An `alpha-` particle enters a magnetic field of 1 T with a velocity `10^6` m/s in a direction perpendicular to the field. The force on `alpha`-particle is :

Options

A. `1.6 xx 10^(-3) N`
B. `3.2 xx 10^(-13) N`
C. `4.8 xx 10^(-13) N`
D. `7.6 xx 10^(-12) N`

Answer

To find the force on an alpha particle entering a magnetic field, we can use the formula for the magnetic force acting on 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, and
- \( \theta \) is the angle between the velocity and the magnetic field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic field, \( B = 1 \, \text{T} \)
- Velocity, \( v = 10^6 \, \text{m/s} \)
- Since the alpha particle enters the magnetic field perpendicularly, \( \theta = 90^\circ \).

2. **Calculate \( \sin(\theta) \):**
- Since \( \theta = 90^\circ \), we have:
\[ \sin(90^\circ) = 1 \]

3. **Determine the Charge of the Alpha Particle:**
- An alpha particle consists of 2 protons and 2 neutrons, thus it has a charge of:
\[ q = 2 \cdot e \]
- Where \( e \) (the charge of a proton) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
- Therefore:
\[ q = 2 \cdot (1.6 \times 10^{-19}) = 3.2 \times 10^{-19} \, \text{C} \]

4. **Substitute the Values into the Force Equation:**
\[ F = q \cdot v \cdot B \cdot \sin(90^\circ) \]
\[ F = (3.2 \times 10^{-19} \, \text{C}) \cdot (10^6 \, \text{m/s}) \cdot (1 \, \text{T}) \]

5. **Calculate the Force:**
\[ F = 3.2 \times 10^{-19} \cdot 10^6 = 3.2 \times 10^{-13} \, \text{N} \]

6. **Conclusion:**
- The force on the alpha particle is:
\[ F = 3.2 \times 10^{-13} \, \text{N} \]