A beam of ions with velocity ` 2xx 10^(5) ` m/s enters normally into a uniform magnetic field of ` 4 xx 10^(-2)` T. if the specific charge to the ions is ` 5xx 10^(7) ` C/kg, the radius of the circular path described will be

Options

A. 5 cm
B. 15 cm
C. 10 cm
D. 30 cm

Answer

To find the radius of the circular path described by a beam of ions entering a magnetic field, we can use the following steps:

### Step 1: Understand the Forces Involved
When charged particles move in a magnetic field, they experience a magnetic force that acts as a centripetal force, causing them to move in a circular path. The magnetic force \( F \) is given by:

\[ F = qvB \]

where:
- \( q \) is the charge of the ion,
- \( v \) is the velocity of the ion,
- \( B \) is the magnetic field strength.

The centripetal force required to keep the ion in circular motion is given by:

\[ F = \frac{mv^2}{r} \]

where:
- \( m \) is the mass of the ion,
- \( r \) is the radius of the circular path.

### Step 2: Set the Forces Equal
Since the magnetic force provides the centripetal force, we can set these two equations equal to each other:

\[ qvB = \frac{mv^2}{r} \]

### Step 3: Rearrange to Find the Radius
Rearranging the equation to solve for the radius \( r \):

\[ r = \frac{mv}{qB} \]

### Step 4: Use the Specific Charge
We are given the specific charge \( \frac{q}{m} = 5 \times 10^7 \, \text{C/kg} \). Thus, we can express \( m \) in terms of \( q \):

\[ m = \frac{q}{\frac{q}{m}} = \frac{q}{5 \times 10^7} \]

Substituting this into the equation for \( r \):

\[ r = \frac{m v}{q B} = \frac{(q / (5 \times 10^7)) v}{q B} \]

The \( q \) cancels out:

\[ r = \frac{v}{5 \times 10^7 B} \]

### Step 5: Substitute the Given Values
Now we can substitute the given values into the equation:

- \( v = 2 \times 10^5 \, \text{m/s} \)
- \( B = 4 \times 10^{-2} \, \text{T} \)

Substituting these values:

\[ r = \frac{2 \times 10^5}{5 \times 10^7 \times 4 \times 10^{-2}} \]

### Step 6: Calculate the Radius
Calculating the denominator:

\[ 5 \times 10^7 \times 4 \times 10^{-2} = 2 \times 10^6 \]

Now substituting back into the equation for \( r \):

\[ r = \frac{2 \times 10^5}{2 \times 10^6} = \frac{1}{10} = 0.1 \, \text{m} \]

### Step 7: Convert to Centimeters
Since the question asks for the radius in centimeters:

\[ r = 0.1 \, \text{m} = 10 \, \text{cm} \]

### Final Answer
Thus, the radius of the circular path described by the ions is:

\[ \boxed{10 \, \text{cm}} \]