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`.
Maximum z-coordinate of the particle is

Options

A. `(sqrt3mv_(0))/(B_(0)q)`
B. `(2sqrt3mv_(0))/(B_(0)q)`
C. `(2mv_(0))/(B_(0)q)`
D. `(mv_(0))/(B_(0)q)`

Answer

To solve the problem of finding the maximum z-coordinate of a charged particle released in a uniform magnetic field, we can follow these steps:

### Step 1: Identify the given information
- Charge of the particle: \( q \)
- Mass of the particle: \( m \)
- Initial velocity: \( \mathbf{v} = v_0 \hat{i} \)
- Magnetic field: \( \mathbf{B} = \frac{B_0}{2} \hat{i} + \frac{\sqrt{3}B_0}{2} \hat{j} \)

### Step 2: Calculate the magnetic force on the particle
The magnetic force \( \mathbf{F} \) on a charged particle moving in a magnetic field is given by:
\[ \mathbf{F} = q \mathbf{v} \times \mathbf{B} \]
Using the cross product, we can set up the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_0 & 0 & 0 \\ \frac{B_0}{2} & \frac{\sqrt{3}B_0}{2} & 0 \end{vmatrix} \]

### Step 3: Calculate the cross product
Calculating the determinant gives:
\[ \mathbf{F} = q \left( 0 - 0 \hat{i} + 0 - 0 \hat{j} + v_0 \cdot \frac{\sqrt{3}B_0}{2} \hat{k} - 0 \right) = q v_0 \frac{\sqrt{3}B_0}{2} \hat{k} \]
Thus, the magnetic force acting on the particle is:
\[ \mathbf{F} = \frac{q v_0 \sqrt{3}B_0}{2} \hat{k} \]

### Step 4: Determine the motion of the particle
The magnetic force acts as a centripetal force, causing the particle to move in a circular path in the x-z plane. The radius \( R \) of this circular motion can be found using the relationship between centripetal force and magnetic force:
\[ F = \frac{mv^2}{R} = qvB \]
Here, \( v \) is the component of velocity perpendicular to the magnetic field. The angle \( \theta \) between the velocity vector and the magnetic field can be found using the components of the magnetic field:
\[ \tan \theta = \frac{\frac{\sqrt{3}B_0}{2}}{\frac{B_0}{2}} = \sqrt{3} \]
This means \( \theta = 60^\circ \).

### Step 5: Find the perpendicular component of the velocity
The perpendicular component of the velocity \( v_\perp \) is:
\[ v_\perp = v_0 \sin(60^\circ) = v_0 \cdot \frac{\sqrt{3}}{2} \]

### Step 6: Calculate the radius of the circular motion
Using the centripetal force equation:
\[ R = \frac{mv_\perp}{qB} \]
Substituting \( v_\perp \):
\[ R = \frac{m \cdot \frac{\sqrt{3}}{2} v_0}{q \cdot \frac{B_0}{2}} = \frac{m \sqrt{3} v_0}{q B_0} \]

### Step 7: Calculate the maximum z-coordinate
The maximum z-coordinate occurs when the particle is at the top of its circular path, which is twice the radius:
\[ z_{\text{max}} = 2R = 2 \cdot \frac{m \sqrt{3} v_0}{q B_0} = \frac{2m \sqrt{3} v_0}{q B_0} \]

### Final Answer
The maximum z-coordinate of the particle is:
\[ z_{\text{max}} = \frac{2m \sqrt{3} v_0}{q B_0} \]