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 step by step, we need to analyze the motion of a charged particle in the presence of electric and magnetic fields. Here’s how we can approach it:

### Step 1: Understand the Forces Acting on the Particle
The particle with charge \( +q \) experiences a force due to the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \). The electric force \( \mathbf{F_E} \) is given by:
\[ \mathbf{F_E} = q \mathbf{E} \]
Since the electric field is directed along the \( \hat{i} \) direction, we have:
\[ \mathbf{E} = E_0 \hat{i} \]
Thus, the electric force becomes:
\[ \mathbf{F_E} = q E_0 \hat{i} \]

### Step 2: Calculate the Change in Kinetic Energy
The particle is released from rest, which means its initial velocity \( V_1 = 0 \). The final velocity \( V_2 \) is given as:
\[ V_2 = 5 \hat{i} + 12 \hat{j} \]
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the magnitude of the velocity. First, we calculate the magnitude of \( V_2 \):
\[ v = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
Thus, the final kinetic energy is:
\[ KE = \frac{1}{2} m (13)^2 = \frac{1}{2} m \cdot 169 = \frac{169m}{2} \]

### Step 3: Relate Work Done to Kinetic Energy
The work done \( W \) on the particle is equal to the change in kinetic energy:
\[ W = KE = \frac{169m}{2} \]
The work done can also be expressed as:
\[ W = \mathbf{F_E} \cdot \mathbf{d} \]
where \( \mathbf{d} \) is the displacement. Since the particle moves in the \( \hat{i} \) direction, the displacement \( d \) can be represented as \( x \hat{i} \). Therefore:
\[ W = q E_0 x \]

### Step 4: Set the Work Done Equal to the Change in Kinetic Energy
Equating the two expressions for work done:
\[ q E_0 x = \frac{169m}{2} \]

### Step 5: Solve for \( x \)
Rearranging the equation to solve for \( x \):
\[ x = \frac{169m}{2q E_0} \]

### Conclusion
Thus, the value of \( x \) is:
\[ x = \frac{169m}{2q E_0} \]