A uniform magnetic field `barB =B_(0)hatj` exists in a 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)` The maximum value v for which the particle does not hit `y-z` plane is .

Options

A. `(2Bq)/(dm)`
B. `(Bqd)/(m)`
C. `(Bq)/(2dm)`
D. `(Bqd)/(2m)`

Answer

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field. Let's break down the solution step by step:

### Step 1: Understand the Problem
We have a uniform magnetic field \(\bar{B} = B_0 \hat{j}\) and a particle with mass \(m\) and charge \(q\) projected towards the negative x-axis from the point \((d, 0, 0)\). We need to find the maximum speed \(v\) such that the particle does not hit the y-z plane.

### Step 2: Identify the Forces Acting on the Particle
When a charged particle moves in a magnetic field, it experiences a magnetic force given by the Lorentz force law:
\[ \bar{F} = q(\bar{v} \times \bar{B}) \]
Here, \(\bar{v}\) is the velocity of the particle and \(\bar{B}\) is the magnetic field.

### Step 3: Determine the Direction of the Force
Since the particle is moving in the negative x-direction (let's denote this as \(-\hat{i}\)), and the magnetic field is in the positive y-direction (\(\hat{j}\)), we can calculate the force:
\[ \bar{F} = q(-\hat{i} \times B_0 \hat{j}) = q(-B_0 \hat{k}) \]
This force acts in the negative z-direction, causing the particle to move in a circular path in the x-y plane.

### Step 4: Calculate the Radius of the Circular Motion
The radius \(r\) of the circular motion of the particle can be determined using the formula:
\[ r = \frac{mv}{qB_0} \]

### Step 5: Condition for Not Hitting the y-z Plane
For the particle to not hit the y-z plane, the distance \(d\) from the origin must be greater than the diameter of the circular path. Therefore, we have the condition:
\[ d > 2r \]
Substituting the expression for \(r\):
\[ d > 2 \left(\frac{mv}{qB_0}\right) \]

### Step 6: Rearranging the Inequality
Rearranging the inequality gives:
\[ d > \frac{2mv}{qB_0} \]
From this, we can express \(v\) in terms of \(d\):
\[ v < \frac{qB_0 d}{2m} \]

### Step 7: Conclusion
The maximum value of \(v\) for which the particle does not hit the y-z plane is:
\[ v_{\text{max}} = \frac{qB_0 d}{2m} \]

### Final Answer
Thus, the maximum speed \(v\) for which the particle does not hit the y-z plane is:
\[ \boxed{\frac{qB_0 d}{2m}} \]