A uniform magnetic ficld `B` ы `B_(0) j` exists in space. A particle of mass `m` and charge `q` is projected towards negative `x` -axis with speed `v` from the a point `(d, 0,0)`. If the maximum value of `v` for which the particle does not hit `y^(-z)` plane is `(4 B_(0) q d)/(alpha m)`, then calculate `alpha`.

Answer

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field. The particle is projected towards the negative x-axis, and we need to determine the maximum speed \( v \) such that the particle does not hit the \( y-z \) plane.

### Step-by-Step Solution:

1. **Understanding the Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts perpendicular to both the velocity of the particle and the magnetic field direction. This causes the particle to move in a circular path.

2. **Magnetic Force**:
The magnetic force \( F \) acting on the particle is given by:
\[ F = q(v \times B) \]
Since the magnetic field \( B = B_0 \hat{j} \) and the velocity \( v = -v \hat{i} \), we can compute the cross product:
\[ v \times B = (-v \hat{i}) \times (B_0 \hat{j}) = -v B_0 (\hat{i} \times \hat{j}) = -v B_0 \hat{k} \]
Thus, the magnetic force becomes:
\[ F = -q v B_0 \hat{k} \]
This force acts in the negative z-direction.

3. **Radius of the Circular Path**:
The radius \( r \) of the circular path followed by the particle can be expressed as:
\[ r = \frac{mv}{qB} \]
Substituting \( B = B_0 \):
\[ r = \frac{mv}{qB_0} \]

4. **Condition for Not Hitting the y-z Plane**:
The particle starts at the point \( (d, 0, 0) \). The particle will not hit the \( y-z \) plane if the radius of the circular path is less than or equal to the distance from the x-axis to the y-z plane, which is \( d \):
\[ r \leq d \]
Therefore, we have:
\[ \frac{mv}{qB_0} \leq d \]

5. **Rearranging for Maximum Speed**:
Rearranging the inequality gives us:
\[ v \leq \frac{qB_0 d}{m} \]
The maximum value of \( v \) for which the particle does not hit the \( y-z \) plane is:
\[ v_{\text{max}} = \frac{qB_0 d}{m} \]

6. **Comparing with Given Expression**:
According to the problem, the maximum speed is given as:
\[ v_{\text{max}} = \frac{4B_0 q d}{\alpha m} \]
Setting the two expressions for \( v_{\text{max}} \) equal gives:
\[ \frac{qB_0 d}{m} = \frac{4B_0 q d}{\alpha m} \]

7. **Solving for \( \alpha \)**:
Canceling common terms from both sides (assuming \( q, B_0, d, m \neq 0 \)):
\[ 1 = \frac{4}{\alpha} \]
Thus, we find:
\[ \alpha = 4 \]

### Final Answer:
\[ \alpha = 4 \]