A charge praticule of sepeific charge (charge/ mass ) `alpha` is realsed from origin at time t=0 with velocity `v= v_(0)(hati+hatj)` in unifrom magnetic fields `B= B_(0)hati`. Co-ordinaties of the particle at time `t = (pi)/(B_(0)alpha)` are

Options

A. `(V_(0)/(2B_(0)alpha),sqrt(2V_(0))/(alphaB_(0)),(-V_(0))/(B_(0)alpha))`
B. `((-V_(0))/(B_(0)alpha),0,0)`
C. `(0,(2V_(0))/(B_(0)alpha),(V_(0)pi)/(2B_(0)alpha))`
D. `((V_(0)pi)/(2B_(0)alpha),0, (-2V_(0))/(B_(0)alpha))`

Answer

To solve the problem, we need to analyze the motion of a charged particle in a uniform magnetic field. The particle has a specific charge \( \alpha \) (where \( \alpha = \frac{q}{m} \)), is released from the origin at time \( t = 0 \) with an initial velocity \( \mathbf{v} = v_0 (\hat{i} + \hat{j}) \), and moves in a magnetic field \( \mathbf{B} = B_0 \hat{i} \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle:**
The magnetic force acting on the particle is given by the Lorentz force equation:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]
Here, \( \mathbf{v} = v_0 (\hat{i} + \hat{j}) \) and \( \mathbf{B} = B_0 \hat{i} \).

2. **Calculate the Cross Product:**
To find the magnetic force, we need to calculate \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} \times \mathbf{B} = v_0 (\hat{i} + \hat{j}) \times B_0 \hat{i} \]
Using the properties of the cross product:
\[ \hat{i} \times \hat{i} = 0 \quad \text{and} \quad \hat{j} \times \hat{i} = -\hat{k} \]
We have:
\[ \mathbf{v} \times \mathbf{B} = v_0 B_0 (-\hat{k}) = -v_0 B_0 \hat{k} \]

3. **Determine the Magnetic Force:**
The magnetic force \( \mathbf{F} \) acting on the particle is:
\[ \mathbf{F} = q (-v_0 B_0 \hat{k}) = -q v_0 B_0 \hat{k} \]

4. **Centripetal Force and Circular Motion:**
The magnetic force acts as a centripetal force, causing the particle to move in a circular path in the \( xz \)-plane. The centripetal force is given by:
\[ F_c = \frac{mv^2}{R} \]
Equating the magnetic force to the centripetal force:
\[ -q v_0 B_0 = \frac{mv^2}{R} \]

5. **Finding the Radius of the Circular Path:**
The radius \( R \) can be expressed as:
\[ R = \frac{mv_0}{qB_0} \]

6. **Determine the Time Period:**
The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi R}{v_0} \]

7. **Calculate the Coordinates at Time \( t = \frac{\pi}{B_0 \alpha} \):**
Since \( \alpha = \frac{q}{m} \), we can express \( q \) as \( q = \alpha m \). The time \( t \) corresponds to half the time period:
\[ t = \frac{\pi}{B_0 \alpha} = \frac{T}{2} \]
At this time, the particle has completed a half-circle in the \( xz \)-plane and has moved downward in the \( z \)-direction.

8. **Calculate the Displacement in the \( z \)-Direction:**
The displacement in the \( z \)-direction at time \( t \) is:
\[ z = -R \]

9. **Calculate the Displacement in the \( x \)-Direction:**
The displacement in the \( x \)-direction at time \( t \) is:
\[ x = R \]

10. **Final Coordinates:**
Therefore, the coordinates of the particle at time \( t = \frac{\pi}{B_0 \alpha} \) are:
\[ (x, y, z) = \left( \frac{mv_0}{qB_0}, 0, -\frac{mv_0}{qB_0} \right) \]

### Final Answer:
The coordinates of the particle at time \( t = \frac{\pi}{B_0 \alpha} \) are:
\[ \left( \frac{mv_0}{qB_0}, 0, -\frac{mv_0}{qB_0} \right) \]