In a certain region of space, there exists a uniform and constant electric field of magnitude E along the positive y-axis of a coordinate system. A charged particle of mass m and charge -q (qgt0) is projected form the origin with speed 2v at an angle of `60^@` with the positive x-axis in x-y plane. When the x-coordinate of particle becomes `sqrt3mv^2//qE` a uniform and constant magnetic field of strength B is also switched on along positive y-axis.
Velocity of the particle just before the magnetic field is switched on is

Options

A. `vhati`
B. `vhati+(sqrt3v)/2 hatj`
C. `vhati-(sqrt3v)/2 hatj`
D. `2vhati-(sqrt3v)/2 hatj`

Explanation

To solve the problem step by step, we will analyze the motion of the charged particle under the influence of the electric field before the magnetic field is switched on.

### Step 1: Analyze the initial conditions
The charged particle is projected from the origin with a speed of \(2v\) at an angle of \(60^\circ\) with the positive x-axis. We can find the initial velocity components in the x and y directions.

- **Initial velocity in the x-direction**:
\[ u_x = 2v \cos(60^\circ) = 2v \cdot \frac{1}{2} = v \]

- **Initial velocity in the y-direction**:
\[ u_y = 2v \sin(60^\circ) = 2v \cdot \frac{\sqrt{3}}{2} = \sqrt{3}v \]

### Step 2: Determine the acceleration due to the electric field
The electric field \(E\) acts along the positive y-axis, and since the charge of the particle is \(-q\), the force acting on it will be in the negative y-direction. The acceleration \(a_y\) can be calculated using Newton's second law:

\[ F = ma \Rightarrow a_y = \frac{F}{m} = -\frac{qE}{m} \]

### Step 3: Find the time when the x-coordinate reaches \(\frac{\sqrt{3}mv^2}{qE}\)
The x-coordinate of the particle as a function of time \(t\) is given by:

\[ x = u_x t \]

Setting \(x = \frac{\sqrt{3}mv^2}{qE}\):

\[ \frac{\sqrt{3}mv^2}{qE} = vt \]

Solving for \(t\):

\[ t = \frac{\sqrt{3}mv^2}{qEv} \]

### Step 4: Calculate the y-velocity just before the magnetic field is switched on
The y-velocity \(v_y\) at time \(t\) can be calculated using the equation of motion:

\[ v_y = u_y + a_y t \]

Substituting the values:

\[ v_y = \sqrt{3}v - \frac{qE}{m} \cdot \frac{\sqrt{3}mv^2}{qEv} \]

Simplifying this expression:

\[ v_y = \sqrt{3}v - \frac{\sqrt{3}mv^2}{m} = \sqrt{3}v - \sqrt{3}v = 0 \]

### Step 5: Determine the total velocity just before the magnetic field is switched on
The total velocity vector \( \vec{v} \) of the particle just before the magnetic field is switched on is given by:

\[ \vec{v} = v_x \hat{i} + v_y \hat{j} \]

Substituting the values we found:

\[ \vec{v} = v \hat{i} + 0 \hat{j} = v \hat{i} \]

### Final Answer
The velocity of the particle just before the magnetic field is switched on is:
\[ \vec{v} = v \hat{i} \]