A charged particle having charge 2q and mass m is projected from origin in X-Y plane with a velocity v inclined at an angle of `45^@` with positive X-axis in a region where a uniform magnetic field B unit exists along +z axis. The coordinates of the centre of the circular path of the particle are

Options

A. (0,0,0)
B. `((-mv)/(sqrt2qB),(mv)/(sqrt2qB),0)`
C. `((-sqrt2mv)/(qB),(sqrt2mv)/(qB),0)`
D. none of the above

Answer

To solve the problem of finding the coordinates of the center of the circular path of a charged particle projected in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a charged particle with charge \(2q\) and mass \(m\).
- It is projected from the origin (0, 0) in the XY plane with a velocity \(v\) at an angle of \(45^\circ\) with the positive X-axis.
- There is a uniform magnetic field \(B\) directed along the positive Z-axis.

2. **Components of Velocity**:
- The velocity \(v\) can be resolved into its components:
\[ v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}}, \quad v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}} \]

3. **Magnetic Force**:
- The magnetic force \(F\) acting on the particle is given by:
\[ F = q(\mathbf{v} \times \mathbf{B}) \]
- Here, the charge \(q = 2q\) and the magnetic field \(B\) is along the Z-axis.
- The magnitude of the magnetic force can be calculated as:
\[ F = 2q \cdot v \cdot B \]

4. **Centripetal Force**:
- This magnetic force acts as the centripetal force required for circular motion:
\[ F = \frac{mv^2}{r} \]
- Setting the two expressions for force equal gives:
\[ 2qvB = \frac{mv^2}{r} \]

5. **Solving for Radius \(r\)**:
- Rearranging the above equation to find the radius \(r\):
\[ r = \frac{mv}{2qB} \]

6. **Coordinates of the Center of the Circular Path**:
- The center of the circular path will be offset from the origin due to the circular motion.
- The coordinates of the center can be determined using the radius and the angle of projection:
- The X-coordinate of the center will be:
\[ x = r \cos(45^\circ) = \frac{mv}{2qB} \cdot \frac{1}{\sqrt{2}} = \frac{mv}{2qB\sqrt{2}} \]
- The Y-coordinate of the center will be:
\[ y = -r \sin(45^\circ) = -\frac{mv}{2qB} \cdot \frac{1}{\sqrt{2}} = -\frac{mv}{2qB\sqrt{2}} \]
- The Z-coordinate is 0 since the motion is confined to the XY plane.

7. **Final Coordinates**:
- Thus, the coordinates of the center of the circular path are:
\[ \left( \frac{mv}{2qB\sqrt{2}}, -\frac{mv}{2qB\sqrt{2}}, 0 \right) \]