An `alpha` - particle with a specific charge of `2.5xx10^7 Ckg^(-1)` moves with a speed of `2xx10^5 ms ^(-1)` in a perpendicular magnetic field of `0.05 T.` Then , the radius of the circular path described by it is

Options

A. 8 cm
B. 4 cm
C. 16 cm
D. 2 cm

Answer

To find the radius of the circular path described by an alpha particle moving in a magnetic field, we can use the formula for the radius of the circular motion due to the magnetic force. The magnetic force acts as the centripetal force that keeps the particle in circular motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Specific charge of the alpha particle, \( \frac{Q}{M} = 2.5 \times 10^7 \, \text{C/kg} \)
- Speed of the alpha particle, \( v = 2 \times 10^5 \, \text{m/s} \)
- Magnetic field strength, \( B = 0.05 \, \text{T} \)

2. **Write the Formula for the Radius:**
The radius \( r \) of the circular path can be calculated using the formula:
\[ r = \frac{mv}{qB} \]
However, since we have the specific charge \( \frac{Q}{M} \), we can rearrange the formula:
\[ r = \frac{v}{\frac{Q}{M} \cdot B} \]

3. **Substitute the Values:**
Substitute the known values into the equation:
\[ r = \frac{2 \times 10^5}{2.5 \times 10^7 \cdot 0.05} \]

4. **Calculate the Denominator:**
First, calculate \( 2.5 \times 10^7 \cdot 0.05 \):
\[ 2.5 \times 0.05 = 0.125 \]
Therefore,
\[ 2.5 \times 10^7 \cdot 0.05 = 0.125 \times 10^7 = 1.25 \times 10^6 \]

5. **Calculate the Radius:**
Now, substitute this back into the equation for \( r \):
\[ r = \frac{2 \times 10^5}{1.25 \times 10^6} \]
Simplifying this gives:
\[ r = \frac{2}{1.25} \times 10^{-1} = 1.6 \times 10^{-1} \, \text{m} = 0.16 \, \text{m} \]

6. **Convert to Centimeters:**
To convert meters to centimeters, multiply by 100:
\[ r = 0.16 \times 100 = 16 \, \text{cm} \]

7. **Final Answer:**
The radius of the circular path described by the alpha particle is:
\[ \boxed{16 \, \text{cm}} \]