In a certain region of space a uniform and constant electric field and a magnetic field parallel to each other are present. A proton is fired from a point A in the field with speed `V=4xx10^(4)m//s` at an angle of `alpha` with the field direction. The proton reaches a point B in the field where its velocity makes an angle `beta` with the field direction. If `(sinalpha)/(sinbeta)=sqrt3.` Find the electric potential difference between the points A and B. Take mp (mass of proton) `=1.6xx10^(-27)`kg and e (magnitude of electronic charge) `=1.6xx10^(-19)`C.

Options

A. 16 V
B. 40 V
C. 90 V
D. 30 V

Answer

To solve the problem, we need to find the electric potential difference between points A and B, given the conditions of the electric and magnetic fields, the velocity of the proton, and the angles involved.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial speed of the proton, \( V_A = 4 \times 10^4 \, \text{m/s} \)
- Mass of the proton, \( m_p = 1.6 \times 10^{-27} \, \text{kg} \)
- Charge of the proton, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- The relationship between angles: \( \frac{\sin \alpha}{\sin \beta} = \sqrt{3} \)

2. **Use the Relationship of Velocities**:
- Since the electric and magnetic fields are parallel, the perpendicular components of the velocity remain unchanged. Therefore, we can use the relationship:
\[ V_A \sin \alpha = V_B \sin \beta \]
- Rearranging gives:
\[ \frac{V_A}{V_B} = \frac{\sin \beta}{\sin \alpha} \]
- Given \( \frac{\sin \alpha}{\sin \beta} = \sqrt{3} \), we can write:
\[ \frac{V_A}{V_B} = \frac{1}{\sqrt{3}} \implies V_B = V_A \cdot \sqrt{3} \]

3. **Calculate \( V_B \)**:
- Substituting \( V_A \):
\[ V_B = 4 \times 10^4 \cdot \sqrt{3} \, \text{m/s} \]

4. **Apply Conservation of Energy**:
- The total mechanical energy at points A and B can be expressed as:
\[ \frac{1}{2} m_p V_A^2 + e \Delta V = \frac{1}{2} m_p V_B^2 \]
- Rearranging gives:
\[ \Delta V = \frac{\frac{1}{2} m_p V_B^2 - \frac{1}{2} m_p V_A^2}{e} \]

5. **Substitute \( V_B \) into the Equation**:
- Calculate \( V_B^2 \):
\[ V_B^2 = (4 \times 10^4 \sqrt{3})^2 = 48 \times 10^8 = 4.8 \times 10^9 \, \text{m}^2/\text{s}^2 \]
- Calculate \( V_A^2 \):
\[ V_A^2 = (4 \times 10^4)^2 = 16 \times 10^8 = 1.6 \times 10^9 \, \text{m}^2/\text{s}^2 \]

6. **Plug Values into the Potential Difference Equation**:
- Now substitute into the equation for \( \Delta V \):
\[ \Delta V = \frac{\frac{1}{2} m_p (4.8 \times 10^9) - \frac{1}{2} m_p (1.6 \times 10^9)}{e} \]
- Simplifying gives:
\[ \Delta V = \frac{1}{2} m_p \left(4.8 \times 10^9 - 1.6 \times 10^9\right) \cdot \frac{1}{e} \]
- This simplifies to:
\[ \Delta V = \frac{1}{2} m_p (3.2 \times 10^9) \cdot \frac{1}{e} \]

7. **Calculate the Final Values**:
- Substitute \( m_p \) and \( e \):
\[ \Delta V = \frac{1}{2} \cdot (1.6 \times 10^{-27}) \cdot (3.2 \times 10^9) \cdot \frac{1}{1.6 \times 10^{-19}} \]
- This results in:
\[ \Delta V = \frac{(1.6 \cdot 3.2) \times 10^{-18}}{2 \cdot 1.6 \times 10^{-19}} = \frac{5.12 \times 10^{-18}}{3.2 \times 10^{-19}} = 16 \, \text{V} \]

### Final Answer:
The electric potential difference between points A and B is \( \Delta V = 16 \, \text{V} \).