In region `x gt 0` a uniform and constant magnetic field `vec(B)_(1) = 2 B_(0) hat(k)` exists. Another uniform and constant magnetic field `vec(B)_(2) = B_(0) hat(k)` exists in region `x lt 0` A positively charged particle of mass m and charge q is crossing origin at time t =0 with a velocity `bar(u) = u_(0) hat(i)`. The particle comes back to its initial position after a time : (`B_(0),u_(0)` are positive constants)

Options

A. a. `(3)/(2) (pi m)/(q B_(0))`
B. b. `(2 pi m)/(q B_(0))`
C. c. `(3 pi m)/(q B_(0))`
D. d. Particle does not come back to its initial position

Explanation

To solve the problem, we need to analyze the motion of a positively charged particle in two different magnetic fields. The particle crosses the origin at time \( t = 0 \) with an initial velocity \( \vec{u} = u_0 \hat{i} \).

### Step-by-Step Solution:

1. **Identify the Magnetic Fields:**
- For \( x > 0 \): The magnetic field is \( \vec{B}_1 = 2B_0 \hat{k} \).
- For \( x < 0 \): The magnetic field is \( \vec{B}_2 = B_0 \hat{k} \).

2. **Determine the Force on the Particle:**
- The magnetic force on a charged particle is given by \( \vec{F} = q \vec{u} \times \vec{B} \).
- At \( t = 0 \), the particle has a velocity \( \vec{u} = u_0 \hat{i} \).
- In the region \( x > 0 \) (where \( \vec{B}_1 \) is present):
\[ \vec{F}_1 = q (u_0 \hat{i}) \times (2B_0 \hat{k}) = 2qB_0 u_0 \hat{j} \]
- This force acts in the positive \( y \)-direction, causing the particle to move in a circular path.

3. **Calculate the Radius of the Circular Motion in \( x > 0 \):**
- The radius \( r_1 \) of the circular path can be found using the formula:
\[ r_1 = \frac{mv}{qB} \]
- Here, \( v = u_0 \) and \( B = 2B_0 \):
\[ r_1 = \frac{mu_0}{q(2B_0)} \]

4. **Determine the Time Taken to Traverse the Circular Path in \( x > 0 \):**
- The time \( t_1 \) to complete a half-circle (from \( x = 0 \) to the maximum \( y \) position) is:
\[ t_1 = \frac{\pi r_1}{u_0} = \frac{\pi \left(\frac{mu_0}{q(2B_0)}\right)}{u_0} = \frac{\pi m}{2qB_0} \]

5. **Analyze the Motion in \( x < 0 \):**
- In the region \( x < 0 \), the magnetic field is \( \vec{B}_2 = B_0 \hat{k} \).
- The radius \( r_2 \) in this region is:
\[ r_2 = \frac{mu_0}{qB_0} \]

6. **Calculate the Time Taken to Traverse the Circular Path in \( x < 0 \):**
- The time \( t_2 \) to complete a half-circle in this region is:
\[ t_2 = \frac{\pi r_2}{u_0} = \frac{\pi \left(\frac{mu_0}{qB_0}\right)}{u_0} = \frac{\pi m}{qB_0} \]

7. **Total Time for the Particle to Return to the Initial Position:**
- The total time \( T \) for the particle to return to its initial position is the sum of the times taken in both regions:
\[ T = t_1 + t_2 + t_1 = \frac{\pi m}{2qB_0} + \frac{\pi m}{qB_0} + \frac{\pi m}{2qB_0} \]
- Simplifying this gives:
\[ T = \frac{2\pi m}{qB_0} \]

### Final Answer:
The total time taken for the particle to return to its initial position is:
\[ T = \frac{2\pi m}{qB_0} \]