Electric and magnetic field are directed as `E_(0) hat(i)` and `B_(0) hat(k)`, a particle of mass m and charge + q is released from position (0,2,0) from rest. The velocity of that particle at (x,5,0) is `(5 hat(i) + 12 hat(j))` the value of x will be

Options

A. `(169 m)/(2qE_(0))`
B. `(25 m)/(2qE_(0))`
C. `(25 m)/(12 qE_(0))`
D. `(144 m)/(12qE_(0))`

Answer

To solve the problem, we will apply the work-energy theorem, which states that the work done by the forces acting on a particle is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Electric field \( \vec{E} = E_0 \hat{i} \)
- Magnetic field \( \vec{B} = B_0 \hat{k} \)
- Initial position of the particle: \( (0, 2, 0) \)
- Initial velocity: \( \vec{v}_0 = 0 \) (released from rest)
- Final position of the particle: \( (x, 5, 0) \)
- Final velocity: \( \vec{v} = 5 \hat{i} + 12 \hat{j} \)

2. **Calculate the Final Kinetic Energy:**
- The magnitude of the final velocity \( v \) is given by:
\[ v = \sqrt{(5^2 + 12^2)} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \text{m/s} \]
- The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} m (13^2) = \frac{1}{2} m (169) = \frac{169m}{2} \]
- The initial kinetic energy \( KE_i \) is:
\[ KE_i = 0 \quad (\text{since the particle is at rest}) \]

3. **Apply the Work-Energy Theorem:**
- The work done by the electric field \( W_E \) and the magnetic field \( W_B \) is given by:
\[ W_E + W_B = KE_f - KE_i \]
- Since the work done by the magnetic field is zero (as it does no work on the charged particle), we have:
\[ W_E = KE_f \]

4. **Calculate the Work Done by the Electric Field:**
- The electric force \( \vec{F_E} \) acting on the particle is:
\[ \vec{F_E} = q \vec{E} = q E_0 \hat{i} \]
- The displacement in the x-direction is \( x - 0 = x \).
- Therefore, the work done by the electric field is:
\[ W_E = \vec{F_E} \cdot \vec{d} = (q E_0 \hat{i}) \cdot (x \hat{i}) = q E_0 x \]

5. **Set Up the Equation:**
- From the work-energy theorem, we have:
\[ q E_0 x = \frac{169m}{2} \]

6. **Solve for \( x \):**
- Rearranging the equation gives:
\[ x = \frac{169m}{2q E_0} \]

### Final Answer:
The value of \( x \) is:
\[ x = \frac{169m}{2q E_0} \]