When a charged particle moving parallel to a magnetic field, then the force F is:

Options

A. qvB
B. qBv/2
C. 0
D. None of these

Explanation

To solve the question, we need to analyze the situation of a charged particle moving parallel to a magnetic field. Let's break it down step by step:

### Step 1: Understanding the Force on a Charged Particle
The force \( F \) experienced by a charged particle moving in a magnetic field is given by the equation:
\[ F = qvB \sin \theta \]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

### Step 2: Identifying the Angle
In this case, the problem states that the charged particle is moving parallel to the magnetic field. This means that the angle \( \theta \) between the velocity of the particle and the magnetic field is:
\[ \theta = 0^\circ \]

### Step 3: Calculating the Sine of the Angle
Now, we need to calculate \( \sin \theta \):
\[ \sin 0^\circ = 0 \]

### Step 4: Substituting into the Force Equation
Substituting \( \sin 0^\circ \) into the force equation:
\[ F = qvB \sin 0^\circ = qvB \times 0 = 0 \]

### Step 5: Conclusion
Thus, the force \( F \) acting on the charged particle when it is moving parallel to the magnetic field is:
\[ F = 0 \, \text{N} \]

### Final Answer
The correct option is **0**.

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