In a region of space, a uniform magnetic field `B` is along positive x-axis. Electrons are ernit.u.d from the origin with speed `u` at different, angles. Show that the paraxial electrons are refocused on the x-axis at a distance `(2pimv)/(Be)`. Here, `m` is the mass of electron and e the charge on it.

Explanation

To solve the problem, we need to analyze the motion of electrons emitted from the origin in a uniform magnetic field directed along the positive x-axis. The goal is to show that paraxial electrons refocus on the x-axis at a distance of \( \frac{2\pi mv}{Be} \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform magnetic field \( \mathbf{B} \) directed along the positive x-axis.
- Electrons are emitted from the origin with speed \( u \) at different angles \( \theta \).

2. **Motion of Electrons in a Magnetic Field**:
- When charged particles (like electrons) move in a magnetic field, they experience a magnetic force given by:
\[ \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \]
- Here, \( q \) is the charge of the electron, \( \mathbf{v} \) is the velocity vector of the electron, and \( \mathbf{B} \) is the magnetic field vector.

3. **Determining the Components of Velocity**:
- The velocity \( \mathbf{u} \) can be decomposed into components:
- \( u_x = u \cos \theta \) (along the x-axis)
- \( u_y = u \sin \theta \) (along the y-axis)

4. **Radius of Circular Motion**:
- The electrons will undergo circular motion in the plane perpendicular to the magnetic field. The radius \( r \) of this circular motion is given by:
\[ r = \frac{mv}{eB} \]
- Here, \( m \) is the mass of the electron, \( v \) is the speed of the electron, \( e \) is the charge of the electron, and \( B \) is the magnetic field strength.

5. **Time Period of Circular Motion**:
- The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi m}{eB} \]

6. **Distance Traveled in the x-direction**:
- The distance traveled along the x-axis during one complete revolution can be calculated as:
\[ \text{Distance} = \text{Velocity in x-direction} \times \text{Time Period} \]
- The velocity in the x-direction is \( u \cos \theta \), so:
\[ \text{Distance} = (u \cos \theta) \times T = (u \cos \theta) \times \frac{2\pi m}{eB} \]

7. **Focusing Condition for Paraxial Electrons**:
- For paraxial electrons, we consider \( \theta \) to be very small (close to zero). Thus, \( \cos \theta \approx 1 \).
- Therefore, the distance becomes:
\[ \text{Distance} = u \times \frac{2\pi m}{eB} \]

8. **Final Distance for Paraxial Electrons**:
- Since we are interested in the distance where the electrons refocus on the x-axis, we can set \( u = v \) (the speed of the electrons):
\[ \text{Distance} = \frac{2\pi mv}{eB} \]

Thus, we have shown that paraxial electrons are refocused on the x-axis at a distance \( \frac{2\pi mv}{eB} \).