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, we need to find the coordinates of the center of the circular path of a charged particle projected in a magnetic field. Let's go through the solution step by step.

### Step 1: Understand the scenario
We have a charged particle with charge \(2q\) and mass \(m\) 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. A uniform magnetic field \(B\) exists along the +Z axis.

### Step 2: Determine the magnetic force
The magnetic force acting on the charged particle is given by the Lorentz force equation:
\[ F = q(\mathbf{v} \times \mathbf{B}) \]
For our case, substituting \(q = 2q\):
\[ F = 2q(\mathbf{v} \times \mathbf{B}) \]
Since the magnetic field is along the +Z direction, we can denote it as \(\mathbf{B} = B\hat{k}\).

### Step 3: Calculate the velocity components
The velocity \(v\) can be broken down into its components:
- \(v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}}\)
- \(v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}}\)

### Step 4: Calculate the magnetic force
Using the right-hand rule and the cross product, the force will be directed towards the center of the circular path. The magnitude of the magnetic force is:
\[ F = 2q \cdot v \cdot B \]

### Step 5: Set up the centripetal force equation
For circular motion, the magnetic force provides the necessary centripetal force:
\[ F_c = \frac{mv^2}{r} \]
Setting the magnetic force equal to the centripetal force:
\[ 2qvB = \frac{mv^2}{r} \]

### Step 6: Solve for the radius \(r\)
Rearranging the equation gives us:
\[ r = \frac{mv}{2qB} \]

### Step 7: Find the coordinates of the center of the circular path
The center of the circular path will be offset from the origin. Since the particle moves in a circular path, the center will lie along the line that bisects the angle of projection. The coordinates of the center can be determined using trigonometric functions:
- The x-coordinate of the center will be \(r \cos(45^\circ)\)
- The y-coordinate of the center will be \(r \sin(45^\circ)\)

Substituting \(r\):
\[ x = r \cos(45^\circ) = \frac{mv}{2qB} \cdot \frac{1}{\sqrt{2}} = \frac{mv}{2qB\sqrt{2}} \]
\[ y = r \sin(45^\circ) = \frac{mv}{2qB} \cdot \frac{1}{\sqrt{2}} = \frac{mv}{2qB\sqrt{2}} \]

Since the particle is moving in the negative y-direction, we take the y-coordinate as negative:
\[ y = -\frac{mv}{2qB\sqrt{2}} \]

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