An electron is launched with velocity `vec v` in a uniform magnetic field `vec B.` The angle `theta` between `vec v and vec B` lies between 0 and `pi /2`. Its velocity vector `vec v` returns to its initial value in a time interval of

Options

A. `(2pi m)/(eB)`
B. `(2xx2pi m)/(eB)`
C. `(pi m)/(eB)`
D. depends upon between `vec v and vec B`

Explanation

To solve the problem, we need to determine the time interval after which the velocity vector of an electron, launched in a uniform magnetic field, returns to its initial value.

### Step-by-Step Solution:

1. **Understanding the Motion**:
When an electron moves in a magnetic field, it experiences a magnetic force that acts perpendicular to its velocity. This causes the electron to move in a circular path. The angle θ between the velocity vector (v) and the magnetic field vector (B) is between 0 and π/2.

2. **Components of Velocity**:
The velocity vector can be decomposed into two components:
- The component parallel to the magnetic field: \( v \cos \theta \)
- The component perpendicular to the magnetic field: \( v \sin \theta \)

3. **Magnetic Force**:
The magnetic force acting on the electron due to the perpendicular component of its velocity is given by:
\[ F = q(v \sin \theta)B \]
where \( q \) is the charge of the electron (denoted as \( e \)).

4. **Centripetal Force**:
This magnetic force provides the necessary centripetal force for circular motion:
\[ F = \frac{mv^2}{r} \]
where \( m \) is the mass of the electron and \( r \) is the radius of the circular path.

5. **Setting Forces Equal**:
By equating the magnetic force to the centripetal force, we have:
\[ e(v \sin \theta)B = \frac{mv^2}{r} \]

6. **Solving for Radius**:
Rearranging the equation gives:
\[ r = \frac{mv}{eB \sin \theta} \]

7. **Finding the Time Period**:
The time period \( T \) for one complete revolution (the time after which the velocity vector returns to its initial value) is given by:
\[ T = \frac{2\pi r}{v_{\text{perpendicular}}} \]
where \( v_{\text{perpendicular}} = v \sin \theta \).

Substituting the expression for \( r \):
\[ T = \frac{2\pi \left(\frac{mv}{eB \sin \theta}\right)}{v \sin \theta} = \frac{2\pi m}{eB \sin \theta} \]

8. **Conclusion**:
The time interval after which the velocity vector of the electron returns to its initial value is:
\[ T = \frac{2\pi m}{eB \sin \theta} \]