A particle having charge `'q'` and mass `'m'` is projected with velocity `(4hati-6hatj+3hatk) m//sec` from the origin in a region occupied by electric field `'E'` and magnetic field `'B'` such that `vecB=B_(0)hati` and `vecE=E_(0)hatj` (take `(qE_(0))/m=2`). Find the time (in sec) when the magnitude of velocity of the charge particle becomes `5sqrt(5) m//sec`. (neglect the gravity)

Explanation

To solve the problem, we need to analyze the motion of the charged particle under the influence of the electric and magnetic fields. The magnetic field does not do work on the particle, so we will only consider the effect of the electric field on the particle's velocity.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- The initial velocity of the particle is given as:
\[ \vec{u} = 4\hat{i} - 6\hat{j} + 3\hat{k} \, \text{m/s} \]
- The electric field is:
\[ \vec{E} = E_0 \hat{j} \]
- The magnetic field is:
\[ \vec{B} = B_0 \hat{i} \]
- Given that \(\frac{qE_0}{m} = 2\), we can denote the acceleration due to the electric field as:
\[ a_y = \frac{qE_0}{m} = 2 \, \text{m/s}^2 \]

2. **Determine the change in the y-component of velocity:**
- The y-component of the velocity changes due to the electric field. The equation of motion for the y-component is:
\[ v_y = u_y + a_y t \]
- Here, \(u_y = -6 \, \text{m/s}\) and \(a_y = 2 \, \text{m/s}^2\). Thus, we have:
\[ v_y = -6 + 2t \]

3. **Calculate the magnitude of the velocity:**
- The total velocity vector is:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \]
- The x-component \(v_x\) and z-component \(v_z\) remain unchanged since the magnetic field does not do work on the particle. Thus:
\[ v_x = 4 \, \text{m/s}, \quad v_z = 3 \, \text{m/s} \]
- The magnitude of the velocity is given by:
\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]
- We want this magnitude to equal \(5\sqrt{5} \, \text{m/s}\):
\[ |\vec{v}| = 5\sqrt{5} \]

4. **Set up the equation for the magnitude:**
- Substitute the known values into the magnitude equation:
\[ 5\sqrt{5} = \sqrt{4^2 + (-6 + 2t)^2 + 3^2} \]
- Squaring both sides gives:
\[ 125 = 16 + (-6 + 2t)^2 + 9 \]
- Simplifying this, we have:
\[ 125 = 25 + (-6 + 2t)^2 \]
\[ 100 = (-6 + 2t)^2 \]

5. **Solve for \(t\):**
- Taking the square root of both sides:
\[ 10 = -6 + 2t \quad \text{or} \quad -10 = -6 + 2t \]
- Solving the first equation:
\[ 2t = 16 \quad \Rightarrow \quad t = 8 \, \text{s} \]
- Solving the second equation:
\[ 2t = -4 \quad \Rightarrow \quad t = -2 \, \text{s} \quad (\text{not physically meaningful}) \]

6. **Final answer:**
- The time when the magnitude of the velocity of the charged particle becomes \(5\sqrt{5} \, \text{m/s}\) is:
\[ t = 8 \, \text{s} \]