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. `(25 mv_(0)^(2))/( qE_0)`
B. `(100 mv_(0)^(2))/(qB_0)`
C. `(50 mV_(0)^(2))/( qE_0)`
D. `(14 mv_(0)^(2))/( q E_0)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in the presence of electric and magnetic fields. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Particle
The particle experiences both electric and magnetic forces. The electric force \( \vec{F}_E \) is given by:
\[ \vec{F}_E = q \vec{E} \]
where \( \vec{E} = E_0 \hat{i} \).

The magnetic force \( \vec{F}_B \) is given by:
\[ \vec{F}_B = q \vec{v} \times \vec{B} \]
where \( \vec{B} = B_0 \hat{k} \) and \( \vec{v} \) is the velocity of the particle.

### Step 2: Determine the Work Done by the Forces
Since the magnetic force is always perpendicular to the velocity, it does no work on the particle. Therefore, the work done on the particle is only from the electric field.

### Step 3: Calculate the Work Done by the Electric Field
The work done by the electric force as the particle moves from the origin to position \( (x, 0, 0) \) is:
\[ W = \vec{F}_E \cdot \vec{d} \]
The displacement vector \( \vec{d} \) is \( x \hat{i} \), so:
\[ W = (q E_0 \hat{i}) \cdot (x \hat{i}) = q E_0 x \]

### Step 4: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done by the electric field is equal to the change in kinetic energy:
\[ W = \Delta KE \]
The initial kinetic energy is zero since the particle starts from rest. The final kinetic energy is:
\[ KE = \frac{1}{2} m v_f^2 \]
where \( v_f \) is the final velocity. Given \( \vec{v} = v_0 (6 \hat{i} + 8 \hat{j}) \), we can find the magnitude of the velocity:
\[ v_f = \sqrt{(6 v_0)^2 + (8 v_0)^2} = v_0 \sqrt{36 + 64} = v_0 \sqrt{100} = 10 v_0 \]
Thus, the final kinetic energy becomes:
\[ KE = \frac{1}{2} m (10 v_0)^2 = 50 m v_0^2 \]

### Step 5: Set the Work Done Equal to the Change in Kinetic Energy
Now, equating the work done to the change in kinetic energy:
\[ q E_0 x = 50 m v_0^2 \]

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

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

---