A particle of mass `m` and charge `q` moves with a constant velocity `v` along the positive `x` direction. It enters a region containing a uniform magnetic field `B` directed along the negative `z` direction, extending from `x = a` to `x = b`. The minimum value of `v` required so that the particle can just enter the region `x gt b` is

Options

A. `qpB//m`
B. `q(b - a)B//m`
C. `qaB//m`
D. `q(b + a)B//2m`

Answer

To solve the problem, we need to determine the minimum velocity \( v \) required for a charged particle to just enter the region beyond \( x = b \) when it moves through a magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the motion of the particle
The particle of mass \( m \) and charge \( q \) is moving along the positive \( x \)-direction with a constant velocity \( v \). When it enters the magnetic field region, which is directed along the negative \( z \)-direction, it will experience a magnetic force.

**Hint:** Recall that the magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force equation:
\[ F = q(\mathbf{v} \times \mathbf{B}) \]

### Step 2: Determine the direction of the magnetic force
Using the right-hand rule, if the velocity \( \mathbf{v} \) is in the positive \( x \)-direction and the magnetic field \( \mathbf{B} \) is in the negative \( z \)-direction, the magnetic force \( \mathbf{F} \) will be directed in the positive \( y \)-direction.

**Hint:** Visualize the right-hand rule: point your thumb in the direction of \( \mathbf{v} \) and your fingers in the direction of \( \mathbf{B} \) to find the direction of \( \mathbf{F} \).

### Step 3: Analyze the particle's trajectory
The particle will undergo circular motion due to the magnetic force acting as the centripetal force. The radius \( r \) of this circular motion can be expressed as:
\[ r = \frac{mv}{qB} \]

**Hint:** Remember that the centripetal force required for circular motion is provided by the magnetic force.

### Step 4: Relate the radius to the magnetic field region
For the particle to just enter the region beyond \( x = b \), the radius of the circular path must be equal to the width of the magnetic field region, which is \( b - a \). Therefore, we set:
\[ \frac{mv}{qB} = b - a \]

**Hint:** This equation relates the radius of the circular path to the dimensions of the magnetic field region.

### Step 5: Solve for the velocity \( v \)
Rearranging the equation gives:
\[ v = \frac{qB(b - a)}{m} \]

This equation gives us the minimum velocity required for the particle to just enter the region \( x > b \).

**Hint:** Ensure you understand how each variable in the equation affects the velocity.

### Final Answer
The minimum value of \( v \) required for the particle to just enter the region \( x > b \) is:
\[ v = \frac{qB(b - a)}{m} \]