A particle of mass m and positive charge q is projected with a speed of `v_0` in y–direction in the presence of electric and magnetic field are in x–direction. Find the instant of time at which the speed of particle becomes double the initial speed.

Options

A. `(sqrt3 m v_0)/(qE)`
B. `(sqrt2 m v_0)/(qE)`
C. `(m v_0)/(qE)`
D. `(sqrt3 m v_0)/(2qE)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in the presence of electric and magnetic fields. The particle is projected in the y-direction, while both fields are aligned in the x-direction. Our goal is to find the time at which the speed of the particle becomes double its initial speed.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- The particle has a mass \( m \) and a charge \( q \).
- It is projected with an initial speed \( v_0 \) in the y-direction.
- The electric field \( E \) and magnetic field \( B \) are both in the x-direction.

2. **Force Analysis**:
- The magnetic force does not do any work on the particle because it acts perpendicular to the direction of motion. Therefore, it does not change the speed of the particle.
- The electric field exerts a force on the particle given by \( F = qE \).
- The acceleration \( a \) of the particle in the x-direction can be expressed as:
\[ a_x = \frac{F}{m} = \frac{qE}{m} \]

3. **Velocity in the x-direction**:
- The initial velocity in the x-direction \( v_{x0} = 0 \).
- The velocity in the x-direction at time \( t \) can be expressed as:
\[ v_x = v_{x0} + a_x t = 0 + \frac{qE}{m} t = \frac{qE}{m} t \]

4. **Velocity in the y-direction**:
- The velocity in the y-direction remains constant:
\[ v_y = v_0 \]

5. **Finding the Resultant Speed**:
- The resultant speed \( v_{net} \) is given by:
\[ v_{net} = \sqrt{v_x^2 + v_y^2} \]
- We want to find the time when the resultant speed is double the initial speed:
\[ v_{net} = 2v_0 \]
- Squaring both sides gives:
\[ (2v_0)^2 = v_x^2 + v_y^2 \]
\[ 4v_0^2 = v_x^2 + v_0^2 \]
- Rearranging gives:
\[ v_x^2 = 4v_0^2 - v_0^2 = 3v_0^2 \]
- Taking the square root:
\[ v_x = \sqrt{3} v_0 \]

6. **Finding Time \( t \)**:
- We already have the expression for \( v_x \):
\[ v_x = \frac{qE}{m} t \]
- Setting this equal to \( \sqrt{3} v_0 \):
\[ \sqrt{3} v_0 = \frac{qE}{m} t \]
- Solving for \( t \):
\[ t = \frac{\sqrt{3} mv_0}{qE} \]

### Final Answer:
The time at which the speed of the particle becomes double the initial speed is:
\[ t = \frac{\sqrt{3} mv_0}{qE} \]