A charge having q/m equal to `10^(8)` 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 of a charged particle moving in a magnetic field, we can use the formula:

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

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

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

### Step 2: Calculate the perpendicular component of velocity
The component of velocity perpendicular to the magnetic field is given by:

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

Substituting the values:

\[ v_{\perp} = 3 \times 10^5 \sin(30^\circ) \]

Since \( \sin(30^\circ) = \frac{1}{2} \):

\[ v_{\perp} = 3 \times 10^5 \times \frac{1}{2} = 1.5 \times 10^5 \, \text{m/s} \]

### Step 3: Substitute into the radius of curvature formula
Now, we can substitute \( v_{\perp} \) into the radius of curvature formula. First, we need to express \( m \) in terms of \( q \) and \( \frac{q}{m} \):

From \( \frac{q}{m} = 10^8 \), we have:

\[ m = \frac{q}{10^8} \]

Now substituting into the formula for \( r \):

\[ r = \frac{m v_{\perp}}{qB} = \frac{\left(\frac{q}{10^8}\right) v_{\perp}}{qB} \]

The \( q \) cancels out:

\[ r = \frac{v_{\perp}}{10^8 B} \]

### Step 4: Substitute the values
Now substitute \( v_{\perp} = 1.5 \times 10^5 \, \text{m/s} \) and \( B = 0.3 \, \text{T} \):

\[ r = \frac{1.5 \times 10^5}{10^8 \times 0.3} \]

Calculating the denominator:

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

Now substituting back:

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

### Step 5: Simplify the expression
\[ r = \frac{1.5}{3} \times 10^{-2} = 0.5 \times 10^{-2} \]

Thus,

\[ r = 0.005 \, \text{m} = 0.5 \, \text{cm} \]

### Conclusion
The radius of curvature is \( 0.5 \, \text{cm} \).