A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

Options

A. a straight line
B. a circle
C. a helix with uniform pitch
D. a helix with nonuniform pitch

Answer

To solve the problem of a charged particle moving in a uniform magnetic field at an acute angle to the field, we can analyze the motion step by step.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charged particle moving with a velocity \( \mathbf{V} \) in a uniform magnetic field \( \mathbf{B} \).
- The angle \( \theta \) between the velocity vector \( \mathbf{V} \) and the magnetic field \( \mathbf{B} \) is acute (i.e., \( 0 < \theta < 90^\circ \)).

**Hint**: Visualize the vectors involved: the velocity vector and the magnetic field vector.

2. **Decomposing the Velocity**:
- The velocity \( \mathbf{V} \) can be decomposed into two components:
- \( V_{\parallel} \): the component of velocity parallel to the magnetic field \( \mathbf{B} \).
- \( V_{\perpendicular} \): the component of velocity perpendicular to the magnetic field \( \mathbf{B} \).
- Mathematically, this can be expressed as:
\[ V_{\parallel} = V \cos(\theta) \quad \text{and} \quad V_{\perpendicular} = V \sin(\theta) \]

**Hint**: Remember that the total velocity is the vector sum of its components.

3. **Force Acting on the Particle**:
- The magnetic force \( \mathbf{F} \) on the charged particle is given by the Lorentz force law:
\[ \mathbf{F} = q \mathbf{V} \times \mathbf{B} \]
- The force due to the component \( V_{\parallel} \) is zero because it is parallel to \( \mathbf{B} \) (no magnetic force acts in this direction).
- The force due to the perpendicular component \( V_{\perpendicular} \) is given by:
\[ F = q V_{\perpendicular} B \]

**Hint**: The direction of the magnetic force is always perpendicular to both the velocity and the magnetic field.

4. **Circular Motion Due to Perpendicular Component**:
- The perpendicular component \( V_{\perpendicular} \) will cause the particle to move in a circular path due to the magnetic force.
- The radius \( r \) of this circular motion can be calculated using:
\[ r = \frac{m V_{\perpendicular}}{qB} \]

**Hint**: Recall that the radius of circular motion depends on the mass, velocity, and charge of the particle.

5. **Combining Motions**:
- While the particle moves in a circular path due to \( V_{\perpendicular} \), it also continues to move in the direction of \( V_{\parallel} \) at a constant speed.
- This combination of circular motion (due to \( V_{\perpendicular} \)) and linear motion (due to \( V_{\parallel} \)) results in a helical path.

**Hint**: Think of the helical path as a spiral staircase where you are moving around while also going up or down.

6. **Conclusion**:
- The path of the charged particle will be a helix with uniform pitch because the component of velocity parallel to the magnetic field remains constant.

**Final Answer**: The path of the particle will be a helix with uniform pitch.