In the above question, the radius of path of the particle will be

A. 12.0 m
B. 1.2 m
C. 0.12 m
D. 0.012 m

Options

A. 12.0 m
B. 1.2 m
C. 0.12 m
D. 0.012m

Explanation

To find the radius of the path of the particle, we can use the formula for the radius of circular motion in a magnetic field, which is given by:

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

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The velocity \( v \) of the particle is given as \( 4 \times 10^5 \) m/s.
- The magnetic field strength \( B \) is given as \( 0.3 \) T.
- The mass \( m \) of the particle (photon) is given as \( 1.6 \times 10^{-27} \) kg.
- The charge \( q \) of the particle is given as \( 1.6 \times 10^{-19} \) C.

2. **Calculate the Perpendicular Component of Velocity**:
- The angle with the magnetic field is \( 60^\circ \).
- The perpendicular component of velocity \( v_{\perp} \) can be calculated using:
\[ v_{\perp} = v \sin(60^\circ) = 4 \times 10^5 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \times 10^5 \text{ m/s} \]

3. **Substitute the Values into the Radius Formula**:
- Now substituting the values into the radius formula:
\[ r = \frac{m v_{\perp}}{B q} = \frac{(1.6 \times 10^{-27}) \times (2\sqrt{3} \times 10^5)}{(0.3) \times (1.6 \times 10^{-19})} \]

4. **Simplify the Expression**:
- First, calculate the numerator:
\[ 1.6 \times 10^{-27} \times 2\sqrt{3} \times 10^5 = 3.2\sqrt{3} \times 10^{-22} \]
- Now calculate the denominator:
\[ 0.3 \times 1.6 \times 10^{-19} = 4.8 \times 10^{-20} \]
- Now substitute these into the radius formula:
\[ r = \frac{3.2\sqrt{3} \times 10^{-22}}{4.8 \times 10^{-20}} = \frac{3.2\sqrt{3}}{4.8} \times 10^{-2} \]

5. **Calculate the Final Value**:
- The numerical value can be approximated:
\[ r \approx 0.012 \text{ m} \]

6. **Conclusion**:
- Therefore, the radius of the path of the particle is approximately \( 0.012 \) m, which corresponds to option D.

### Final Answer:
**D. 0.012 m**