substituting the charge and the maximum electric field:
\[ F_{\text{max}} = Q_{\text{alpha}} \cdot E_0 = (3.2 \times 10^{-19} \, \text{C}) \cdot (1.56 \, \text{V/m}) \]
- Calculating F_max:
\[ F_{\text{max}} = 3.2 \times 1.56 \times 10^{-19} = 4.992 \times 10^{-19} \, \text{N} \approx 5 \times 10^{-19} \, \text{N} \]

### Final Answers:
- The maximum electric field \( E_0 \) is \( 1.56 \, \text{V/m} \).
- The maximum electric force \( F_{\text{max}} \) on the alpha particle is \( 5 \times 10^{-19} \, \text{N} \).