A charged particle with charge to mass ratio `((q)/(m)) = (10)^(3)/(19) Ckg^(-1)` enters a uniform magnetic field `vec(B) = 20 hat(i) + 30 hat(j) + 50 hat(k) T` at time t = 0 with velocity `vec(V) = (20 hat(i) + 50 hat(j) + 30 hat(k)) m//s`. Assume that magnetic field exists in large space.
During the further motion of the particle in the magnetic field, the angle between the magnetic field and velocity of the particle

Options

A. remains constant
B. increases
C. decreases
D. may increase or decrease

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a uniform magnetic field. The key points to consider are the relationship between the velocity of the particle and the magnetic field, and how the angle between them affects the path of the particle.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Charge to mass ratio: \(\frac{q}{m} = \frac{10^3}{19} \, \text{C/kg}\)
- Magnetic field: \(\vec{B} = 20 \hat{i} + 30 \hat{j} + 50 \hat{k} \, \text{T}\)
- Initial velocity: \(\vec{V} = 20 \hat{i} + 50 \hat{j} + 30 \hat{k} \, \text{m/s}\)

2. **Determine the Angle Between Velocity and Magnetic Field:**
- The angle \(\theta\) between two vectors can be found using the dot product formula:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]
- Here, \(\vec{A} = \vec{V}\) and \(\vec{B}\) is the magnetic field.

3. **Calculate the Magnitudes of \(\vec{V}\) and \(\vec{B}\):**
- Magnitude of \(\vec{V}\):
\[ |\vec{V}| = \sqrt{(20)^2 + (50)^2 + (30)^2} = \sqrt{400 + 2500 + 900} = \sqrt{3800} = 10\sqrt{38} \, \text{m/s} \]
- Magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{(20)^2 + (30)^2 + (50)^2} = \sqrt{400 + 900 + 2500} = \sqrt{3800} = 10\sqrt{38} \, \text{T} \]

4. **Calculate the Dot Product \(\vec{V} \cdot \vec{B}\):**
\[ \vec{V} \cdot \vec{B} = (20)(20) + (50)(30) + (30)(50) = 400 + 1500 + 1500 = 3400 \]

5. **Use the Dot Product to Find \(\cos(\theta)\):**
\[ \cos(\theta) = \frac{\vec{V} \cdot \vec{B}}{|\vec{V}| |\vec{B}|} = \frac{3400}{(10\sqrt{38})(10\sqrt{38})} = \frac{3400}{3800} = \frac{17}{19} \]

6. **Determine the Angle \(\theta\):**
- To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{17}{19}\right) \]

7. **Conclusion:**
- The angle \(\theta\) between the magnetic field and the velocity of the particle remains constant throughout the motion of the particle in the magnetic field.