A particle of charge per unit mass `alpha` is released from origin with a velocity `vecv=v_(0)hati` uniform magnetic field `vecB=-B_(0)hatk`. If the particle passes through `(0,y,0)`, then `y` is equal to

Options

A. `-(2v_(0))/(B_(0)alpha)`
B. `(v_(0))/(B_(0)alpha)`
C. `(2v_(0))/(B_(0)alpha)`
D. `-(v_(0))/(B_(0)alpha)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a uniform magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the motion of the charged particle
The particle is released from the origin with an initial velocity \( \vec{v} = v_0 \hat{i} \) and is subjected to a uniform magnetic field \( \vec{B} = -B_0 \hat{k} \). The magnetic field is directed along the negative z-axis.

### Step 2: Determine the force acting on the particle
The force acting on a charged particle in a magnetic field is given by the Lorentz force law:
\[ \vec{F} = q(\vec{v} \times \vec{B}) \]
Since we have the charge per unit mass \( \alpha \), we can express the charge \( q \) in terms of mass \( m \) as \( q = \alpha m \).

### Step 3: Calculate the cross product \( \vec{v} \times \vec{B} \)
Substituting the values of \( \vec{v} \) and \( \vec{B} \):
\[ \vec{F} = \alpha m (v_0 \hat{i} \times (-B_0 \hat{k})) \]
Calculating the cross product:
\[ \hat{i} \times \hat{k} = \hat{j} \quad \Rightarrow \quad \vec{F} = -\alpha m v_0 B_0 \hat{j} \]
This force acts in the negative y-direction.

### Step 4: Determine the radius of the circular motion
The particle will undergo circular motion due to the magnetic force. The radius \( r \) of the circular path can be determined using the formula:
\[ r = \frac{mv}{qB} \]
Substituting \( q = \alpha m \):
\[ r = \frac{mv_0}{\alpha m B_0} = \frac{v_0}{\alpha B_0} \]

### Step 5: Find the position of the particle when it passes through \( (0, y, 0) \)
In circular motion, the particle will complete a semicircle to reach the point \( (0, y, 0) \). The distance \( y \) will be equal to the diameter of the circular path:
\[ y = 2r = 2 \left(\frac{v_0}{\alpha B_0}\right) = \frac{2v_0}{\alpha B_0} \]

### Final Answer
Thus, the value of \( y \) when the particle passes through \( (0, y, 0) \) is:
\[ y = \frac{2v_0}{\alpha B_0} \]

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