A proton moving with a velocity `3xx10^(5)m//s` enters a magnetic field of 0.3 tesla at angle of `30^@` with the field. The radius of curvature of its path will be (e/m for proton `=10^(8)C//kg`)

Options

A. 2 cm
B. `0.5 cm`
C. `0.02 m `
D. `1.25 cm`

Explanation

To find the radius of curvature of a proton moving in a magnetic field, we can use the formula derived from the motion of charged particles in a magnetic field. The formula for the radius \( r \) of the circular path of a charged particle moving in a magnetic field is given by:

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

Where:
- \( m \) 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,
- \( \theta \) is the angle between the velocity and the magnetic field.

### Step 1: Identify the given values
- Velocity \( v = 3 \times 10^5 \, \text{m/s} \)
- Magnetic field \( B = 0.3 \, \text{T} \)
- Angle \( \theta = 30^\circ \)
- Charge-to-mass ratio \( \frac{e}{m} = 10^8 \, \text{C/kg} \)

### Step 2: Convert the charge-to-mass ratio to mass-to-charge ratio
From the given charge-to-mass ratio \( \frac{e}{m} \), we can find \( \frac{m}{e} \):

\[ \frac{m}{e} = \frac{1}{\frac{e}{m}} = \frac{1}{10^8} \, \text{kg/C} \]

### Step 3: Calculate \( \sin \theta \)
Since \( \theta = 30^\circ \):

\[ \sin 30^\circ = \frac{1}{2} \]

### Step 4: Substitute values into the formula
Now, substituting the values into the radius formula:

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

We can express \( q \) in terms of \( e \):

\[ r = \frac{m \cdot v}{e \cdot B \cdot \sin \theta} \]

Substituting \( \frac{m}{e} \):

\[ r = \frac{(m/e) \cdot e \cdot v}{e \cdot B \cdot \sin \theta} = \frac{m}{e} \cdot \frac{v}{B \cdot \sin \theta} \]

Now substituting the known values:

\[ r = \left(\frac{1}{10^8} \, \text{kg/C}\right) \cdot \frac{3 \times 10^5 \, \text{m/s}}{0.3 \, \text{T} \cdot \frac{1}{2}} \]

### Step 5: Simplify the expression
Calculating the denominator:

\[ 0.3 \cdot \frac{1}{2} = 0.15 \, \text{T} \]

Now substituting this back:

\[ r = \left(\frac{1}{10^8}\right) \cdot \frac{3 \times 10^5}{0.15} \]

Calculating \( \frac{3 \times 10^5}{0.15} \):

\[ \frac{3 \times 10^5}{0.15} = 2 \times 10^6 \]

Now substituting this value:

\[ r = \left(\frac{1}{10^8}\right) \cdot (2 \times 10^6) = \frac{2 \times 10^6}{10^8} = 2 \times 10^{-2} \, \text{m} \]

### Step 6: Final result
Thus, the radius of curvature \( r \) is:

\[ r = 0.02 \, \text{m} = 2 \, \text{cm} \]