Choose the correct option:
A particle of charge per unit mass `alpha` is released from origin with a velocity `vecv=v_(0)hati` in a magnetic field
`vec(B)=-B_(0)hatk` for `xle(sqrt(3))/2 (v_(0))/(B_(0)alpha)`
and `vec(B)=0` for `xgt(sqrt(3))/2 (v_(0))/(B_(0)alpha)`
The `x`-coordinate of the particle at time `t((pi)/(3B_(0)alpha))` would be

Options

A. (a) `sqrt(3)/(2)(V_(0))/(B_(0)alpha)+sqrt(3)/(2)v_(0)(t-(pi)/(B_(0)alpha))`
B. (b) `sqrt(3)/(2)(V_(0))/(B_(0)alpha)+v_(0)(t-(pi)/(B_(0)alpha))`
C. (c) `sqrt(3)/(2)(V_(0))/(B_(0)alpha)+(v_(0))/(2)(t-(pi)/(B_(0)alpha))`
D. (d) `sqrt(3)/(2)(V_(0))/(B_(0)alpha)+(v_(0)t)/(2)`

Answer

To solve the problem, we need to determine the x-coordinate of a charged particle moving in a magnetic field at a specific time. The particle is released from the origin with an initial velocity and experiences different magnetic field conditions based on its position.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle is released from the origin (0, 0) with an initial velocity \( \vec{v} = v_0 \hat{i} \). The magnetic field \( \vec{B} \) is given as \( -B_0 \hat{k} \) for \( x \leq \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} \) and \( \vec{B} = 0 \) for \( x > \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} \).

2. **Finding the Time**:
We are interested in the position of the particle at time \( t = \frac{\pi}{3 B_0 \alpha} \). The particle will initially move under the influence of the magnetic field until it reaches the boundary \( x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} \).

3. **Determining the Angle**:
The position \( x \) can be related to the angle \( \theta \) in the circular motion induced by the magnetic field. The sine of the angle can be calculated as:
\[ \frac{x}{r} = \sin \theta \quad \Rightarrow \quad \sin \theta = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \theta = 60^\circ \]

4. **Calculating Time for Motion in Magnetic Field**:
The time taken to reach the point \( A \) (where \( x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} \)) can be calculated using the relationship:
\[ t = \frac{\theta}{\omega} = \frac{\frac{\pi}{3}}{B_0 \alpha} \]

5. **Position After Leaving the Magnetic Field**:
After reaching \( x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} \), the particle moves in a straight line since the magnetic field is zero. The x-coordinate at any time \( t \) after this is given by:
\[ x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} + v_0 \left(t - \frac{\pi}{3 B_0 \alpha}\right) \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} + v_0 \left(t - \frac{\pi}{3 B_0 \alpha}\right) \cdot \frac{1}{2} \]

6. **Final Expression**:
Substituting \( t = \frac{\pi}{3 B_0 \alpha} \):
\[ x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} + \frac{v_0}{2} \left(t - \frac{\pi}{3 B_0 \alpha}\right) \]

7. **Identifying the Correct Option**:
Comparing the derived expression with the given options, we find that it matches with option (c):
\[ x = \frac{\sqrt{3}}{2} \frac{v_0}{B_0 \alpha} + \frac{v_0}{2} \left(t - \frac{\pi}{3 B_0 \alpha}\right) \]

### Conclusion:
The x-coordinate of the particle at time \( t = \frac{\pi}{3 B_0 \alpha} \) is given by option (c).