A charged particle (q.m) released from origin with velocity `v=v_(0)hati` in a uniform magnetic field `B=(B_(0))/(2)hati+(sqrt3B_(0))/(2)hatJ`.
Z-component of velocity is `(sqrt3v_(0))/(2)` after in t=……….

Options

A. `(2pim)/(B_(0)q)`
B. `(pim)/(B_(0)q)`
C. `(pim)/(2B_(0)q)`
D. `(2pim)/(4B_(0)q)`

Answer

To solve the problem, we need to determine the time \( t \) at which the z-component of the velocity of a charged particle becomes \( \frac{\sqrt{3}v_0}{2} \) when it is released from the origin with an initial velocity \( \vec{v} = v_0 \hat{i} \) in a uniform magnetic field \( \vec{B} = \frac{B_0}{2} \hat{i} + \frac{\sqrt{3}B_0}{2} \hat{j} \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Charge of the particle: \( q \)
- Mass of the particle: \( m \)
- Initial velocity: \( \vec{v} = v_0 \hat{i} \)
- Magnetic field: \( \vec{B} = \frac{B_0}{2} \hat{i} + \frac{\sqrt{3}B_0}{2} \hat{j} \)

2. **Calculate the Angle Between Velocity and Magnetic Field:**
- The magnetic field components are:
- \( B_x = \frac{B_0}{2} \)
- \( B_y = \frac{\sqrt{3}B_0}{2} \)
- The angle \( \theta \) can be found using:
\[ \tan(\theta) = \frac{B_y}{B_x} = \frac{\frac{\sqrt{3}B_0}{2}}{\frac{B_0}{2}} = \sqrt{3} \]
- Therefore, \( \theta = 60^\circ \).

3. **Find the Components of Velocity:**
- The component of velocity in the direction of the magnetic field:
\[ v_{\parallel} = v_0 \cos(60^\circ) = v_0 \cdot \frac{1}{2} = \frac{v_0}{2} \]
- The component of velocity perpendicular to the magnetic field:
\[ v_{\perpendicular} = v_0 \sin(60^\circ) = v_0 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}v_0}{2} \]

4. **Determine the Motion in the Magnetic Field:**
- The perpendicular component of velocity will cause the particle to undergo circular motion in the plane perpendicular to the magnetic field.
- The z-component of velocity will be influenced by this motion.

5. **Calculate the Time Period of Circular Motion:**
- The time period \( T \) of the circular motion of the charged particle in a magnetic field is given by:
\[ T = \frac{2\pi m}{qB} \]
- Here, the magnitude of the magnetic field \( B \) can be calculated as:
\[ B = |\vec{B}| = \sqrt{\left(\frac{B_0}{2}\right)^2 + \left(\frac{\sqrt{3}B_0}{2}\right)^2} = \sqrt{\frac{B_0^2}{4} + \frac{3B_0^2}{4}} = \sqrt{B_0^2} = B_0 \]
- Therefore, the time period becomes:
\[ T = \frac{2\pi m}{qB_0} \]

6. **Determine the Time \( t \) for the z-component of Velocity:**
- The z-component of velocity reaches \( \frac{\sqrt{3}v_0}{2} \) after a quarter of the time period:
\[ t = \frac{T}{4} = \frac{1}{4} \cdot \frac{2\pi m}{qB_0} = \frac{\pi m}{2qB_0} \]

### Final Answer:
\[ t = \frac{\pi m}{2qB_0} \]