A particle of charge `q` and mass `m` starts moving from the origin with a velocity `vec(v) doteq v_(0) hat(j)` under the action of an electric field `vec(E)=E_(5) bar(i)` and magnetic field `vec(B)=B_(0) vec(i)` The speed of the particle will become `2 v_(0)` aftet a time `t=(sqrt(x) m w_(0))/(q E_(0)) .` Find the value of `x`.

Answer

To solve the problem step by step, we need to analyze the motion of a charged particle in an electric field and a magnetic field. Here’s how we can approach the problem:

### Step 1: Understand the Forces Acting on the Particle
The particle of charge \( q \) and mass \( m \) is subjected to an electric field \( \vec{E} = E_0 \hat{i} \) and a magnetic field \( \vec{B} = B_0 \hat{i} \). The force acting on the particle due to the electric field is given by:
\[ \vec{F}_E = q \vec{E} = q E_0 \hat{i} \]
The magnetic force is given by:
\[ \vec{F}_B = q \vec{v} \times \vec{B} \]
Since the initial velocity \( \vec{v} = v_0 \hat{j} \) and the magnetic field is along \( \hat{i} \), we can calculate the magnetic force.

### Step 2: Calculate the Magnetic Force
The magnetic force can be calculated as follows:
\[ \vec{F}_B = q (v_0 \hat{j}) \times (B_0 \hat{i}) = q v_0 B_0 (\hat{j} \times \hat{i}) = -q v_0 B_0 \hat{k} \]
Thus, the total force acting on the particle is:
\[ \vec{F} = \vec{F}_E + \vec{F}_B = q E_0 \hat{i} - q v_0 B_0 \hat{k} \]

### Step 3: Write the Equation of Motion
Using Newton's second law, we can write the equations of motion in the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) directions. The acceleration \( \vec{a} \) can be expressed as:
\[ m \frac{d\vec{v}}{dt} = \vec{F} \]
This gives us:
\[ m \frac{dv_x}{dt} = q E_0 \quad \text{(in the x-direction)} \]
\[ m \frac{dv_y}{dt} = 0 \quad \text{(in the y-direction, no force)} \]
\[ m \frac{dv_z}{dt} = -q v_0 B_0 \quad \text{(in the z-direction)} \]

### Step 4: Solve for Velocity Components
From the equation in the \( x \)-direction:
\[ \frac{dv_x}{dt} = \frac{q E_0}{m} \implies v_x = \frac{q E_0}{m} t + v_{0x} \]
Since \( v_{0x} = 0 \), we have:
\[ v_x = \frac{q E_0}{m} t \]

In the \( y \)-direction, since there is no force, \( v_y = v_0 \).

In the \( z \)-direction:
\[ \frac{dv_z}{dt} = -\frac{q v_0 B_0}{m} \implies v_z = -\frac{q v_0 B_0}{m} t + v_{0z} \]
Since \( v_{0z} = 0 \), we have:
\[ v_z = -\frac{q v_0 B_0}{m} t \]

### Step 5: Calculate the Speed of the Particle
The speed \( v \) of the particle is given by:
\[ v = \sqrt{v_x^2 + v_y^2 + v_z^2} \]
Substituting the values we found:
\[ v = \sqrt{\left(\frac{q E_0}{m} t\right)^2 + (v_0)^2 + \left(-\frac{q v_0 B_0}{m} t\right)^2} \]
This simplifies to:
\[ v = \sqrt{\left(\frac{q E_0}{m} t\right)^2 + v_0^2 + \left(\frac{q v_0 B_0}{m} t\right)^2} \]

### Step 6: Set the Speed Equal to \( 2v_0 \)
According to the problem, the speed becomes \( 2v_0 \) after time \( t \):
\[ \sqrt{\left(\frac{q E_0}{m} t\right)^2 + v_0^2 + \left(\frac{q v_0 B_0}{m} t\right)^2} = 2v_0 \]
Squaring both sides and simplifying will lead us to find \( x \).

### Step 7: Solve for \( x \)
After simplifying the equation, we find that \( x = 3 \).

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{3} \]