A particle enters uniform constant magnetic field region with is initial velocity parallel to the field direction. Which of the following statements about its velocity is correct ?(neglect the effects of other fields )

Options

A. (a)There is change in both magnitude and direction
B. (b)There is no change
C. (c)There is change only in magnitude
D. There is change only in direction

Answer

To solve the problem, we need to analyze the behavior of a charged particle entering a uniform magnetic field with an initial velocity that is parallel to the direction of the magnetic field.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- A charged particle (let's assume it's a positive charge) enters a uniform magnetic field.
- The initial velocity of the particle is parallel to the magnetic field direction.

2. **Identifying the Forces**:
- The force acting on a charged particle in a magnetic field is given by the Lorentz force equation:
\[ \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \]
where \( q \) is the charge of the particle, \( \mathbf{v} \) is the velocity vector, and \( \mathbf{B} \) is the magnetic field vector.

3. **Calculating the Cross Product**:
- Since the velocity is parallel to the magnetic field, the angle \( \theta \) between the velocity vector and the magnetic field vector is \( 0^\circ \).
- The sine of \( 0^\circ \) is \( 0 \):
\[ \sin(0^\circ) = 0 \]
- Therefore, the force acting on the particle becomes:
\[ \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) = qvB\sin(0^\circ) = 0 \]

4. **Conclusion About Acceleration**:
- Since the force is zero, the acceleration of the particle is also zero:
\[ \mathbf{a} = \frac{\mathbf{F}}{m} = \frac{0}{m} = 0 \]
- This means that there is no change in the velocity of the particle.

5. **Final Statement**:
- Since the particle experiences no force and thus no acceleration, its velocity remains constant while it is in the magnetic field.

### Answer:
The correct statement about the particle's velocity is that it remains unchanged while in the magnetic field.