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.
x-coordinate of the particle as a function of time after the magnetic field is switched on is

Options

A. `(sqrt3mv^2)/(qE)-(mv)/(qB) sin ((qB)/mt)`
B. `(sqrt3mv^2)/(qE)+(mv)/(qB) sin ((qB)/mt)`
C. `(sqrt3mv^2)/(qE)-(mv)/(qB) cos ((qB)/mt)`
D. `(sqrt3mv^2)/(qE)+(mv)/(qB) cos ((qB)/mt)`

Explanation

To solve the problem step by step, we will analyze the motion of the charged particle under the influence of the electric field and then the magnetic field.

### Step 1: Analyze the initial conditions
The charged particle of mass \( m \) and charge \( -q \) is projected from the origin with speed \( 2v \) at an angle of \( 60^\circ \) with the positive x-axis.

**Components of the initial velocity:**
- The x-component of the velocity:
\[ v_x = 2v \cos(60^\circ) = 2v \cdot \frac{1}{2} = v \]
- The y-component of the velocity:
\[ v_y = 2v \sin(60^\circ) = 2v \cdot \frac{\sqrt{3}}{2} = v\sqrt{3} \]

### Step 2: Determine the effect of the electric field
The electric field \( E \) is directed along the positive y-axis. Since the charge of the particle is negative, the force \( F \) on the particle due to the electric field is directed downwards (negative y-direction).

**Force due to electric field:**
\[ F = -qE \]

**Acceleration of the particle:**
Using Newton's second law, the acceleration \( a \) in the y-direction is:
\[ a_y = \frac{F}{m} = \frac{-qE}{m} \]

### Step 3: Calculate the time to reach a specific x-coordinate
We need to find the time \( t \) when the x-coordinate of the particle becomes:
\[ x = \frac{\sqrt{3}mv^2}{qE} \]

Since the x-component of the velocity is constant (no force in the x-direction):
\[ x = v_x \cdot t \implies t = \frac{x}{v_x} = \frac{\frac{\sqrt{3}mv^2}{qE}}{v} = \frac{\sqrt{3}mv}{qE} \]

### Step 4: Analyze the motion in the y-direction at time \( t \)
At time \( t \), the y-coordinate can be calculated using the kinematic equation:
\[ y = v_y \cdot t + \frac{1}{2} a_y t^2 \]
Substituting the values:
\[ y = v\sqrt{3} \cdot \frac{\sqrt{3}mv}{qE} + \frac{1}{2} \left(-\frac{qE}{m}\right) \left(\frac{\sqrt{3}mv}{qE}\right)^2 \]

Calculating each term:
1. The first term:
\[ y_1 = v\sqrt{3} \cdot \frac{\sqrt{3}mv}{qE} = \frac{3mv^2}{qE} \]
2. The second term:
\[ y_2 = \frac{1}{2} \left(-\frac{qE}{m}\right) \cdot \frac{3m^2v^2}{q^2E^2} = -\frac{3mv^2}{2qE} \]

Combining both terms gives:
\[ y = \frac{3mv^2}{qE} - \frac{3mv^2}{2qE} = \frac{3mv^2}{2qE} \]

### Step 5: Switch on the magnetic field
When the magnetic field \( B \) is switched on, it acts along the positive y-axis. The velocity of the particle in the x-direction remains \( v \) (since there is no force acting in the x-direction).

### Step 6: Determine the circular motion in the magnetic field
The magnetic force will cause the particle to move in a circular path. The radius \( R \) of the circular motion is given by:
\[ R = \frac{mv}{|q|B} \]

### Step 7: Write the x-coordinate as a function of time
The x-coordinate of the particle at any time \( t \) after the magnetic field is switched on can be expressed as:
\[ x(t) = \frac{\sqrt{3}mv^2}{qE} + R \sin(\omega t) \]
where \( \omega = \frac{|q|B}{m} \).

### Final Expression
Thus, the x-coordinate of the particle as a function of time after the magnetic field is switched on is:
\[ x(t) = \frac{\sqrt{3}mv^2}{qE} + \frac{mv}{|q|B} \sin\left(\frac{|q|B}{m} t\right) \]