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. (a)8 cm
B. (b)4 cm
C. (c)16 cm
D. (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 following steps:

### Step 1: Understand the formula for the radius of the circular path
The radius \( r \) of the circular path of a charged particle moving in a magnetic field is given by the formula:
\[ r = \frac{mv}{qB} \]
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

### Step 2: Use specific charge to express mass
The specific charge \( \frac{q}{m} \) is given as \( 2.5 \times 10^7 \, \text{C/kg} \). This means:
\[ \frac{q}{m} = 2.5 \times 10^7 \, \text{C/kg} \]
From this, we can express \( m \) in terms of \( q \):
\[ m = \frac{q}{2.5 \times 10^7} \]

### Step 3: Substitute mass into the radius formula
Substituting \( m \) into the radius formula:
\[ r = \frac{qv}{qB \cdot (1/(2.5 \times 10^7))} = \frac{v}{B \cdot (2.5 \times 10^7)} \]

### Step 4: Substitute the given values
We are given:
- \( v = 2 \times 10^5 \, \text{m/s} \)
- \( B = 0.05 \, \text{T} \)

Substituting these values into the formula:
\[ r = \frac{2 \times 10^5}{0.05 \times 2.5 \times 10^7} \]

### Step 5: Simplify the expression
Calculating the denominator:
\[ 0.05 \times 2.5 \times 10^7 = 0.125 \times 10^7 = 1.25 \times 10^6 \]
Now substituting this back into the radius formula:
\[ r = \frac{2 \times 10^5}{1.25 \times 10^6} \]

### Step 6: Perform the division
\[ r = \frac{2}{1.25} \times 10^{-1} = 1.6 \times 10^{-1} \, \text{m} = 0.16 \, \text{m} \]

### Step 7: Convert to centimeters
To convert meters to centimeters:
\[ r = 0.16 \, \text{m} = 16 \, \text{cm} \]

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