A proton beam enters magnetic field of `10^(-4) "Wb//m^(2)` normally. If the specific charge of the proton is `10^(11) C//kg` and its velocity is `10^(9)m//s` then the radius of the circle described will be

Options

A. 100 m
B. 0.1 m
C. 1 m
D. 10 m

Answer

To find the radius of the circular path described by a proton beam entering a magnetic field, we can use the formula for the radius of the circular motion of a charged particle in a magnetic field. The formula is given by:

\[ r = \frac{mv}{|q|B} \]

Where:
- \( r \) is the radius of the circular path,
- \( m \) is the mass of the proton,
- \( v \) is the velocity of the proton,
- \( q \) is the charge of the proton,
- \( B \) is the magnetic field strength.

However, we can also express the specific charge \( \frac{q}{m} \) as given in the problem, which allows us to simplify the formula:

\[ r = \frac{v}{B \cdot \frac{q}{m}} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic field \( B = 10^{-4} \, \text{Wb/m}^2 \)
- Specific charge \( \frac{q}{m} = 10^{11} \, \text{C/kg} \)
- Velocity \( v = 10^{9} \, \text{m/s} \)

2. **Substitute the values into the formula:**
\[ r = \frac{v}{B \cdot \frac{q}{m}} = \frac{10^{9} \, \text{m/s}}{10^{-4} \, \text{Wb/m}^2 \cdot 10^{11} \, \text{C/kg}} \]

3. **Calculate the denominator:**
- First, calculate \( B \cdot \frac{q}{m} \):
\[ B \cdot \frac{q}{m} = 10^{-4} \cdot 10^{11} = 10^{7} \, \text{Wb/(m}^2 \cdot \text{C/kg)} \]

4. **Now substitute back into the radius formula:**
\[ r = \frac{10^{9}}{10^{7}} = 10^{2} \, \text{m} \]

5. **Final answer:**
\[ r = 100 \, \text{m} \]

### Conclusion:
The radius of the circle described by the proton beam in the magnetic field is \( 100 \, \text{m} \).