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 x 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 step by step, we will use the work-energy theorem and the properties of electric and magnetic fields.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- A particle of charge \( q \) and mass \( m \) starts moving from the origin \( (0, 0, 0) \).
- The electric field is given by \( \vec{E} = E_0 \hat{i} \).
- The magnetic field is given by \( \vec{B} = B_0 \hat{k} \).
- The velocity of the particle at the point \( (x, 0, 0) \) is \( \vec{v}_0 = 6 \hat{i} + 8 \hat{j} \).

2. **Applying the Work-Energy Theorem**:
- The work-energy theorem states that the change in kinetic energy (\( \Delta KE \)) is equal to the work done by the forces acting on the particle.
- The work done by the magnetic field is zero because the magnetic force is always perpendicular to the displacement.
- Therefore, we have:
\[ \Delta KE = W_{\text{electric}}. \]

3. **Calculating the Change in Kinetic Energy**:
- The initial kinetic energy (\( KE_i \)) at the origin (where the particle starts from rest) is:
\[ KE_i = 0. \]
- The final kinetic energy (\( KE_f \)) at the point \( (x, 0, 0) \) is given by:
\[ KE_f = \frac{1}{2} m v^2, \]
where \( v = |\vec{v}_0| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).
- Thus,
\[ KE_f = \frac{1}{2} m (10^2) = \frac{1}{2} m (100) = 50m. \]

4. **Calculating the Work Done by the Electric Field**:
- The work done by the electric field is given by:
\[ W_{\text{electric}} = F \cdot d = (q \vec{E}) \cdot \vec{d}, \]
where \( \vec{E} = E_0 \hat{i} \) and \( \vec{d} = x \hat{i} \).
- Therefore,
\[ W_{\text{electric}} = q E_0 x. \]

5. **Setting Up the Equation**:
- From the work-energy theorem, we have:
\[ \Delta KE = W_{\text{electric}} \implies 50m = q E_0 x. \]

6. **Solving for \( x \)**:
- Rearranging the equation gives:
\[ x = \frac{50m}{q E_0}. \]

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