A proton moving withh a velocity `2.5xx10^(7)m//s` enters a magnetic field of intensity `2.5T` making an angle `30^(@)` with the magnetic field. The force on the proton is

Options

A. `3xx10^(-12)N`
B. `5xx10^(-12)N`
C. `6xx10^(-12)N`
D. `9xx10^(-12)N`

Answer

To find the force on a proton moving in a magnetic field, we can use the formula for the magnetic force experienced by a charged particle:

\[ F = Q \cdot v \cdot B \cdot \sin(\theta) \]

Where:
- \( F \) is the magnetic force,
- \( Q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

### Step-by-Step Solution:

1. **Identify the given values:**
- Velocity of the proton, \( v = 2.5 \times 10^7 \, \text{m/s} \)
- Magnetic field strength, \( B = 2.5 \, \text{T} \)
- Angle, \( \theta = 30^\circ \)
- Charge of a proton, \( Q = 1.6 \times 10^{-19} \, \text{C} \)

2. **Calculate \( \sin(\theta) \):**
- For \( \theta = 30^\circ \), we know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

3. **Substitute the values into the formula:**
\[ F = Q \cdot v \cdot B \cdot \sin(\theta) \]
\[ F = (1.6 \times 10^{-19} \, \text{C}) \cdot (2.5 \times 10^7 \, \text{m/s}) \cdot (2.5 \, \text{T}) \cdot \left(\frac{1}{2}\right) \]

4. **Perform the multiplication:**
- First, calculate \( v \cdot B \):
\[ v \cdot B = (2.5 \times 10^7) \cdot (2.5) = 6.25 \times 10^7 \]
- Now, substitute this back into the force equation:
\[ F = (1.6 \times 10^{-19}) \cdot (6.25 \times 10^7) \cdot \left(\frac{1}{2}\right) \]
\[ F = (1.6 \times 10^{-19}) \cdot (3.125 \times 10^7) \]

5. **Calculate the final force:**
\[ F = 5.0 \times 10^{-12} \, \text{N} \]

### Final Answer:
The force on the proton is \( F = 5.0 \times 10^{-12} \, \text{N} \).