In the previous problem, the maximum range of movement of the centre of the part of the circle from line AD in which charged particle of charge Q moves with a velocity v when `(theta)` is positive to when `(theta)` is negative is given by

Options

A. `+-(mv)/(2QB)`
B. `+-(mv)/(QB)`
C. `+-(2mv)/(QB)`
D. `+-(2mv)/(3QB)`

Explanation

To solve the problem of finding the maximum range of movement of the center of the path of a charged particle in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
The charged particle of charge \( Q \) moves with a velocity \( v \) in a magnetic field. The motion of the charged particle will be circular due to the Lorentz force acting on it.

2. **Identifying the Angle**:
The problem states that we are considering the angles \( \theta \) as positive and negative. When \( \theta \) is positive, it indicates one direction of motion, and when \( \theta \) is negative, it indicates the opposite direction.

3. **Determining the Radius of Circular Motion**:
The radius \( r \) of the circular path traced by the charged particle in a magnetic field is given by the formula:
\[ r = \frac{mv}{QB} \]
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( Q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

4. **Calculating Maximum Distance**:
When the angle \( \theta \) is \( +90^\circ \), the center of the circular path will be at a distance \( r \) from the line AD. Conversely, when \( \theta \) is \( -90^\circ \), the center will also be at a distance \( r \) but in the opposite direction. Therefore, the maximum range of movement of the center from line AD can be expressed as:
\[ \text{Maximum Range} = r \text{ (positive direction)} + r \text{ (negative direction)} = r + (-r) = 2r \]

5. **Final Expression**:
Substituting the expression for \( r \):
\[ \text{Maximum Range} = 2 \left(\frac{mv}{QB}\right) \]

### Conclusion:
Thus, the maximum range of movement of the center of the part of the circle from line AD is given by:
\[ \text{Maximum Range} = \frac{2mv}{QB} \]