A charge having q/m equal to `10^(g)` c/kg and with velocity `3 xx 10^(5)` m/s enters into a uniform magnetic field B = 0.3 tesla at an angle `30^(@)` with direction of field. Then radius of curvature will be:

Options

A. 0.01 cm
B. 0.5 cm
C. 1 cm
D. 2 cm

Explanation

To find the radius of curvature \( r \) of a charged particle moving in a magnetic field, we can use the formula:

\[ r = \frac{mv_{\perp}}{qB} \]

where:
- \( m \) is the mass of the particle,
- \( v_{\perp} \) is the component of the velocity perpendicular to the magnetic field,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

### Step 1: Calculate the perpendicular component of the velocity
The velocity \( v \) of the particle is given as \( 3 \times 10^5 \) m/s, and it enters the magnetic field at an angle of \( 30^\circ \). The perpendicular component of the velocity \( v_{\perp} \) can be calculated using:

\[ v_{\perp} = v \sin(\theta) \]

Substituting the values:

\[ v_{\perp} = 3 \times 10^5 \, \text{m/s} \times \sin(30^\circ) \]

Since \( \sin(30^\circ) = 0.5 \):

\[ v_{\perp} = 3 \times 10^5 \, \text{m/s} \times 0.5 = 1.5 \times 10^5 \, \text{m/s} \]

### Step 2: Substitute the known values into the radius formula
We know:
- \( \frac{q}{m} = 10^8 \, \text{C/kg} \)
- \( B = 0.3 \, \text{T} \)

Now substituting \( v_{\perp} \) into the radius formula:

\[ r = \frac{mv_{\perp}}{qB} \]

We can express \( m \) in terms of \( \frac{q}{m} \):

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

Substituting the values:

\[ r = \frac{1.5 \times 10^5 \, \text{m/s}}{10^8 \, \text{C/kg} \cdot 0.3 \, \text{T}} \]

### Step 3: Calculate the radius
Calculating the denominator:

\[ 10^8 \cdot 0.3 = 3 \times 10^7 \]

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

\[ r = \frac{1.5 \times 10^5}{3 \times 10^7} \]

Calculating this gives:

\[ r = 0.05 \, \text{m} = 5 \times 10^{-2} \, \text{m} \]

### Step 4: Convert to centimeters
To convert meters to centimeters:

\[ r = 5 \times 10^{-2} \, \text{m} = 5 \, \text{cm} \]

### Final Answer
Thus, the radius of curvature \( r \) is:

\[ \boxed{5 \, \text{cm}} \]