A particle of charge q and mass m starts moving from the origin under the action of an electric field `vec E = E_0 hat i and vec B = B_0 hat i` with a velocity `vec v = v_0 hat j`. The speed of the particle will become `2v_0` after a time.

Options

A. `(sqrt2m upsilon_(0))/(qE_(0))`
B. `(2m upsilon_(0))/(qE_(0))`
C. `(3m upsilon_(0))/(qE_(0))`
D. `(sqrt3m upsilon_(0))/(qE_(0))`

Answer

To solve the problem, we will analyze the motion of a charged particle in the presence of electric and magnetic fields. The particle starts from the origin with an initial velocity and experiences forces due to the electric and magnetic fields.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Charge of the particle: \( q \)
- Mass of the particle: \( m \)
- Electric field: \( \vec{E} = E_0 \hat{i} \)
- Magnetic field: \( \vec{B} = B_0 \hat{i} \)
- Initial velocity: \( \vec{v} = v_0 \hat{j} \)
- Final speed: \( 2v_0 \)

2. **Determine the Forces Acting on the Particle:**
- The electric force acting on the particle is given by:
\[ \vec{F}_E = q \vec{E} = q E_0 \hat{i} \]
- The magnetic force acting on the particle is given by:
\[ \vec{F}_B = q \vec{v} \times \vec{B} \]
- Since \( \vec{v} = v_0 \hat{j} \) and \( \vec{B} = B_0 \hat{i} \), we can calculate the magnetic force:
\[ \vec{F}_B = q (v_0 \hat{j}) \times (B_0 \hat{i}) = q v_0 B_0 (\hat{j} \times \hat{i}) = -q v_0 B_0 \hat{k} \]

3. **Analyze the Motion:**
- The electric force \( \vec{F}_E \) acts in the \( \hat{i} \) direction, while the magnetic force \( \vec{F}_B \) acts in the \( -\hat{k} \) direction.
- The net force acting on the particle is:
\[ \vec{F}_{net} = \vec{F}_E + \vec{F}_B = q E_0 \hat{i} - q v_0 B_0 \hat{k} \]

4. **Calculate the Acceleration:**
- The acceleration \( \vec{a} \) can be found using Newton's second law:
\[ \vec{a} = \frac{\vec{F}_{net}}{m} = \frac{q E_0}{m} \hat{i} - \frac{q v_0 B_0}{m} \hat{k} \]

5. **Determine the Velocity Components:**
- The velocity in the \( \hat{i} \) direction (x-direction) after time \( t \) is:
\[ v_x = a_x t = \frac{q E_0}{m} t \]
- The velocity in the \( \hat{j} \) direction (y-direction) remains constant:
\[ v_y = v_0 \]
- The velocity in the \( \hat{k} \) direction (z-direction) due to the magnetic force will also change, but for speed calculation, we will focus on the resultant speed.

6. **Calculate the Resultant Speed:**
- The speed of the particle can be expressed as:
\[ v_{net} = \sqrt{v_x^2 + v_y^2} \]
- Setting this equal to the final speed \( 2v_0 \):
\[ \sqrt{\left(\frac{q E_0}{m} t\right)^2 + v_0^2} = 2v_0 \]
- Squaring both sides gives:
\[ \left(\frac{q E_0}{m} t\right)^2 + v_0^2 = 4v_0^2 \]
- Rearranging yields:
\[ \left(\frac{q E_0}{m} t\right)^2 = 3v_0^2 \]

7. **Solve for Time \( t \):**
- Taking the square root:
\[ \frac{q E_0}{m} t = \sqrt{3} v_0 \]
- Thus, solving for \( t \):
\[ t = \frac{m \sqrt{3} v_0}{q E_0} \]

### Final Answer:
The time \( t \) taken for the speed of the particle to become \( 2v_0 \) is:
\[ t = \frac{m \sqrt{3} v_0}{q E_0} \]