A particle of charge q, mass starts moving from origin under the action of an electric field `vecE=E_(0)hati` and magnetic field `vecB=B_(0)hatk` . Its velocity at `(x,0,0)` is `v_(0)(6hati+8hatj)`. The value of is

Options

A. `(25mv_(0)^(2))/(qE_(0))`
B. `(100mv_(0)^(2))/(qB_(0))`
C. `(50mv_(0)^(2))/(qE_(0))`
D. `(14mv_(0)^(2))/(qE_(0))`

Explanation

To solve the problem, we need to analyze the motion of a charged particle under the influence of an electric field and a magnetic field. Let's break down the steps:

### Step 1: Understand the Forces Acting on the Particle
The particle experiences two forces:
1. Electric Force (\( \vec{F}_E \)): Given by \( \vec{F}_E = q \vec{E} \)
2. Magnetic Force (\( \vec{F}_B \)): Given by \( \vec{F}_B = q \vec{v} \times \vec{B} \)

Where:
- \( \vec{E} = E_0 \hat{i} \)
- \( \vec{B} = B_0 \hat{k} \)
- \( \vec{v} = v_0(6 \hat{i} + 8 \hat{j}) \)

### Step 2: Calculate the Magnetic Force
The magnetic force can be calculated using the cross product:
\[ \vec{F}_B = q \vec{v} \times \vec{B} \]
Calculating \( \vec{v} \times \vec{B} \):
\[ \vec{v} = v_0(6 \hat{i} + 8 \hat{j}), \quad \vec{B} = B_0 \hat{k} \]
\[ \vec{v} \times \vec{B} = v_0(6 \hat{i} + 8 \hat{j}) \times B_0 \hat{k} \]
Using the determinant method for the cross product:
\[ \vec{v} \times \vec{B} = v_0 B_0 \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 8 & 0 \\ 0 & 0 & 1 \end{vmatrix} = v_0 B_0 (8 \hat{i} - 6 \hat{j}) \]
Thus,
\[ \vec{F}_B = q v_0 B_0 (8 \hat{i} - 6 \hat{j}) \]

### Step 3: Write the Equation of Motion
The net force acting on the particle is:
\[ \vec{F}_{net} = \vec{F}_E + \vec{F}_B = q E_0 \hat{i} + q v_0 B_0 (8 \hat{i} - 6 \hat{j}) \]
This results in:
\[ \vec{F}_{net} = q \left( E_0 + 8 v_0 B_0 \right) \hat{i} - q (6 v_0 B_0) \hat{j} \]

### Step 4: Use Work-Energy Theorem
The work done by the electric force is equal to the change in kinetic energy:
\[ W = \Delta KE = \frac{1}{2} m v^2 - 0 = \frac{1}{2} m v^2 \]
Where \( v \) is the magnitude of the final velocity:
\[ v = \sqrt{(6 v_0)^2 + (8 v_0)^2} = v_0 \sqrt{36 + 64} = v_0 \sqrt{100} = 10 v_0 \]
Thus,
\[ \Delta KE = \frac{1}{2} m (10 v_0)^2 = 50 m v_0^2 \]

### Step 5: Work Done by Electric Field
The work done by the electric field is:
\[ W = q E_0 x \]
Setting the work done equal to the change in kinetic energy:
\[ q E_0 x = 50 m v_0^2 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x = \frac{50 m v_0^2}{q E_0} \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = \frac{50 m v_0^2}{q E_0} \]